Reason for (type) in (type)function() if the output of function() is ignored
0
I was looking through the Linux kernel source code and in the file init/main.c I saw the line
(void) ksys_dup(0);
I was wondering why (void) is there. I've seen code like
char *str = (char *)malloc(len);
This makes sense, as malloc is declared to return a void pointer but the memory it allocates here needs to be pointed to by a char pointer. This does not explain the ksys_dup() line, as its return value isn't recorded.
ksys_dup() returns an integer, and there are function calls around that line whose returns also aren't recorded but don't have a type specified.
c types
asked Nov 23 '18 at 4:09
Billy
91219
add a comment |
0
I was looking through the Linux kernel source code and in the file init/main.c I saw the line
(void) ksys_dup(0);
I was wondering why (void) is there. I've seen code like
char *str = (char *)malloc(len);
This makes sense, as malloc is declared to return a void pointer but the memory it allocates here needs to be pointed to by a char pointer. This does not explain the ksys_dup() line, as its return value isn't recorded.
ksys_dup() returns an integer, and there are function calls around that line whose returns also aren't recorded but don't have a type specified.
c types
asked Nov 23 '18 at 4:09
Billy
91219
3
it is used to say explicitly that the return value is ignored
– shan
Nov 23 '18 at 4:13
1
This makes sense, as malloc is declared to return a void pointer but the memory it allocates here needs to be pointed to by a char pointer. – no, especially with*alloc()it makes no sense sincevoid*is implicitly convertible to all other pointer types. It just adds noise with no function.
– Swordfish
Nov 23 '18 at 4:26
See Should I cast the result ofmalloc()? for diverging opinions on the merits or otherwise of the cast for a memory allocation function. Suffice to say, there are strong disagreements, but the majority is probably on the "don't cast" side (but the "it might hide that you've not declaredmalloc()by including<stdlib.h>" excuse is pathetic, and has been ever since C99 mandated that functions be declared before use — that's a problem with using compilers that only conform to archaic versions of the C standard).
– Jonathan Leffler
Nov 23 '18 at 7:02
@Swordfish I've never casted malloc() in my own code, but I've seen other developers do that and I was just using it as an example.
– Billy
Nov 23 '18 at 15:59
Precisely, the value returned is not used. This is the use of the(void)cast, to avoid some compilers warning about a function returning a value that is dropped. The case ofmalloc()I'm afraid is different, you are hidding the actual type returned with a cast (the type of malloc will be assumed by the compiler to be anintif you make the mistake of not including the proper header) and that will hide the error shutting down the compiler warning (which makes the warning useles) See other comments for Why should you never cast the value returned by malloc(3).
– Luis Colorado
Nov 27 '18 at 8:38
add a comment |
0
0
0
I was looking through the Linux kernel source code and in the file init/main.c I saw the line
(void) ksys_dup(0);
I was wondering why (void) is there. I've seen code like
char *str = (char *)malloc(len);
This makes sense, as malloc is declared to return a void pointer but the memory it allocates here needs to be pointed to by a char pointer. This does not explain the ksys_dup() line, as its return value isn't recorded.
ksys_dup() returns an integer, and there are function calls around that line whose returns also aren't recorded but don't have a type specified.
c types
asked Nov 23 '18 at 4:09
Billy
91219
I was looking through the Linux kernel source code and in the file init/main.c I saw the line
(void) ksys_dup(0);
I was wondering why (void) is there. I've seen code like
char *str = (char *)malloc(len);
This makes sense, as malloc is declared to return a void pointer but the memory it allocates here needs to be pointed to by a char pointer. This does not explain the ksys_dup() line, as its return value isn't recorded.
ksys_dup() returns an integer, and there are function calls around that line whose returns also aren't recorded but don't have a type specified.
c types
c types
asked Nov 23 '18 at 4:09
Billy
91219
asked Nov 23 '18 at 4:09
Billy
91219
asked Nov 23 '18 at 4:09
Billy
91219
asked Nov 23 '18 at 4:09
Billy
91219
asked Nov 23 '18 at 4:09
Billy
91219
91219
3
it is used to say explicitly that the return value is ignored
– shan
Nov 23 '18 at 4:13
1
This makes sense, as malloc is declared to return a void pointer but the memory it allocates here needs to be pointed to by a char pointer. – no, especially with*alloc()it makes no sense sincevoid*is implicitly convertible to all other pointer types. It just adds noise with no function.
