Is it possible to catch an exception of lambda type?
While it is good practice to throw only exceptions of types derived from std::exception
class, C++ makes it possible to throw anything. All below examples are valid C++:
throw "foo"; // throws an instance of const char*
throw 5; // throws an instance of int
struct {} anon;
throw anon; // throws an instance of not-named structure
throw {}; // throws a lambda!
The last example is interesting, as it potentially allows passing some code to execute at catch site without having to define a separate class or function.
But is it at all possible to catch a lambda (or a closure)? catch ({} e)
does not work.
c++ exception lambda
|
show 1 more comment
While it is good practice to throw only exceptions of types derived from std::exception
class, C++ makes it possible to throw anything. All below examples are valid C++:
throw "foo"; // throws an instance of const char*
throw 5; // throws an instance of int
struct {} anon;
throw anon; // throws an instance of not-named structure
throw {}; // throws a lambda!
The last example is interesting, as it potentially allows passing some code to execute at catch site without having to define a separate class or function.
But is it at all possible to catch a lambda (or a closure)? catch ({} e)
does not work.
c++ exception lambda
I think we can assumecatch(...){}
is not desired.
– Joshua
Dec 5 '18 at 21:33
14
catch (...) { asm("call %rax"); }
unsafe problems call for unsafe solutions
– sudo rm -rf slash
Dec 5 '18 at 22:27
@Joshuacatch(...)
is not desired if only for not giving access to the exception object being caught.
– Krzysiek Karbowiak
Dec 6 '18 at 7:06
It turns out someone was already curious about lambda exceptions: stackoverflow.com/questions/42338470/catching-lambda-exception
– Krzysiek Karbowiak
Dec 6 '18 at 7:37
@sudo rm -rf slash: I don't think you tried it. You got the x64 calling convention wrong.
– Joshua
Dec 6 '18 at 14:56
|
show 1 more comment
While it is good practice to throw only exceptions of types derived from std::exception
class, C++ makes it possible to throw anything. All below examples are valid C++:
throw "foo"; // throws an instance of const char*
throw 5; // throws an instance of int
struct {} anon;
throw anon; // throws an instance of not-named structure
throw {}; // throws a lambda!
The last example is interesting, as it potentially allows passing some code to execute at catch site without having to define a separate class or function.
But is it at all possible to catch a lambda (or a closure)? catch ({} e)
does not work.
c++ exception lambda
While it is good practice to throw only exceptions of types derived from std::exception
class, C++ makes it possible to throw anything. All below examples are valid C++:
throw "foo"; // throws an instance of const char*
throw 5; // throws an instance of int
struct {} anon;
throw anon; // throws an instance of not-named structure
throw {}; // throws a lambda!
The last example is interesting, as it potentially allows passing some code to execute at catch site without having to define a separate class or function.
But is it at all possible to catch a lambda (or a closure)? catch ({} e)
does not work.
c++ exception lambda
c++ exception lambda
edited Dec 5 '18 at 9:53
GPhilo
5,78412243
5,78412243
asked Dec 5 '18 at 9:51
Krzysiek KarbowiakKrzysiek Karbowiak
27827
27827
I think we can assumecatch(...){}
is not desired.
– Joshua
Dec 5 '18 at 21:33
14
catch (...) { asm("call %rax"); }
unsafe problems call for unsafe solutions
– sudo rm -rf slash
Dec 5 '18 at 22:27
@Joshuacatch(...)
is not desired if only for not giving access to the exception object being caught.
– Krzysiek Karbowiak
Dec 6 '18 at 7:06
It turns out someone was already curious about lambda exceptions: stackoverflow.com/questions/42338470/catching-lambda-exception
– Krzysiek Karbowiak
Dec 6 '18 at 7:37
@sudo rm -rf slash: I don't think you tried it. You got the x64 calling convention wrong.
– Joshua
Dec 6 '18 at 14:56
|
show 1 more comment
I think we can assumecatch(...){}
is not desired.
– Joshua
Dec 5 '18 at 21:33
14
catch (...) { asm("call %rax"); }
unsafe problems call for unsafe solutions
– sudo rm -rf slash
Dec 5 '18 at 22:27
@Joshuacatch(...)
is not desired if only for not giving access to the exception object being caught.
– Krzysiek Karbowiak
Dec 6 '18 at 7:06
It turns out someone was already curious about lambda exceptions: stackoverflow.com/questions/42338470/catching-lambda-exception
– Krzysiek Karbowiak
Dec 6 '18 at 7:37
@sudo rm -rf slash: I don't think you tried it. You got the x64 calling convention wrong.
