c89

问题 I wrote such sample code: #include <stdio.h> #include <sys/types.h> #include <sys/stat.h> #include <fcntl.h> #include <errno.h> char* print_errno_msg(int value); int main(void){ struct stat buffer; int status; status = stat("./main.c", &buffer); char *msg = print_errno_msg(errno); /* syntax error : missing ';' before 'type' */ printf("status = %i; errno = %i; %s\n", status, errno, msg); /* 'msg' : undeclared identifier */ errno = 0; int ch = getchar(); /* syntax error : missing ';' before

问题 Answers to this and this question say that function pointers of the form return-type (*pointer)() are pointers to a function which takes any number of arguments, though the latter says they obsolesced in C11. On an i386 system with GCC, “extra” arguments passed in a call to an empty-parentheses-type’d function pointer are ignored, because of how stack frames work; e.g., /* test.c */ #include <stdio.h> int foo(int arg) { return arg; } int main(void) { int (*fp)() = foo; printf("%d\n", fp

来源： https://stackoverflow.com/questions/44590043/why-is-generic-keyword-supported-in-c99-or-c90-modes

来源： https://stackoverflow.com/questions/62785780/how-do-i-implement-a-modulus-operator-for-double-variables-using-frexp

来源： https://stackoverflow.com/questions/62785780/how-do-i-implement-a-modulus-operator-for-double-variables-using-frexp

来源： https://stackoverflow.com/questions/62785780/how-do-i-implement-a-modulus-operator-for-double-variables-using-frexp

问题 In my current project, which uses the MISRA 2004 standard, we use three GCC compilers, versions 3.2.3, 4.4.2 and 5.4.0. We run build checks with the pedantic switch and c89 standard and a load of other restrictions. One of the restrictions is that all data must be initialised at declaration. I have a problem in that on GCC 3.2.3, the universal zero initialiser {0} only compiles for arrays of the basic unitary types. If I have an array of structs, then I get a missing braces warning and the

问题 When looking for evidence of unsigned long being enough to hold size_t for the purpose of being argument to printf I ran into two fact(oid)s. First there's this answer stating that long is indeed not guaranteed to be large enough for size_t . On the other hand I saw this answer suggesting to use printf("%lu", (unsigned long)x) in pre C99, x being of size_t . So the question is could you assume or were you guaranteed that long were enough to hold size_t in pre C99 . The other question is

问题 When looking for evidence of unsigned long being enough to hold size_t for the purpose of being argument to printf I ran into two fact(oid)s. First there's this answer stating that long is indeed not guaranteed to be large enough for size_t . On the other hand I saw this answer suggesting to use printf("%lu", (unsigned long)x) in pre C99, x being of size_t . So the question is could you assume or were you guaranteed that long were enough to hold size_t in pre C99 . The other question is

问题 When looking for evidence of unsigned long being enough to hold size_t for the purpose of being argument to printf I ran into two fact(oid)s. First there's this answer stating that long is indeed not guaranteed to be large enough for size_t . On the other hand I saw this answer suggesting to use printf("%lu", (unsigned long)x) in pre C99, x being of size_t . So the question is could you assume or were you guaranteed that long were enough to hold size_t in pre C99 . The other question is