Question

I have the following code:

#include <stdlib.h>
#include <stdio.h>
#include <errno.h>

void main(void)
{
     int data;
     char * tmp;
     data = strtol("23ef23",&tmp,10);
     printf("%d",errno);
     getchar();
}

output is 0 …

why?

i am using visual studio 2010 C++
code must be C89 compatible.

Answer 1

strtol only sets errno for overflow conditions, not to indicate parsing failures. For that purpose, you have to check the value of the end pointer, but you need to store a pointer to the original string:

char const * const str = "blah";
char const * endptr;

int n = strtol(str, &endptr, 0);

if (endptr == str) { /* no conversion was performed */ }

else if (*endptr == '\0') { /* the entire string was converted */ }

else { /* the unconverted rest of the string starts at endptr */ }

I think the only required error values are for underflow and overflow.

Conversely, if the entire string has been consumed in the conversion, you have *endptr = '\0', which may be an additional thing you might want to check.