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?

dest is a local variable, so any changes you make to it will not be visible outside the function. Make it a double pointer instead if you want to change the value of the pointer.

Code:

void my_strcpy(char **dest, char *src)
{
    ...
    *dest = malloc(...);
}

Try passing char** dest and allocating for *dest.

Originally Posted by creek23

Saying there is none and saying you can simulate it is a contradicting statement. Just as saying there no pointers in Java but you can fake one.

Of course, I'm aware that C uses pointers to fake passing by reference. "reference" is actually synanymous to "pointer".


It's not like I put random codes like:

PHP Code:

[code]
int char * () {
  include <&var> = "helloworld";
  echo **var;
}
[/code] 


Of course, I'm reading some reference and tutorials. But much of what I want to know are very specific and most of the tutorials and books are discussing very general thought. Like when I wanted to know how to write resizeable array of resizeable strings, it will be very hard to find a tutorial that discuss that specific topic. It's why I'm asking in this forum for help. And I'm thankful that there are people responding to my inqueries.

(I may sound like I'm ranting but I'm not )

Actually neither C nor Java are "faking" anything. They are both pass-by-value languages.

Originally Posted by creek23

Of course, I'm aware that C uses pointers to fake passing by reference. "reference" is actually synanymous to "pointer".

Ultimately, ALL forms of "pass by reference" (in any language) use this method -- a reference is the location of a data structure. Logically, there is no other way to do this.

So pass-by reference vs. value discourse is somewhat chimeric and rhetorical. It's essentially about normative nomenclature. In C it's "fake", since a pointer is a distinct datatype of it's own, whereas a reference (which C doesn't have) is more just a syntactical method used in other languages (and many of those, nb, use interpreters written in C, so C can be used to implement references, they are just not part of the syntax).

I wanted to know how to write resizeable array of resizeable strings, it will be very hard to find a tutorial that discuss that specific topic.

You need to learn about memory allocation and re-allocation -- malloc(), realloc(), free(). And get comfy with pointers. At that point it will be self-explanatory.