CODEWITHSUNDEEP
cmallocstrcpy
toni
toni

Reputation: 152

How to copy string without allocating memory - C

I just started in C and this is a beginner question. But is there any way to copy a string without allocating memory beforehand? I want to reproduce the strcopy function inbuilt in C with the same structure (char *strcpy(char *src, char *dest);). I got it to work with declaring the char pointer in the function itself and then returning it. It works also with the original structure as strcpy if I allocate the memory beforehand.

#include <stdlib.h>

char *strcopy(char *src, char *dest)
{
    int i = 0;
    while (src[i] != '\0')
    {
        dest[i] = src[i];
        i++;
    }
    dest[i] = '\0';
    return (dest);
}

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

int main(int argc, char **argv)
{
    char *src = argv[1];
    char *dest = (char*)malloc(str_size(src));
    dest = strcopy(src, dest);
    return (0);
}

So is it somehow possible to copy a string without allocating memory beforehand and without changing my function-head?

Upvotes: 0

Views: 1368

Answers (1)

quent
quent

Reputation: 2185

If you want to avoid dynamic allocation, use static allocation:

#include <stdlib.h>
#include <string.h>

#define BUFFER_SIZE        1024

char *strcopy(char *src, char *dest)
{
    int i = 0;
    while (src[i] != '\0')
    {
        dest[i] = src[i];
        i++;
    }
    dest[i] = '\0';
    return (dest);
}

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

int main(int argc, char **argv)
{
    char *src = argv[1];
    char dest[BUFFER_SIZE];
    
    if(strlen(src) >= BUFFER_SIZE){
        return -1;
    }
    else{
        dest = strcopy(src, dest);
    }

    return 0;
}

Upvotes: 1

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