– Swordfish
Nov 23 '18 at 4:26
See Should I cast the result ofmalloc()? for diverging opinions on the merits or otherwise of the cast for a memory allocation function. Suffice to say, there are strong disagreements, but the majority is probably on the "don't cast" side (but the "it might hide that you've not declaredmalloc()by including<stdlib.h>" excuse is pathetic, and has been ever since C99 mandated that functions be declared before use — that's a problem with using compilers that only conform to archaic versions of the C standard).
– Jonathan Leffler
Nov 23 '18 at 7:02
@Swordfish I've never casted malloc() in my own code, but I've seen other developers do that and I was just using it as an example.
– Billy
Nov 23 '18 at 15:59
Precisely, the value returned is not used. This is the use of the(void)cast, to avoid some compilers warning about a function returning a value that is dropped. The case ofmalloc()I'm afraid is different, you are hidding the actual type returned with a cast (the type of malloc will be assumed by the compiler to be anintif you make the mistake of not including the proper header) and that will hide the error shutting down the compiler warning (which makes the warning useles) See other comments for Why should you never cast the value returned by malloc(3).
– Luis Colorado
Nov 27 '18 at 8:38
add a comment |
3
it is used to say explicitly that the return value is ignored
– shan
Nov 23 '18 at 4:13
1
This makes sense, as malloc is declared to return a void pointer but the memory it allocates here needs to be pointed to by a char pointer. – no, especially with*alloc()it makes no sense sincevoid*is implicitly convertible to all other pointer types. It just adds noise with no function.
– Swordfish
Nov 23 '18 at 4:26
See Should I cast the result ofmalloc()? for diverging opinions on the merits or otherwise of the cast for a memory allocation function. Suffice to say, there are strong disagreements, but the majority is probably on the "don't cast" side (but the "it might hide that you've not declaredmalloc()by including<stdlib.h>" excuse is pathetic, and has been ever since C99 mandated that functions be declared before use — that's a problem with using compilers that only conform to archaic versions of the C standard).
– Jonathan Leffler
Nov 23 '18 at 7:02
@Swordfish I've never casted malloc() in my own code, but I've seen other developers do that and I was just using it as an example.
– Billy
Nov 23 '18 at 15:59
Precisely, the value returned is not used. This is the use of the(void)cast, to avoid some compilers warning about a function returning a value that is dropped. The case ofmalloc()I'm afraid is different, you are hidding the actual type returned with a cast (the type of malloc will be assumed by the compiler to be anintif you make the mistake of not including the proper header) and that will hide the error shutting down the compiler warning (which makes the warning useles) See other comments for Why should you never cast the value returned by malloc(3).