– Joshua
Dec 6 '18 at 14:56
I think we can assume
catch(...){}
is not desired.– Joshua
Dec 5 '18 at 21:33
I think we can assume
catch(...){}
is not desired.– Joshua
Dec 5 '18 at 21:33
14
14
catch (...) { asm("call %rax"); }
unsafe problems call for unsafe solutions– sudo rm -rf slash
Dec 5 '18 at 22:27
catch (...) { asm("call %rax"); }
unsafe problems call for unsafe solutions– sudo rm -rf slash
Dec 5 '18 at 22:27
@Joshua
catch(...)
is not desired if only for not giving access to the exception object being caught.– Krzysiek Karbowiak
Dec 6 '18 at 7:06
@Joshua
catch(...)
is not desired if only for not giving access to the exception object being caught.– Krzysiek Karbowiak
Dec 6 '18 at 7:06
It turns out someone was already curious about lambda exceptions: stackoverflow.com/questions/42338470/catching-lambda-exception
– Krzysiek Karbowiak
Dec 6 '18 at 7:37
It turns out someone was already curious about lambda exceptions: stackoverflow.com/questions/42338470/catching-lambda-exception
– Krzysiek Karbowiak
Dec 6 '18 at 7:37
@sudo rm -rf slash: I don't think you tried it. You got the x64 calling convention wrong.
– Joshua
Dec 6 '18 at 14:56
@sudo rm -rf slash: I don't think you tried it. You got the x64 calling convention wrong.
– Joshua
Dec 6 '18 at 14:56
|
show 1 more comment
4 Answers
4
active
oldest
votes
Exception handlers are matched based on type, and the implicit conversions done to match an exception object to a handler are more limited than in other contexts.
Each lambda expression introduces a closure type that is unique to the surrounding scope. So your naive attempt cannot work, for {}
has an entirely different type in the throw expression and the handler!
But you are correct. C++ allows you to throw any object. So if you explicitly convert the lambda before-hand to a type that matches an exception handler, it will allow you to call that arbitrary callable. For instance:
try {
throw std::function<void()>{ {} }; // Note the explicit conversion
} catch(std::function<void()> const& f) {
f();
}
This may have interesting utility, but I'd caution against throwing things not derived from std::exception
. A better option would probably be to create a type that derives from std::exception
and can hold a callable.
7
Sure, I would not use this in production code. Rather, I was exploring intricacies of the language. And while I did try to catch by pointer-to-function, it did not occur to me to throw and catch asstd::function
.
– Krzysiek Karbowiak
Dec 5 '18 at 10:25
4
@KrzysiekKarbowiak - If the type is designed carefully I don't see why you can't do it in production. It may have interesting utility, like I noted. Ingenuity is after all taking what is known already and using it in a novel way :)
– StoryTeller
Dec 5 '18 at 10:27
2
The exact rules used to matchcatch
expressions are available in this answer. Sadly, it looks like there's no way to catch the lambda if it's thrown asthrow {};
, as there's no public base class and the type is not a pointer so the pointer rules don't apply.
– Silvio Mayolo
Dec 5 '18 at 19:07
add a comment |
C++ allows you to throw anything. And It allows you to catch whatever you throw. You can, of course, throw a lambda. The only problem is that, to catch something, you need to know the type or at least a parent type of that something. Since lambdas do not derive from a common base, you have to know the type of your lambda to catch a lambda. The main issue with that is that every lambda expression will give you an rvalue of a distinct type. That means that both your throw and your catch need to be based on the same lambda expression (note: the same expression, not just some expression that looks exactly the same). One way I can think of to make this work to some degree would be to encapsulate the creation of the lambda to throw in a function. That way, you can call the function in your throw
expression, and use the return type of the function to deduce the type to catch
:
#include <utility>
auto makeMyLambda(int some_arg)
{
return [some_arg](int another_arg){ return some_arg + another_arg; };
}
void f()
{
throw makeMyLambda(42);
}
int main()
{
try
{
f();
}
catch (const decltype(makeMyLambda(std::declval<int>()))& l)
{
return l(23);
}
}
Try it out here.
You could also just use std::function
like suggested in some of the other answers, which is potentially a more practical approach. The downsides of that, however, would be
- It means you don't actually throw a lambda. You throw an
std::function
, which is not really what you asked for 😉 - The creation of an
std::function
object from a lambda can throw an exception
add a comment |
You can throw and catch a std::function
:
#include <iostream>
#include <functional>
void f() {
throw std::function<void(void)>({std::cout << "lambdan"; });
}
int main()
{
try{ f(); }
catch( std::function<void(void)> &e)
{
e();
std::cout << "catchn";
}
}
Output:
lambda
catch
add a comment |
A lambda is a unique anonymous type. The only way to name a lambda instance's type is to store it in a variable, then do a decltype
on that variable type.
