Updating the string using function in c

Question

I have written following c code to update the char array rb but it is printing garbage value

#include <stdio.h>
void update(char* buff){
    char word[] = "HII_O";
    buff = word;
    return;
}

int main(){
    char rb[6];
    update(rb);
    rb[5] = '\0';
    printf("[%s]\n",rb);
    return 0;
}

The restriction is we can't use any other library. So how to solve this

Answer 1

Within the function update the parameter buff is a local variable of the function that will not be alive after exiting the function.

You can imagine the function call the following way

update( rb );
//...
void update( /*char* buff*/){
    char *buff = rb;
    char word[] = "HII_O";
    buff = word;
    return;
}

As you see the original array was not changed.

That is at first the pointer buff was initialized by the address of the first element of the source array rb.

    char *buff = rb;

and then this pointer was reassigned with the address of the first element of the local character array word

    buff = word;

What you need is to copy characters of the string literal "HII_O" into the source array rb using standard string function strcpy or strncpy.

For example

#include <string.h>
#include <stdio.h>

void update(char* buff){
    strcpy( buff, "HII_O" );
}

int main(){
    char rb[6];
    update(rb);
    printf("[%s]\n",rb);
    return 0;
}