convert a computed string to uint32_t without coping it into a char[] buffer in c

Question

I am trying to convert a substring to integer but I am getting this output

    int main()
{
    char buf[50] = "210-567-12-2040-34567890.txt";
    char* pStart = buf;
    char* pCurrent = buf;
    while(*pCurrent != '\0')
    {
        if (*pCurrent == '-' || *pCurrent == '.') 
        {
            uint32_t val = strtoul(pStart, NULL, 10); 
            pStart = pCurrent+1;
            printf("%ul\n",val);
        }
        ++pCurrent;
    }
    return 0;
}

I am getting this output

Why is there a l in it?

Literally nothing in `buf` is a numeric value. `strtoul` won't work on it no matter what. What sort of end result are you expecting? — ShadowRanger, Apr 17 '18 at 00:35
what wrong with `strtoul(pStart, NULL, 10);` ? Note that your code isn't not very clear. — Stargateur, Apr 17 '18 at 00:35

score 0 · Answer 1 · answered Apr 17 '18 at 00:42

0

Some of the answers here are what you're looking for: What is the difference between char s[] and char *s?

Short answer is there's no difference between the two as a function argument. You can pass the char* pStart variable into strtoul the same as you'd pass the char str[20] variable into it. This will make it read the string directly from whatever memory address pStart points to, no need to copy into an array first.

answered Apr 17 '18 at 00:42

sudo

5,604
5
40
78

But, I need to convert it upto "pCurrent - pStart", how do I do that? – sephora Apr 17 '18 at 00:46

convert a computed string to uint32_t without coping it into a char[] buffer in c

1 Answers1