c89

This question already has an answer here: Reliably determine the number of elements in an array 2 answers The usual approach to getting an array's element count in C in something like this: #define COUNTOF(arr) (sizeof(arr) / sizeof(arr[0])) This results in an integral-constant expression, which is a very nice plus as well. The problem is that it isn't type-safe: int* i; COUNTOF(i); /* compiles :( */ . In practice, this should come up rarely, but for the sake of correctness it would be nice to make this type-safe. In C++03 this is easy (and in C++11 it's even easier, left as an exercise for

I've read that C89 does not support variable-length arrays, but the following experiment seems to disprove that: #include <stdio.h> int main() { int x; printf("Enter a number: "); scanf("%d", &x); int a[x]; a[0] = 1; // ... return 0; } When I compile as such (assuming filename is va_test.c ): gcc va_test.c -std=c89 -o va_test It works... What am I missing? :-) GCC always supported variable length arrays AFAIK. Setting -std to C89 doesn't turn off GCC extensions ... Edit: In fact if you check here: http://gcc.gnu.org/onlinedocs/gcc/C-Dialect-Options.html#C-Dialect-Options Under -std= you will

i wanted to convert double to float in C, but wanted to preserve the decimal point exactly as possible without any changes... for example, let's say i have double d = 0.1108; double dd = 639728.170000; double ddd = 345.2345678 now correct me if i am wrong, i know that floating point precision is about 5 numbers after the dot. can i get those five numbers after the dot exactly as the double had it? so that above results as follows: float f = x(d); float ff = x(dd); float fff = x(ddd); printf("%f\n%f\n%f\n", f, ff, fff); it should print 0.1108 639728.17000 345.23456 all digits after the

Consider this C program: int main() { puts("Hello world!"); return 0; } This compiles and runs fine and as far as I understand, is legal C89. However, I'm not 100% sure about that. Compiling in C99 mode with clang informs me that implicit declaration of function 'puts' is invalid in C99 (which makes me think that the C standard must have changed in C99 to make implicit function declaration illegal, which is what I'm trying to confirm). Is implicit function declaration legal in C89? (even if it's a bad idea to do it (unless your in an obfuscated C code challenge)) Is implicit function

I'm very curious to know why exactly C89 compilers will dump on you when you try to mix variable declarations and code, like this for example: rutski@imac:~$ cat test.c #include <stdio.h> int main(void) { printf("Hello World!\n"); int x = 7; printf("%d!\n", x); return 0; } rutski@imac:~$ gcc -std=c89 -pedantic test.c test.c: In function ‘main’: test.c:7: warning: ISO C90 forbids mixed declarations and code rutski@imac:~$ Yes, you can avoid this sort of thing by staying away from -pedantic. But then your code is no longer standards compliant. And as anybody capable of answering this post

#include <stdio.h> #include <stdlib.h> float values[] = { 4, 1, 10, 9, 2, 5, -1, -9, -2,10000,-0.05,-3,-1.1 }; int compare (const void * a, const void * b) { return ( (int) (*(float*)a - *(float*)b) ); } int main () { int i; qsort (values, 13, sizeof(float), compare); for (i = 0; i < 13; i++) { printf ("%f ",values[ i ]); } putchar('\n'); return 0; } The result is: -9.000000 -3.000000 -2.000000 -1.000000 -1.100000 -0.050000 1.000000 2.000000 4.000000 5.000000 9.000000 10.000000 10000.000000 It's wrong because the order of -1 and -1.1 is changed. I believe it is happening because my "compare"