Allocate vector size with list initialization (curly braces)
How can I do the equivelant of:
#include <vector>
size_t bufferSize = 1024 * 1024;
std::vector<unsigned char> buffer(bufferSize, ' ');
With list (curly braced) initialization?
When I try to do the following:
#include <vector>
size_t bufferSize = 1024 * 1024;
std::vector<unsigned char> buffer {bufferSize, ' '};
It wrongly interprets bufferSize as the value to be stored in the first index of the container (i.e. calls the wrong std::vector constructor), and fails to compile due to invalid narrowing conversion from unsigned int (size_t) to unsigned char.
c++ c++11 list-initialization
asked Nov 29 at 15:04
not an alien
215110
|
show 7 more comments
How can I do the equivelant of:
#include <vector>
size_t bufferSize = 1024 * 1024;
std::vector<unsigned char> buffer(bufferSize, ' ');
With list (curly braced) initialization?
When I try to do the following:
#include <vector>
size_t bufferSize = 1024 * 1024;
std::vector<unsigned char> buffer {bufferSize, ' '};
It wrongly interprets bufferSize as the value to be stored in the first index of the container (i.e. calls the wrong std::vector constructor), and fails to compile due to invalid narrowing conversion from unsigned int (size_t) to unsigned char.
c++ c++11 list-initialization
asked Nov 29 at 15:04
not an alien
215110
9
Why do you insist on doing this with "curly brace initialization"?
– Max Langhof
Nov 29 at 15:08
5
@MaxLanghof Well, it is called uniform initialization. One could be lead to believe it should be the preferred way to initialize an object ;)
– NathanOliver
Nov 29 at 15:10
3
@NathanOliver Related xkcd.com/927
– liliscent
Nov 29 at 15:12
2
@OP Here is a very good talk on the nightmare of C++ initialization: youtube.com/watch?v=7DTlWPgX6zs
– NathanOliver
Nov 29 at 15:12
1
@MaxLanghof because some recommend to use it everywhere as it is "safer" (does not have most vexing parse issue). In reality it is more dangerous.
– Slava
Nov 29 at 15:13
|
show 7 more comments
How can I do the equivelant of:
#include <vector>
size_t bufferSize = 1024 * 1024;
std::vector<unsigned char> buffer(bufferSize, ' ');
With list (curly braced) initialization?
When I try to do the following:
#include <vector>
size_t bufferSize = 1024 * 1024;
std::vector<unsigned char> buffer {bufferSize, ' '};
It wrongly interprets bufferSize as the value to be stored in the first index of the container (i.e. calls the wrong std::vector constructor), and fails to compile due to invalid narrowing conversion from unsigned int (size_t) to unsigned char.
c++ c++11 list-initialization
asked Nov 29 at 15:04
not an alien
215110
How can I do the equivelant of:
#include <vector>
size_t bufferSize = 1024 * 1024;
std::vector<unsigned char> buffer(bufferSize, ' ');
With list (curly braced) initialization?
When I try to do the following:
#include <vector>
size_t bufferSize = 1024 * 1024;
std::vector<unsigned char> buffer {bufferSize, ' '};
It wrongly interprets bufferSize as the value to be stored in the first index of the container (i.e. calls the wrong std::vector constructor), and fails to compile due to invalid narrowing conversion from unsigned int (size_t) to unsigned char.
c++ c++11 list-initialization
c++ c++11 list-initialization
asked Nov 29 at 15:04
not an alien
215110
asked Nov 29 at 15:04
not an alien
215110
asked Nov 29 at 15:04
not an alien
215110
asked Nov 29 at 15:04
not an alien
215110
asked Nov 29 at 15:04
not an alien
215110
215110
9
Why do you insist on doing this with "curly brace initialization"?
– Max Langhof
Nov 29 at 15:08
5
@MaxLanghof Well, it is called uniform initialization. One could be lead to believe it should be the preferred way to initialize an object ;)
– NathanOliver
Nov 29 at 15:10
3
@NathanOliver Related xkcd.com/927
– liliscent
Nov 29 at 15:12
2
@OP Here is a very good talk on the nightmare of C++ initialization: youtube.com/watch?v=7DTlWPgX6zs
– NathanOliver
Nov 29 at 15:12
1
@MaxLanghof because some recommend to use it everywhere as it is "safer" (does not have most vexing parse issue). In reality it is more dangerous.
– Slava
Nov 29 at 15:13
|
show 7 more comments
9
Why do you insist on doing this with "curly brace initialization"?