There are a few ways you can catch a thrown lambda.
try {
throw {};
} catch(...) {
}
in this case you cannot use it, other than throwing it again.
try {
throw +{};
} catch(void(*f)()) {
}
a stateless lambda can be converted to a function pointer.
try {
throw std::function<void()>({});
} catch(std::function<void()> f) {
}
you can convert it to a std::function
. The downside with std::function
is that it heap allocates for larger lambdas, which could in theory cause it to throw.
We can eliminate that heap allocation:
template<class Sig>
struct callable;
template<class R, class...Args>
struct callable<R(Args...)> {
void* state = nullptr;
R(*action)(void*, Args&&...) = nullptr;
R operator()(Args...args) const {
return action( state, std::forward<Args>(args)... );
}
};
template<class Sig, class F>
struct lambda_wrapper;
template<class R, class...Args, class F>
struct lambda_wrapper<R(Args...), F>
:
F,
callable<R(Args...)>
{
lambda_wrapper( F fin ):
F(std::move(fin)),
callable<R(Args...)>{
static_cast<F*>(this),
(void* self, Args&&...args)->R {
return static_cast<R>( (*static_cast<F*>(self))( std::forward<Args>(args)... ) );
}
}
{}
lambda_wrapper(lambda_wrapper && o):
F(static_cast<F&&>(o)),
callable<R(Args...)>( o )
{
this->state = static_cast<F*>(this);
}
lambda_wrapper& operator=(lambda_wrapper && o)
{
static_cast<F&>(*this) = (static_cast<F&&>(o));
static_cast<callable<R(Args...)>&>(*this) = static_cast<callable<R(Args...)>&>( o );
this->state = static_cast<F*>(this);
}
};
template<class Sig, class F>
lambda_wrapper<Sig, F> wrap_lambda( F fin ) {
return std::move(fin);
}
now you can do:
try {
throw wrap_lambda<void()>({});
} catch( callable<void()> const& f ) {
}
callable
is "lighter weight" type erasure than std::function
as it cannot cause new heap memory to be allocated.
Live example.
1
@krzy you are missing the+
; I never said implicitly.throw +{};
– Yakk - Adam Nevraumont
Dec 11 '18 at 11:54
Whoa, that's cool! And sorry for the comment, I was sure the+
was a typo...
– Krzysiek Karbowiak
Dec 12 '18 at 7:45
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Exception handlers are matched based on type, and the implicit conversions done to match an exception object to a handler are more limited than in other contexts.
Each lambda expression introduces a closure type that is unique to the surrounding scope. So your naive attempt cannot work, for {}
has an entirely different type in the throw expression and the handler!
But you are correct. C++ allows you to throw any object. So if you explicitly convert the lambda before-hand to a type that matches an exception handler, it will allow you to call that arbitrary callable. For instance:
try {
throw std::function<void()>{ {} }; // Note the explicit conversion
} catch(std::function<void()> const& f) {
f();
}
This may have interesting utility, but I'd caution against throwing things not derived from std::exception
. A better option would probably be to create a type that derives from std::exception
and can hold a callable.
7
Sure, I would not use this in production code. Rather, I was exploring intricacies of the language. And while I did try to catch by pointer-to-function, it did not occur to me to throw and catch asstd::function
.
– Krzysiek Karbowiak
Dec 5 '18 at 10:25
4
@KrzysiekKarbowiak - If the type is designed carefully I don't see why you can't do it in production. It may have interesting utility, like I noted. Ingenuity is after all taking what is known already and using it in a novel way :)
– StoryTeller
Dec 5 '18 at 10:27
2
The exact rules used to matchcatch
expressions are available in this answer. Sadly, it looks like there's no way to catch the lambda if it's thrown asthrow {};
, as there's no public base class and the type is not a pointer so the pointer rules don't apply.
– Silvio Mayolo
Dec 5 '18 at 19:07
add a comment |
Exception handlers are matched based on type, and the implicit conversions done to match an exception object to a handler are more limited than in other contexts.
Each lambda expression introduces a closure type that is unique to the surrounding scope. So your naive attempt cannot work, for {}
has an entirely different type in the throw expression and the handler!
But you are correct. C++ allows you to throw any object. So if you explicitly convert the lambda before-hand to a type that matches an exception handler, it will allow you to call that arbitrary callable. For instance:
try {
throw std::function<void()>{ {} }; // Note the explicit conversion
} catch(std::function<void()> const& f) {
f();
}
This may have interesting utility, but I'd caution against throwing things not derived from std::exception
. A better option would probably be to create a type that derives from std::exception
and can hold a callable.