– Luis Colorado
Nov 27 '18 at 8:38
3
3
it is used to say explicitly that the return value is ignored
– shan
Nov 23 '18 at 4:13
it is used to say explicitly that the return value is ignored
– shan
Nov 23 '18 at 4:13
1
1
This makes sense, as malloc is declared to return a void pointer but the memory it allocates here needs to be pointed to by a char pointer. – no, especially with
*alloc() it makes no sense since void* is implicitly convertible to all other pointer types. It just adds noise with no function.– Swordfish
Nov 23 '18 at 4:26
This makes sense, as malloc is declared to return a void pointer but the memory it allocates here needs to be pointed to by a char pointer. – no, especially with
*alloc() it makes no sense since void* is implicitly convertible to all other pointer types. It just adds noise with no function.– Swordfish
Nov 23 '18 at 4:26
See Should I cast the result of
malloc()? for diverging opinions on the merits or otherwise of the cast for a memory allocation function. Suffice to say, there are strong disagreements, but the majority is probably on the "don't cast" side (but the "it might hide that you've not declared malloc() by including <stdlib.h>" excuse is pathetic, and has been ever since C99 mandated that functions be declared before use — that's a problem with using compilers that only conform to archaic versions of the C standard).– Jonathan Leffler
Nov 23 '18 at 7:02
See Should I cast the result of
malloc()? for diverging opinions on the merits or otherwise of the cast for a memory allocation function. Suffice to say, there are strong disagreements, but the majority is probably on the "don't cast" side (but the "it might hide that you've not declared malloc() by including <stdlib.h>" excuse is pathetic, and has been ever since C99 mandated that functions be declared before use — that's a problem with using compilers that only conform to archaic versions of the C standard).– Jonathan Leffler
Nov 23 '18 at 7:02
@Swordfish I've never casted malloc() in my own code, but I've seen other developers do that and I was just using it as an example.
– Billy
Nov 23 '18 at 15:59
@Swordfish I've never casted malloc() in my own code, but I've seen other developers do that and I was just using it as an example.
– Billy
Nov 23 '18 at 15:59
Precisely, the value returned is not used. This is the use of the
(void) cast, to avoid some compilers warning about a function returning a value that is dropped. The case of malloc() I'm afraid is different, you are hidding the actual type returned with a cast (the type of malloc will be assumed by the compiler to be an int if you make the mistake of not including the proper header) and that will hide the error shutting down the compiler warning (which makes the warning useles) See other comments for Why should you never cast the value returned by malloc(3).– Luis Colorado
Nov 27 '18 at 8:38
Precisely, the value returned is not used. This is the use of the
(void) cast, to avoid some compilers warning about a function returning a value that is dropped. The case of malloc() I'm afraid is different, you are hidding the actual type returned with a cast (the type of malloc will be assumed by the compiler to be an int if you make the mistake of not including the proper header) and that will hide the error shutting down the compiler warning (which makes the warning useles) See other comments for Why should you never cast the value returned by malloc(3).– Luis Colorado
Nov 27 '18 at 8:38
add a comment |
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3
it is used to say explicitly that the return value is ignored
– shan
Nov 23 '18 at 4:13
1
This makes sense, as malloc is declared to return a void pointer but the memory it allocates here needs to be pointed to by a char pointer. – no, especially with
*alloc()it makes no sense sincevoid*is implicitly convertible to all other pointer types. It just adds noise with no function.– Swordfish
Nov 23 '18 at 4:26
See Should I cast the result of
malloc()? for diverging opinions on the merits or otherwise of the cast for a memory allocation function. Suffice to say, there are strong disagreements, but the majority is probably on the "don't cast" side (but the "it might hide that you've not declaredmalloc()by including<stdlib.h>" excuse is pathetic, and has been ever since C99 mandated that functions be declared before use — that's a problem with using compilers that only conform to archaic versions of the C standard).– Jonathan Leffler
Nov 23 '18 at 7:02
@Swordfish I've never casted malloc() in my own code, but I've seen other developers do that and I was just using it as an example.
– Billy
Nov 23 '18 at 15:59
Precisely, the value returned is not used. This is the use of the
(void)cast, to avoid some compilers warning about a function returning a value that is dropped. The case ofmalloc()I'm afraid is different, you are hidding the actual type returned with a cast (the type of malloc will be assumed by the compiler to be anintif you make the mistake of not including the proper header) and that will hide the error shutting down the compiler warning (which makes the warning useles) See other comments for Why should you never cast the value returned by malloc(3).– Luis Colorado
Nov 27 '18 at 8:38