I was learning C with a book called "C in Linux" by David Haskins but there's a problem. When i try to compile this code:
#include <stdio.h>
#include <string.h>
int main (int argc, char *argv[], char *env[]) {
printf("Content-type:text/html\n\n<html><body bgcolor=#23abe2>\n");
char value[256] = "";
strncpy(value,(char *) getenv("QUERY_STRING"), 255);
printf("QUERY_STRING:%s<BR>\n", value );
printf("<form>\n");
printf("<input type=\"TEXT\" name=\"ITEM1\"> \n");
printf("<input type=\"TEXT\" name=\"ITEM2\"> \n");
printf("<input type=\"SUBMIT\">");
printf("</form></body></html>\n");
return 0;
}
The terminal shows this warning!
chapter4_1.c: In function ‘main’:
chapter4_1.c:14:16: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]
You forgot to
#include <stdlib.h>. This means thatgetenv()isn't declared anywhere, so it's assumed to return anintby default, which you're casting tochar *. On a 64-bit machine,int(32 bits) andchar *(64 bits) have different sizes, hence the warning.As an aside, the cast to
char *is not necessary, sincegetenv()already returns achar *. The cast only serves to mask errors (i.e. without it, the program would have given you a clear error message about passing anintto achar *).