Thread: how to retain an allocated memory

I'm trying to make my own strcpy. A version of strcpy that will allow this,

PHP Code:

[code]
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void) {
    char *content1;
    char *content2 = "hello world";
    
    printf("%s\n", content2);
    
    strcpy(content1, content2);
    
    printf("%s\n", content1);
    
    return 0;
}
[/code] 


Of course the app crashes upon strcpy attempt because content1 doesn't have memory allocated in it. It's why I wrote this function.

PHP Code:

[code]
void my_strcpy(char *dest, const char *src) {
    dest = (char *) malloc(sizeof(strlen(src)));
    
    while(*src) {
        *dest++ = *src++;
    }
    *dest = '\0';
}
[/code] 


where I was hoping my_strcpy(content1, content2); would work as expected. But no. content1 is NULL;

How do I retain the allocated memory in my_strcpy's dest parameter?

One solution is to make my_strcpy a non-void function.

PHP Code:

[code]
char *my_strcpy(char *dest, const char *src) {
    dest = (char *) malloc(sizeof(strlen(src)));
    char *destcpy = dest;
    
    while(*src) {
        *dest++ = *src++;
    }
    *dest = '\0';
    
    return destcpy;
}
[/code] 


But I would have to use content1 = my_strcpy(content1, content2);, and not as simple as my_strcpy(content1, content2); like strcpy does.

How do I pass-by-reference and retain allocated memory?