add char to string without help of functions

Question

for my program I need to add char(char) to string(char *) without using standard library or IO functions. For example:

char *s = "This is GOO";
char c = 'D';

s = append(s, c);

and s would produce string "This is GOOD". Is there some proper way to manipulate arrays in order to achieve this? Also it would be sufficient to produce string out of number of characters. I'm pretty sure I can use malloc, not positive though...

char * app(char* s, char c){
    char *copy;
    int l = strlen_(s);
    copy = malloc(l+1);
    copy = s;
    copy[l] = c;
    copy[l+1] = '\0';
    return copy;
}

can't use strcpy

John Kugelman · Accepted Answer · 2013-03-14 03:25:19Z

1

Without giving away the answer, since this sounds like classwork, here's what you want to do at a high level:

Determine the length of the string, i.e. find the '\0' terminator.
Allocate a new char array which is one character longer.
Copy the old string to the new string.
Add the new character on at the end.
Make sure there's a '\0' terminator at the end of the new string.

(If you're allowed to modify the existing string then you could possibly skip steps 2 and 3. But in your example char *s = "This is GOO" has s pointing to an unmodifiable string literal, meaning you can't modify it in place and have to work with a copy.)

Comments on the code you posted:

char * app(char* s, char c) {
    char *copy;
    int l = strlen_(s);
    copy = malloc(l+1);    /* should be +2: +1 for the extra character and +1 for \0 */
    copy = s;              /* arrays must be copied item by item. need a for loop */
    copy[l] = c;
    copy[l+1] = '\0';
    return copy;
}

edited Mar 14, 2013 at 3:25

answered Mar 14, 2013 at 2:59

John Kugelman

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4 Comments

Jim Balter Over a year ago

s = append(s, c); implies that s is not modified by append, unless append has a truly wretched specification.

Medardas Over a year ago

' char * app(char* s, char c){ char *copy; int l = strlen_(s); copy = malloc(l+1); copy = s; copy[l] = c; copy[l+1] = '\0'; return copy; } ' segmentation fault =/

Medardas Over a year ago

how to make this format like a code here...? Anyway, how it is possible to copy old string to the new one without strcpy

Medardas Over a year ago

@duskast gave me an answer and you gave me an understanding ,thanks a lot

user1149862 · Accepted Answer · 2013-03-14 03:14:27Z

1

#include <stdlib.h>

char *append(char *s, char c)
{
    int i = 0, j = 0;
    char *tmp;
    while (s[i] != '\0')
        i++;
    tmp = malloc((i+2) * sizeof(char));
    while (j < i)
    {
        tmp[j] = s[j];
        j++;
    }
    tmp[j++] = c;
    tmp[j] = '\0';
    return tmp;
}

int main(void)
{
    char *s = "This is Goo";
    char c = 'D';
    s = append(s, c);
    return 0;
}

answered Mar 14, 2013 at 3:14

user1149862

2 Comments

Jim Balter Over a year ago

a) use strlen b) sizeof(char) is always 1 c) use memcpy or strcpy

Jim Balter Over a year ago

Oops, sorry, missed the requirement that strcpy can't be used. Still, nothing said about strlen or memcpy.

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