7
Sure, I would not use this in production code. Rather, I was exploring intricacies of the language. And while I did try to catch by pointer-to-function, it did not occur to me to throw and catch asstd::function
.
– Krzysiek Karbowiak
Dec 5 '18 at 10:25
4
@KrzysiekKarbowiak - If the type is designed carefully I don't see why you can't do it in production. It may have interesting utility, like I noted. Ingenuity is after all taking what is known already and using it in a novel way :)
– StoryTeller
Dec 5 '18 at 10:27
2
The exact rules used to matchcatch
expressions are available in this answer. Sadly, it looks like there's no way to catch the lambda if it's thrown asthrow {};
, as there's no public base class and the type is not a pointer so the pointer rules don't apply.
– Silvio Mayolo
Dec 5 '18 at 19:07
add a comment |
Exception handlers are matched based on type, and the implicit conversions done to match an exception object to a handler are more limited than in other contexts.
Each lambda expression introduces a closure type that is unique to the surrounding scope. So your naive attempt cannot work, for {}
has an entirely different type in the throw expression and the handler!
But you are correct. C++ allows you to throw any object. So if you explicitly convert the lambda before-hand to a type that matches an exception handler, it will allow you to call that arbitrary callable. For instance:
try {
throw std::function<void()>{ {} }; // Note the explicit conversion
} catch(std::function<void()> const& f) {
f();
}
This may have interesting utility, but I'd caution against throwing things not derived from std::exception
. A better option would probably be to create a type that derives from std::exception
and can hold a callable.
Exception handlers are matched based on type, and the implicit conversions done to match an exception object to a handler are more limited than in other contexts.
Each lambda expression introduces a closure type that is unique to the surrounding scope. So your naive attempt cannot work, for {}
has an entirely different type in the throw expression and the handler!
But you are correct. C++ allows you to throw any object. So if you explicitly convert the lambda before-hand to a type that matches an exception handler, it will allow you to call that arbitrary callable. For instance:
try {
throw std::function<void()>{ {} }; // Note the explicit conversion
} catch(std::function<void()> const& f) {
f();
}
This may have interesting utility, but I'd caution against throwing things not derived from std::exception
. A better option would probably be to create a type that derives from std::exception
and can hold a callable.
answered Dec 5 '18 at 10:04
StoryTellerStoryTeller
94.4k12191254
94.4k12191254
7
Sure, I would not use this in production code. Rather, I was exploring intricacies of the language. And while I did try to catch by pointer-to-function, it did not occur to me to throw and catch asstd::function
.
– Krzysiek Karbowiak
Dec 5 '18 at 10:25
4
@KrzysiekKarbowiak - If the type is designed carefully I don't see why you can't do it in production. It may have interesting utility, like I noted. Ingenuity is after all taking what is known already and using it in a novel way :)
– StoryTeller
Dec 5 '18 at 10:27
2
The exact rules used to matchcatch
expressions are available in this answer. Sadly, it looks like there's no way to catch the lambda if it's thrown asthrow {};
, as there's no public base class and the type is not a pointer so the pointer rules don't apply.
– Silvio Mayolo
Dec 5 '18 at 19:07
add a comment |
7
Sure, I would not use this in production code. Rather, I was exploring intricacies of the language. And while I did try to catch by pointer-to-function, it did not occur to me to throw and catch asstd::function
.
– Krzysiek Karbowiak
Dec 5 '18 at 10:25
4
@KrzysiekKarbowiak - If the type is designed carefully I don't see why you can't do it in production. It may have interesting utility, like I noted. Ingenuity is after all taking what is known already and using it in a novel way :)
– StoryTeller
Dec 5 '18 at 10:27
2
The exact rules used to matchcatch
expressions are available in this answer. Sadly, it looks like there's no way to catch the lambda if it's thrown asthrow {};
, as there's no public base class and the type is not a pointer so the pointer rules don't apply.
