How to remove leading spaces in a dynamic char array?

Question

I need to remove leading spaces from dynamic character array in C. My application almost works, but it leaves just one space at the beginning. How to get rid of all leading spaces from the given text? I should not use functions from string.h. Heres my code:

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

int mystrlen(char *tab)
{
    int i = 0;
    while(tab[i] != '\0')
    {
        i++;
    }
    return i;
}

char* ex6(char *tab)
{
    int spaces = 0, i, j = 0, s = 0;
    for(i=0; tab[i] != '\0'; i++)
    {
        if (tab[i] == ' ')
            spaces ++;
        else
            break;
    }

    spaces -= 1;

    char *w = (char*) malloc (sizeof(char) * (mystrlen(tab) - spaces + 1));
    for(i=0; tab[i] != '\0'; i++)
    {
        if (tab[i] == ' ')
        {
            ++ s;
        }
        if(s > spaces)
        {
            w[j] = tab[i];
            j ++;
        }
    }
    w[j] = 0;
    return w;
}

int main()
{
    char txt[] = "        Hello World";

    char *w = ex6(txt);
    printf("%s\n", w);

    free(w);
    w = NULL;

    return 0;
}

[Thou shalt not cast the result of malloc in C](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc). — Paul R, Jan 05 '15 at 15:51
You're already using `strlen()`, which is defined in ``, so you need to decide exactly what the constraints are here. — Paul R, Jan 05 '15 at 15:53

score 3 · Answer 1 · answered Jan 05 '15 at 16:03

Modifying the string in-place allows stripping leading spaces in a fairly compact manner since you don't need to calculate the length of the result string:

/* No includes needed for ltrim. */

void ltrim( char *s )
{
    const char *t = s;
    while ( *t && *t == ' ' )
        ++t;

    while ( ( *s++ = *t++ ) )
        ;
}

#include <stdio.h>

int main()
{
    char txt[] = "        Hello World";

    ltrim(txt);
    printf("%s\n", txt);

    return 0;
}

score 2 · Answer 2 · answered Jan 05 '15 at 15:48

2

The problem is in spaces -= 1 line. You go back 1 space left.

answered Jan 05 '15 at 15:48

i486

6,491
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1

whe I removed this, I got only ` World` word – Brian Brown Jan 05 '15 at 15:50

Iharob Al Asimi · Answer 3 · 2015-01-05T16:07:02.227

1

You can just move the tab pointer forward using pointer arithmetic, and then count the remaining characters in the string, then allocate space for the new string and copy each character to the new allocated space.

This is how to do it without strings.h

char* ex6(char *tab)
{
    int   i;
    char *w;
    while ((*tab == ' ') && (*tab != '\0'))
        tab++;
    w = malloc(mystrlen(tab) + 1);
    i = 0;
    while (*tab != '\0')
        w[i++] = *tab++;
    w[i] = '\0';
    return w;
}

edited Jan 05 '15 at 16:07

answered Jan 05 '15 at 15:51

Iharob Al Asimi

52,653
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Both `memmove` as well as `strdup` are from `string.h` which is not available to the OP. – Frerich Raabe Jan 05 '15 at 15:59
@FrerichRaabe I know, but he used `strlen` in his code, and didn't mention that `string.h` was forbidden. – Iharob Al Asimi Jan 05 '15 at 16:01

How to remove leading spaces in a dynamic char array?

3 Answers3