– Max Langhof
Nov 29 at 15:08
5
@MaxLanghof Well, it is called uniform initialization. One could be lead to believe it should be the preferred way to initialize an object ;)
– NathanOliver
Nov 29 at 15:10
3
@NathanOliver Related xkcd.com/927
– liliscent
Nov 29 at 15:12
2
@OP Here is a very good talk on the nightmare of C++ initialization: youtube.com/watch?v=7DTlWPgX6zs
– NathanOliver
Nov 29 at 15:12
1
@MaxLanghof because some recommend to use it everywhere as it is "safer" (does not have most vexing parse issue). In reality it is more dangerous.
– Slava
Nov 29 at 15:13
9
9
Why do you insist on doing this with "curly brace initialization"?
– Max Langhof
Nov 29 at 15:08
Why do you insist on doing this with "curly brace initialization"?
– Max Langhof
Nov 29 at 15:08
5
5
@MaxLanghof Well, it is called uniform initialization. One could be lead to believe it should be the preferred way to initialize an object ;)
– NathanOliver
Nov 29 at 15:10
@MaxLanghof Well, it is called uniform initialization. One could be lead to believe it should be the preferred way to initialize an object ;)
– NathanOliver
Nov 29 at 15:10
3
3
@NathanOliver Related xkcd.com/927
– liliscent
Nov 29 at 15:12
@NathanOliver Related xkcd.com/927
– liliscent
Nov 29 at 15:12
2
2
@OP Here is a very good talk on the nightmare of C++ initialization: youtube.com/watch?v=7DTlWPgX6zs
– NathanOliver
Nov 29 at 15:12
@OP Here is a very good talk on the nightmare of C++ initialization: youtube.com/watch?v=7DTlWPgX6zs
– NathanOliver
Nov 29 at 15:12
1
1
@MaxLanghof because some recommend to use it everywhere as it is "safer" (does not have most vexing parse issue). In reality it is more dangerous.
– Slava
Nov 29 at 15:13
@MaxLanghof because some recommend to use it everywhere as it is "safer" (does not have most vexing parse issue). In reality it is more dangerous.
– Slava
Nov 29 at 15:13
|
show 7 more comments
2 Answers
2
active
oldest
votes
Short answer: you don't.
This is not a problem with uniform initialization per se, but with std::initializer_list. There is a special rule in overload resolution that always gives priority to constructors taking std::initializer_list if list-initialization is used, regardless of the existence of other constructors which might require less implicit conversions.
I would suggest using
std::vector<unsigned char> buffer(bufferSize, ' ');
or, if you really want to use list-initialization, create your wrapper around std::vector that provides constructor overloads that do the right thing.
answered Nov 29 at 15:19
Vittorio Romeo
56.8k17152292
add a comment |
The two relevant overload of std::vector are:
explicit vector( size_type count,
const T& value = T(),
const Allocator& alloc = Allocator()); //(1)
vector( std::initializer_list<T> init,
const Allocator& alloc = Allocator() ); // (2)
These two overload has clear meaning, where the second is used to initialize the vector with the elements of the std::initializer_list.
Overload resolution prefer initializer-list constructors when list-initialization is used.
Narrowing conversions are not allowed with list-initialization, you're trying to create a std::vector with T=unsigned char but the deduced T for the std::initializer_list parameter is T= unsigned long which will involve a narrowing conversion (not allowed).
answered Nov 29 at 15:21
Jans
7,26422233
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Short answer: you don't.
This is not a problem with uniform initialization per se, but with std::initializer_list. There is a special rule in overload resolution that always gives priority to constructors taking std::initializer_list if list-initialization is used, regardless of the existence of other constructors which might require less implicit conversions.
I would suggest using
std::vector<unsigned char> buffer(bufferSize, ' ');
or, if you really want to use list-initialization, create your wrapper around std::vector that provides constructor overloads that do the right thing.
answered Nov 29 at 15:19
Vittorio Romeo
56.8k17152292
add a comment |
Short answer: you don't.
This is not a problem with uniform initialization per se, but with std::initializer_list. There is a special rule in overload resolution that always gives priority to constructors taking std::initializer_list if list-initialization is used, regardless of the existence of other constructors which might require less implicit conversions.
I would suggest using
std::vector<unsigned char> buffer(bufferSize, ' ');
or, if you really want to use list-initialization, create your wrapper around std::vector that provides constructor overloads that do the right thing.
answered Nov 29 at 15:19
Vittorio Romeo
56.8k17152292
add a comment |
Short answer: you don't.
This is not a problem with uniform initialization per se, but with std::initializer_list. There is a special rule in overload resolution that always gives priority to constructors taking std::initializer_list if list-initialization is used, regardless of the existence of other constructors which might require less implicit conversions.