– Silvio Mayolo
Dec 5 '18 at 19:07
7
7
Sure, I would not use this in production code. Rather, I was exploring intricacies of the language. And while I did try to catch by pointer-to-function, it did not occur to me to throw and catch as
std::function
.– Krzysiek Karbowiak
Dec 5 '18 at 10:25
Sure, I would not use this in production code. Rather, I was exploring intricacies of the language. And while I did try to catch by pointer-to-function, it did not occur to me to throw and catch as
std::function
.– Krzysiek Karbowiak
Dec 5 '18 at 10:25
4
4
@KrzysiekKarbowiak - If the type is designed carefully I don't see why you can't do it in production. It may have interesting utility, like I noted. Ingenuity is after all taking what is known already and using it in a novel way :)
– StoryTeller
Dec 5 '18 at 10:27
@KrzysiekKarbowiak - If the type is designed carefully I don't see why you can't do it in production. It may have interesting utility, like I noted. Ingenuity is after all taking what is known already and using it in a novel way :)
– StoryTeller
Dec 5 '18 at 10:27
2
2
The exact rules used to match
catch
expressions are available in this answer. Sadly, it looks like there's no way to catch the lambda if it's thrown as throw {};
, as there's no public base class and the type is not a pointer so the pointer rules don't apply.– Silvio Mayolo
Dec 5 '18 at 19:07
The exact rules used to match
catch
expressions are available in this answer. Sadly, it looks like there's no way to catch the lambda if it's thrown as throw {};
, as there's no public base class and the type is not a pointer so the pointer rules don't apply.– Silvio Mayolo
Dec 5 '18 at 19:07
add a comment |
C++ allows you to throw anything. And It allows you to catch whatever you throw. You can, of course, throw a lambda. The only problem is that, to catch something, you need to know the type or at least a parent type of that something. Since lambdas do not derive from a common base, you have to know the type of your lambda to catch a lambda. The main issue with that is that every lambda expression will give you an rvalue of a distinct type. That means that both your throw and your catch need to be based on the same lambda expression (note: the same expression, not just some expression that looks exactly the same). One way I can think of to make this work to some degree would be to encapsulate the creation of the lambda to throw in a function. That way, you can call the function in your throw
expression, and use the return type of the function to deduce the type to catch
:
#include <utility>
auto makeMyLambda(int some_arg)
{
return [some_arg](int another_arg){ return some_arg + another_arg; };
}
void f()
{
throw makeMyLambda(42);
}
int main()
{
try
{
f();
}
catch (const decltype(makeMyLambda(std::declval<int>()))& l)
{
return l(23);
}
}
Try it out here.
You could also just use std::function
like suggested in some of the other answers, which is potentially a more practical approach. The downsides of that, however, would be
- It means you don't actually throw a lambda. You throw an
std::function
, which is not really what you asked for 😉 - The creation of an
std::function
object from a lambda can throw an exception
add a comment |
C++ allows you to throw anything. And It allows you to catch whatever you throw. You can, of course, throw a lambda. The only problem is that, to catch something, you need to know the type or at least a parent type of that something. Since lambdas do not derive from a common base, you have to know the type of your lambda to catch a lambda. The main issue with that is that every lambda expression will give you an rvalue of a distinct type. That means that both your throw and your catch need to be based on the same lambda expression (note: the same expression, not just some expression that looks exactly the same). One way I can think of to make this work to some degree would be to encapsulate the creation of the lambda to throw in a function. That way, you can call the function in your throw
expression, and use the return type of the function to deduce the type to catch
:
#include <utility>
auto makeMyLambda(int some_arg)
{
return [some_arg](int another_arg){ return some_arg + another_arg; };
}
void f()
{
throw makeMyLambda(42);
}
int main()
{
try
{
f();
}
catch (const decltype(makeMyLambda(std::declval<int>()))& l)
{
return l(23);
}
}
Try it out here.
You could also just use std::function
like suggested in some of the other answers, which is potentially a more practical approach. The downsides of that, however, would be
- It means you don't actually throw a lambda. You throw an
std::function
, which is not really what you asked for 😉 - The creation of an
std::function
object from a lambda can throw an exception
add a comment |
C++ allows you to throw anything. And It allows you to catch whatever you throw. You can, of course, throw a lambda. The only problem is that, to catch something, you need to know the type or at least a parent type of that something. Since lambdas do not derive from a common base, you have to know the type of your lambda to catch a lambda. The main issue with that is that every lambda expression will give you an rvalue of a distinct type. That means that both your throw and your catch need to be based on the same lambda expression (note: the same expression, not just some expression that looks exactly the same). One way I can think of to make this work to some degree would be to encapsulate the creation of the lambda to throw in a function. That way, you can call the function in your throw
expression, and use the return type of the function to deduce the type to catch
:
#include <utility>
auto makeMyLambda(int some_arg)
{
return [some_arg](int another_arg){ return some_arg + another_arg; };
}
void f()
{
throw makeMyLambda(42);
}
int main()
{
try
{
f();
}
catch (const decltype(makeMyLambda(std::declval<int>()))& l)
{
return l(23);
}
}
Try it out here.
