c99

Why does not GCC compile the C99 by default? I mean why is it necessary to add --std=c99 flag everytime a code in C99 is written? Edit: As of GCC 5, -std=gnu11 is the default. See Porting to GCC 5 . See C Dialect Options , gnu89 is the default. `gnu89' GNU dialect of ISO C90 (including some C99 features). This is the default for C code. As @tsv mentioned, ISO C99 is not fully supported yet: `c99' `c9x' `iso9899:1999' `iso9899:199x' ISO C99. Note that this standard is not yet fully supported; see http://gcc.gnu.org/c99status.html for more information. The names `c9x' and `iso9899:199x' are

I just did a experiment yesterday, and find something confusing: #include <stdio.h> int main() { int j; scanf("%d",&j); const int i = j; int arr[i]; return 0; } The number j is read from keyboard and it’s used to allocate the array arr on the stack. The compiler does not even know the size of the array at compile time (initializes j to 0?), but there is no compilation error. How is it possible? Variable length arrays were added to C99. It's described in the C99 rationale: 6.7.5.2 Array declarators C99 adds a new array type called a variable length array type. The inability to declare arrays

The following code compiles and runs just fine in C (at least according to 'gcc -std=gnu99'), but it fails to compile under C++, giving "line 5: error: cannot convert 'double' to 'double complex ' in initialization". Does anybody know why? #include "/usr/include/complex.h" #include <stdio.h> int main(int argc, char * argv[]) { double complex a = 3; // ERROR ON THIS LINE printf("%lf\n", creal(a)); return 0; } I realize there is another way of doing complex numbers in C++, but I have to use C complex numbers in C++, because that is how the legacy code I was given does things. Thanks if you can

I would like to compute the sum, rounded up, of two IEEE 754 binary64 numbers. To that end I wrote the C99 program below: #include <stdio.h> #include <fenv.h> #pragma STDC FENV_ACCESS ON int main(int c, char *v[]){ fesetround(FE_UPWARD); printf("%a\n", 0x1.0p0 + 0x1.0p-80); } However, if I compile and run my program with various compilers: $ gcc -v … gcc version 4.2.1 (Apple Inc. build 5664) $ gcc -Wall -std=c99 add.c && ./a.out add.c:3: warning: ignoring #pragma STDC FENV_ACCESS 0x1p+0 $ clang -v Apple clang version 1.5 (tags/Apple/clang-60) Target: x86_64-apple-darwin10 Thread model: posix $

what is the best way to compare int arrays b and c with a: int a[] = {0,1,0,0,1}; int b[] = {0,1,0,0,1}; int c[] = {1,1,0,0,1}; b and c are just examples, assume they can be any combination of 0s and 1s. I am trying to detect arrays identical to a. I have googled this for a while and have not found a satisfactory answer. This is a beginners question I realise, thank you for your patience. Use the standard memcmp function from <string.h> . memcmp(a, b, sizeof(a)) == 0 whenever a and b are equal. If you mean int a[] = {0,1,0,0,1}; int b[] = {0,1,0,0,1}; int c[] = {1,1,0,0,1}; then memcmp(a, b,