I would suggest using
std::vector<unsigned char> buffer(bufferSize, ' ');
or, if you really want to use list-initialization, create your wrapper around std::vector that provides constructor overloads that do the right thing.
answered Nov 29 at 15:19
Vittorio Romeo
56.8k17152292
Short answer: you don't.
This is not a problem with uniform initialization per se, but with std::initializer_list. There is a special rule in overload resolution that always gives priority to constructors taking std::initializer_list if list-initialization is used, regardless of the existence of other constructors which might require less implicit conversions.
I would suggest using
std::vector<unsigned char> buffer(bufferSize, ' ');
or, if you really want to use list-initialization, create your wrapper around std::vector that provides constructor overloads that do the right thing.
answered Nov 29 at 15:19
Vittorio Romeo
56.8k17152292
answered Nov 29 at 15:19
Vittorio Romeo
56.8k17152292
answered Nov 29 at 15:19
Vittorio Romeo
56.8k17152292
answered Nov 29 at 15:19
Vittorio Romeo
56.8k17152292
56.8k17152292
add a comment |
add a comment |
The two relevant overload of std::vector are:
explicit vector( size_type count,
const T& value = T(),
const Allocator& alloc = Allocator()); //(1)
vector( std::initializer_list<T> init,
const Allocator& alloc = Allocator() ); // (2)
These two overload has clear meaning, where the second is used to initialize the vector with the elements of the std::initializer_list.
Overload resolution prefer initializer-list constructors when list-initialization is used.
Narrowing conversions are not allowed with list-initialization, you're trying to create a std::vector with T=unsigned char but the deduced T for the std::initializer_list parameter is T= unsigned long which will involve a narrowing conversion (not allowed).
answered Nov 29 at 15:21
Jans
7,26422233
add a comment |
The two relevant overload of std::vector are:
explicit vector( size_type count,
const T& value = T(),
const Allocator& alloc = Allocator()); //(1)
vector( std::initializer_list<T> init,
const Allocator& alloc = Allocator() ); // (2)
These two overload has clear meaning, where the second is used to initialize the vector with the elements of the std::initializer_list.
Overload resolution prefer initializer-list constructors when list-initialization is used.
Narrowing conversions are not allowed with list-initialization, you're trying to create a std::vector with T=unsigned char but the deduced T for the std::initializer_list parameter is T= unsigned long which will involve a narrowing conversion (not allowed).
answered Nov 29 at 15:21
Jans
7,26422233
add a comment |
The two relevant overload of std::vector are:
explicit vector( size_type count,
const T& value = T(),
const Allocator& alloc = Allocator()); //(1)
vector( std::initializer_list<T> init,
const Allocator& alloc = Allocator() ); // (2)
These two overload has clear meaning, where the second is used to initialize the vector with the elements of the std::initializer_list.
Overload resolution prefer initializer-list constructors when list-initialization is used.
Narrowing conversions are not allowed with list-initialization, you're trying to create a std::vector with T=unsigned char but the deduced T for the std::initializer_list parameter is T= unsigned long which will involve a narrowing conversion (not allowed).
answered Nov 29 at 15:21
Jans
7,26422233
The two relevant overload of std::vector are:
explicit vector( size_type count,
const T& value = T(),
const Allocator& alloc = Allocator()); //(1)
vector( std::initializer_list<T> init,
const Allocator& alloc = Allocator() ); // (2)
These two overload has clear meaning, where the second is used to initialize the vector with the elements of the std::initializer_list.
Overload resolution prefer initializer-list constructors when list-initialization is used.
Narrowing conversions are not allowed with list-initialization, you're trying to create a std::vector with T=unsigned char but the deduced T for the std::initializer_list parameter is T= unsigned long which will involve a narrowing conversion (not allowed).
answered Nov 29 at 15:21
Jans
7,26422233
edited Nov 29 at 15:27
answered Nov 29 at 15:21
Jans
7,26422233
answered Nov 29 at 15:21
Jans
7,26422233
answered Nov 29 at 15:21
Jans
7,26422233
7,26422233
add a comment |
add a comment |
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9
Why do you insist on doing this with "curly brace initialization"?
– Max Langhof
Nov 29 at 15:08
5
@MaxLanghof Well, it is called uniform initialization. One could be lead to believe it should be the preferred way to initialize an object ;)
– NathanOliver
Nov 29 at 15:10
3
@NathanOliver Related xkcd.com/927
– liliscent
Nov 29 at 15:12
2
@OP Here is a very good talk on the nightmare of C++ initialization: youtube.com/watch?v=7DTlWPgX6zs
– NathanOliver
Nov 29 at 15:12
1
@MaxLanghof because some recommend to use it everywhere as it is "safer" (does not have most vexing parse issue). In reality it is more dangerous.
– Slava
Nov 29 at 15:13