You could also just use std::function
like suggested in some of the other answers, which is potentially a more practical approach. The downsides of that, however, would be
- It means you don't actually throw a lambda. You throw an
std::function
, which is not really what you asked for 😉 - The creation of an
std::function
object from a lambda can throw an exception
C++ allows you to throw anything. And It allows you to catch whatever you throw. You can, of course, throw a lambda. The only problem is that, to catch something, you need to know the type or at least a parent type of that something. Since lambdas do not derive from a common base, you have to know the type of your lambda to catch a lambda. The main issue with that is that every lambda expression will give you an rvalue of a distinct type. That means that both your throw and your catch need to be based on the same lambda expression (note: the same expression, not just some expression that looks exactly the same). One way I can think of to make this work to some degree would be to encapsulate the creation of the lambda to throw in a function. That way, you can call the function in your throw
expression, and use the return type of the function to deduce the type to catch
:
#include <utility>
auto makeMyLambda(int some_arg)
{
return [some_arg](int another_arg){ return some_arg + another_arg; };
}
void f()
{
throw makeMyLambda(42);
}
int main()
{
try
{
f();
}
catch (const decltype(makeMyLambda(std::declval<int>()))& l)
{
return l(23);
}
}
Try it out here.
You could also just use std::function
like suggested in some of the other answers, which is potentially a more practical approach. The downsides of that, however, would be
- It means you don't actually throw a lambda. You throw an
std::function
, which is not really what you asked for 😉 - The creation of an
std::function
object from a lambda can throw an exception
edited Dec 5 '18 at 16:05
answered Dec 5 '18 at 10:12
Michael KenzelMichael Kenzel
4,043719
4,043719
add a comment |
add a comment |
You can throw and catch a std::function
:
#include <iostream>
#include <functional>
void f() {
throw std::function<void(void)>({std::cout << "lambdan"; });
}
int main()
{
try{ f(); }
catch( std::function<void(void)> &e)
{
e();
std::cout << "catchn";
}
}
Output:
lambda
catch
add a comment |
You can throw and catch a std::function
:
#include <iostream>
#include <functional>
void f() {
throw std::function<void(void)>({std::cout << "lambdan"; });
}
int main()
{
try{ f(); }
catch( std::function<void(void)> &e)
{
e();
std::cout << "catchn";
}
}
Output:
lambda
catch
add a comment |
You can throw and catch a std::function
:
#include <iostream>
#include <functional>
void f() {
throw std::function<void(void)>({std::cout << "lambdan"; });
}
int main()
{
try{ f(); }
catch( std::function<void(void)> &e)
{
e();
std::cout << "catchn";
}
}
Output:
lambda
catch
You can throw and catch a std::function
:
#include <iostream>
#include <functional>
void f() {
throw std::function<void(void)>({std::cout << "lambdan"; });
}
int main()
{
try{ f(); }
catch( std::function<void(void)> &e)
{
e();
std::cout << "catchn";
}
}
Output:
lambda
catch
answered Dec 5 '18 at 10:04
davedave
4718
4718
add a comment |
add a comment |
A lambda is a unique anonymous type. The only way to name a lambda instance's type is to store it in a variable, then do a decltype
on that variable type.
There are a few ways you can catch a thrown lambda.
try {
throw {};
} catch(...) {
}
in this case you cannot use it, other than throwing it again.
try {
throw +{};
} catch(void(*f)()) {
}
a stateless lambda can be converted to a function pointer.
try {
throw std::function<void()>({});
} catch(std::function<void()> f) {
}
you can convert it to a std::function
. The downside with std::function
is that it heap allocates for larger lambdas, which could in theory cause it to throw.
We can eliminate that heap allocation:
template<class Sig>
struct callable;
template<class R, class...Args>
struct callable<R(Args...)> {
void* state = nullptr;
R(*action)(void*, Args&&...) = nullptr;
R operator()(Args...args) const {
return action( state, std::forward<Args>(args)... );
}
};
template<class Sig, class F>
struct lambda_wrapper;
template<class R, class...Args, class F>
struct lambda_wrapper<R(Args...), F>
:
F,
callable<R(Args...)>
{
lambda_wrapper( F fin ):
F(std::move(fin)),
callable<R(Args...)>{
static_cast<F*>(this),
(void* self, Args&&...args)->R {
return static_cast<R>( (*static_cast<F*>(self))( std::forward<Args>(args)... ) );
}
}
{}
lambda_wrapper(lambda_wrapper && o):
F(static_cast<F&&>(o)),
callable<R(Args...)>( o )
{
this->state = static_cast<F*>(this);
}
lambda_wrapper& operator=(lambda_wrapper && o)
{
static_cast<F&>(*this) = (static_cast<F&&>(o));
static_cast<callable<R(Args...)>&>(*this) = static_cast<callable<R(Args...)>&>( o );
this->state = static_cast<F*>(this);
}
};
template<class Sig, class F>
lambda_wrapper<Sig, F> wrap_lambda( F fin ) {
return std::move(fin);
}
now you can do:
try {
throw wrap_lambda<void()>({});
} catch( callable<void()> const& f ) {
}
callable
is "lighter weight" type erasure than std::function
as it cannot cause new heap memory to be allocated.
