How to pass array "by reference" in C?

Question

This is what I'm trying to do but my code is either not compiling or giving me an unexpected output "BC" instead of just "B".

#include <stdio.h>

void removeFirstAndLastChar(char** string) {
    *string += 1; // Removes the first character
    int i = 0;
    for (; *string[i] != '\0'; i++);
    *string[i - 1] = '\0';
}

int main(void) {
    char* title = "ABC";
    removeFirstAndLastChar(&title);
    printf("%s", title);
    // Expected output: B
    return 0;
}

I looked through a lot of answers here related to passing pointers by reference but none of them seemed to contain the operations that I want to do in my removeFirstAndLastChar() function.

Undefined behavior for attempting to modify a string literal. — EOF, Jan 02 '17 at 12:55
You should change `char* title = "ABC";` to `char title[] = "ABC";` — mch, Jan 02 '17 at 12:55
@mch I don't want to do that because this example is taken from bigger code where I want it to be a pointer — hb20007, Jan 02 '17 at 12:56
@hb20007 Then dynamically allocate memory for the string using `malloc`. — Spikatrix, Jan 02 '17 at 12:57
The best thing you can do is read a book of C. SO is not a tutorial site. There are hundred of questions about that. — Stargateur, Jan 02 '17 at 12:57
@Nidhoegger using a pointer to the literal is not correct at all, since the function modifies it. — Quentin, Jan 02 '17 at 12:57
@KerrekSB I tried that but I still get the incorrect output BC — hb20007, Jan 02 '17 at 12:58
"_my code is either not compiling or giving me an unexpected output_" - Err, What? — Spikatrix, Jan 02 '17 at 13:00
Sure, your code is still broken, but as far as effecting reference semantics in C that seemed like the biggest issue, at least for other readers. — Kerrek SB, Jan 02 '17 at 13:00
@BLUEPIXY You don't even need a compound literal. You could just do `char storage[] = "ABC"; char *title = storage;`. — melpomene, Jan 02 '17 at 13:02
@melpomene: That would require a variable with no other use than having its address taken. The compound literal seems like an elegant way to achieve the same thing. — Kerrek SB, Jan 02 '17 at 13:05
C does not support pass-by-reference. It is **strictly** pass-by-value. — too honest for this site, Jan 02 '17 at 13:10
```char* title = "ABC"; ``` Modifying a literal string in C is an undefined behavior. It can lead to the result you want as much as it can lead to a segfault. A good compiler will notify it to you with a warning like `error: assignment of read-only location 'name'`. You can prevent it in C by using `const char * ` for literals. — informaticienzero, Jan 02 '17 at 13:47

score 2 · Accepted Answer · answered Jan 02 '17 at 13:18

I do not judge your algorithm or C conventions, friends who comment on your problem are totally right. But if you still do it in this way you can use this approach.

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

void removeFirstAndLastChar(char* string) {
    memmove(string,string+1,strlen(string));
    string[strlen(string)-1]=0;
}

int main(void) {
    char title[] = "ABC";
    removeFirstAndLastChar(title);
    printf("%s", title);
    // Expected output: B
    return 0;
}

How to pass array "by reference" in C?

1 Answers1