strtol not working as expected

Question

I am unable to convert a string to a long using strtol. Having a leading "." before the number in the string returns 0. Without the "." strtol returns 3456 as expected.

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

int main ()
{                                                                   
    char str[20] = " . 3456\r\n";                                                                    

    long ret = strtol(str, NULL, 10);                                            
    printf("ret is %ld\n",ret);

    return(0);                                                                   
}

Here is [the documentation](http://en.cppreference.com/w/cpp/string/byte/strtol) — Mike Kinghan, Mar 29 '18 at 11:27
@KarstenKoop Ok that is my misunderstanding of the function. Is there a similar function I can use that will fit this use case — homeGrown, Mar 29 '18 at 11:29
what is the usecase? Do you want a long with the value 3456 or a double with value 0.3456? — Karsten Koop, Mar 29 '18 at 11:30
@KarstenKoop A long with 3456. That `.` is not a decimal value it is just a char returned from a sensor so I want to ignore it. — homeGrown, Mar 29 '18 at 11:31
@zwol it doesn't report negative values. I only asked about ignoring the `.` — homeGrown, Mar 29 '18 at 11:34
@homeGrown But you're using `strtol` and `long` instead of `strtoul` and `unsigned long`, so I didn't know that. — zwol, Mar 29 '18 at 11:38

score 3 · Accepted Answer · answered Mar 29 '18 at 11:45

The strto* library functions will only skip over leading whitespace. If you want to skip over other text, you need to do it by hand. The isxxx functions from ctype.h can help with this:

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

int main (int argc, char **argv)
{
    char *p, *endp;
    unsigned long ret;
    int fail = 1;

    if (argc != 2) {
        fprintf(stderr, "usage: %s number-to-parse\n", argv[0]);
        return 2;
    }
    p = argv[1];
    while (*p && !isdigit(*p)) p++;

    errno = 0;
    ret = strtoul(p, &endp, 10);     
    if (endp == p)
        printf("'%s': no number found\n", str);
    else if (*endp && !isspace(*endp))
        printf("'%s': junk on line after number\n", str);
    else if (errno)
        printf("'%s': %s\n", str, strerror(errno));
    else {
        printf("'%s': parsed as %lu\n", str, ret);
        fail = 0;
    }
    return fail;
}

strtol not working as expected

1 Answers1