Live example.
1
@krzy you are missing the+
; I never said implicitly.throw +{};
– Yakk - Adam Nevraumont
Dec 11 '18 at 11:54
Whoa, that's cool! And sorry for the comment, I was sure the+
was a typo...
– Krzysiek Karbowiak
Dec 12 '18 at 7:45
add a comment |
A lambda is a unique anonymous type. The only way to name a lambda instance's type is to store it in a variable, then do a decltype
on that variable type.
There are a few ways you can catch a thrown lambda.
try {
throw {};
} catch(...) {
}
in this case you cannot use it, other than throwing it again.
try {
throw +{};
} catch(void(*f)()) {
}
a stateless lambda can be converted to a function pointer.
try {
throw std::function<void()>({});
} catch(std::function<void()> f) {
}
you can convert it to a std::function
. The downside with std::function
is that it heap allocates for larger lambdas, which could in theory cause it to throw.
We can eliminate that heap allocation:
template<class Sig>
struct callable;
template<class R, class...Args>
struct callable<R(Args...)> {
void* state = nullptr;
R(*action)(void*, Args&&...) = nullptr;
R operator()(Args...args) const {
return action( state, std::forward<Args>(args)... );
}
};
template<class Sig, class F>
struct lambda_wrapper;
template<class R, class...Args, class F>
struct lambda_wrapper<R(Args...), F>
:
F,
callable<R(Args...)>
{
lambda_wrapper( F fin ):
F(std::move(fin)),
callable<R(Args...)>{
static_cast<F*>(this),
(void* self, Args&&...args)->R {
return static_cast<R>( (*static_cast<F*>(self))( std::forward<Args>(args)... ) );
}
}
{}
lambda_wrapper(lambda_wrapper && o):
F(static_cast<F&&>(o)),
callable<R(Args...)>( o )
{
this->state = static_cast<F*>(this);
}
lambda_wrapper& operator=(lambda_wrapper && o)
{
static_cast<F&>(*this) = (static_cast<F&&>(o));
static_cast<callable<R(Args...)>&>(*this) = static_cast<callable<R(Args...)>&>( o );
this->state = static_cast<F*>(this);
}
};
template<class Sig, class F>
lambda_wrapper<Sig, F> wrap_lambda( F fin ) {
return std::move(fin);
}
now you can do:
try {
throw wrap_lambda<void()>({});
} catch( callable<void()> const& f ) {
}
callable
is "lighter weight" type erasure than std::function
as it cannot cause new heap memory to be allocated.
Live example.
1
@krzy you are missing the+
; I never said implicitly.throw +{};
– Yakk - Adam Nevraumont
Dec 11 '18 at 11:54
Whoa, that's cool! And sorry for the comment, I was sure the+
was a typo...
– Krzysiek Karbowiak
Dec 12 '18 at 7:45
add a comment |
A lambda is a unique anonymous type. The only way to name a lambda instance's type is to store it in a variable, then do a decltype
on that variable type.
There are a few ways you can catch a thrown lambda.
try {
throw {};
} catch(...) {
}
in this case you cannot use it, other than throwing it again.
try {
throw +{};
} catch(void(*f)()) {
}
a stateless lambda can be converted to a function pointer.
try {
throw std::function<void()>({});
} catch(std::function<void()> f) {
}
you can convert it to a std::function
. The downside with std::function
is that it heap allocates for larger lambdas, which could in theory cause it to throw.
We can eliminate that heap allocation:
template<class Sig>
struct callable;
template<class R, class...Args>
struct callable<R(Args...)> {
void* state = nullptr;
R(*action)(void*, Args&&...) = nullptr;
R operator()(Args...args) const {
return action( state, std::forward<Args>(args)... );
}
};
template<class Sig, class F>
struct lambda_wrapper;
template<class R, class...Args, class F>
struct lambda_wrapper<R(Args...), F>
:
F,
callable<R(Args...)>
{
lambda_wrapper( F fin ):
F(std::move(fin)),
callable<R(Args...)>{
static_cast<F*>(this),
(void* self, Args&&...args)->R {
return static_cast<R>( (*static_cast<F*>(self))( std::forward<Args>(args)... ) );
}
}
{}
lambda_wrapper(lambda_wrapper && o):
F(static_cast<F&&>(o)),
callable<R(Args...)>( o )
{
this->state = static_cast<F*>(this);
}
lambda_wrapper& operator=(lambda_wrapper && o)
{
static_cast<F&>(*this) = (static_cast<F&&>(o));
static_cast<callable<R(Args...)>&>(*this) = static_cast<callable<R(Args...)>&>( o );
this->state = static_cast<F*>(this);
}
};
template<class Sig, class F>
lambda_wrapper<Sig, F> wrap_lambda( F fin ) {
return std::move(fin);
}
now you can do:
try {
throw wrap_lambda<void()>({});
} catch( callable<void()> const& f ) {
}
callable
is "lighter weight" type erasure than std::function
as it cannot cause new heap memory to be allocated.
Live example.
A lambda is a unique anonymous type. The only way to name a lambda instance's type is to store it in a variable, then do a decltype
on that variable type.
There are a few ways you can catch a thrown lambda.
try {
throw {};
} catch(...) {
}
in this case you cannot use it, other than throwing it again.
try {
throw +{};
} catch(void(*f)()) {
}
a stateless lambda can be converted to a function pointer.
try {
throw std::function<void()>({});
} catch(std::function<void()> f) {
}
you can convert it to a std::function
. The downside with std::function
is that it heap allocates for larger lambdas, which could in theory cause it to throw.
We can eliminate that heap allocation:
template<class Sig>
struct callable;
template<class R, class...Args>
struct callable<R(Args...)> {
void* state = nullptr;
R(*action)(void*, Args&&...) = nullptr;
R operator()(Args...args) const {
return action( state, std::forward<Args>(args)... );
}
};
template<class Sig, class F>
struct lambda_wrapper;
template<class R, class...Args, class F>
struct lambda_wrapper<R(Args...), F>
:
F,
callable<R(Args...)>
{
lambda_wrapper( F fin ):
F(std::move(fin)),
callable<R(Args...)>{
static_cast<F*>(this),
(void* self, Args&&...args)->R {
return static_cast<R>( (*static_cast<F*>(self))( std::forward<Args>(args)... ) );
}
}
{}
lambda_wrapper(lambda_wrapper && o):
F(static_cast<F&&>(o)),
callable<R(Args...)>( o )
{
this->state = static_cast<F*>(this);
}
lambda_wrapper& operator=(lambda_wrapper && o)
{
static_cast<F&>(*this) = (static_cast<F&&>(o));
static_cast<callable<R(Args...)>&>(*this) = static_cast<callable<R(Args...)>&>( o );
this->state = static_cast<F*>(this);
}
};
template<class Sig, class F>
lambda_wrapper<Sig, F> wrap_lambda( F fin ) {
return std::move(fin);
}
now you can do:
try {
throw wrap_lambda<void()>({});
} catch( callable<void()> const& f ) {
}
callable
is "lighter weight" type erasure than std::function
as it cannot cause new heap memory to be allocated.
Live example.
answered Dec 10 '18 at 15:36
Yakk - Adam NevraumontYakk - Adam Nevraumont
183k19189374
183k19189374
1
@krzy you are missing the+
; I never said implicitly.throw +{};
– Yakk - Adam Nevraumont
Dec 11 '18 at 11:54
Whoa, that's cool! And sorry for the comment, I was sure the+
was a typo...
– Krzysiek Karbowiak
Dec 12 '18 at 7:45
add a comment |
1
@krzy you are missing the+
; I never said implicitly.throw +{};
– Yakk - Adam Nevraumont
Dec 11 '18 at 11:54
Whoa, that's cool! And sorry for the comment, I was sure the+
was a typo...
– Krzysiek Karbowiak
Dec 12 '18 at 7:45
1
1
@krzy you are missing the
+
; I never said implicitly. throw +{};
– Yakk - Adam Nevraumont
Dec 11 '18 at 11:54
@krzy you are missing the
+
; I never said implicitly. throw +{};
– Yakk - Adam Nevraumont
Dec 11 '18 at 11:54
Whoa, that's cool! And sorry for the comment, I was sure the
+
was a typo...– Krzysiek Karbowiak
Dec 12 '18 at 7:45
Whoa, that's cool! And sorry for the comment, I was sure the
+
was a typo...– Krzysiek Karbowiak
Dec 12 '18 at 7:45
add a comment |
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I think we can assume
catch(...){}
is not desired.– Joshua
Dec 5 '18 at 21:33
14
catch (...) { asm("call %rax"); }
unsafe problems call for unsafe solutions– sudo rm -rf slash
Dec 5 '18 at 22:27
@Joshua
catch(...)
is not desired if only for not giving access to the exception object being caught.– Krzysiek Karbowiak
Dec 6 '18 at 7:06
It turns out someone was already curious about lambda exceptions: stackoverflow.com/questions/42338470/catching-lambda-exception
– Krzysiek Karbowiak
Dec 6 '18 at 7:37
@sudo rm -rf slash: I don't think you tried it. You got the x64 calling convention wrong.
– Joshua
Dec 6 '18 at 14:56