Correctly pass a char array and char pointer to function by reference in C

Question

Is there a right way to call a char array and a char pointer to go to a function but it's pass by reference where it will also be manipulated?

Something like this:

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

void manipulateStrings(char *string1, char *string2[])
{

    strcpy (string1, "Apple");
    strcpy (string2, "Banana");

    printf ("2 string1: %s", string1);
    printf ("2 string2: %s", &string2);

}


int main ()
{
    char *stringA;
    char stringB[1024];

    stringA = (char *) malloc ( 1024 + 1 );

    strcpy (stringA, "Alpha");
    strcpy (stringB, "Bravo");
    printf ("1 stringA: %s", stringA);
    printf ("1 stringB: %s", stringB);

    manipulateStrings(stringA, stringB);

    printf ("3 stringA: %s", stringA);
    printf ("3 stringB: %s", stringB);


    return 0;
}

I am not sure if I'm understanding correctly how to pass such variables to a function and change the values of those variables who happen to be char / strings

Edit: My question is - How would you be able to change the values of the two strings in the function?

What is the question?

Andrey
– Andrey

2013-03-25 03:03:25 +00:00
Commented Mar 25, 2013 at 3:03 — Andrey
– Andrey, Commented Mar 25, 2013 at 3:03
@Andrey - added an edit to clarify my question

AisIceEyes
– AisIceEyes

2013-03-25 03:06:10 +00:00
Commented Mar 25, 2013 at 3:06 — AisIceEyes
– AisIceEyes, Commented Mar 25, 2013 at 3:06

Ed Swangren · Accepted Answer · 2013-03-25 03:03:35Z

15

There is no such thing as pass by reference in C. Everything in C is passed by value. This leads to the solution you need; add another level of indirection.

However, your code has other problems. You don't need to pass a pointer to pointer (or pointer to array) because you are not mutating the input, only what it refers to. You want to copy a string. Great. All you need for that is a pointer to char initialized to point to a sufficient amount of memory.

In the future, if you need to mutate the input (i.e., assign a new value to it), then use a pointer to pointer.

int mutate(char **input) 
{
    assert(input);
    *input = malloc(some_size);
}

int main(void)
{
    /* p is an uninitialized pointer */
    char *p;
    mutate(&p);
    /* p now points to a valid chunk of memory */
    free(p);
    return 0;
}

answered Mar 25, 2013 at 3:03

Ed Swangren

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11 Comments

SevenBits Over a year ago

There most certainly is pass by reference in C. You do so with the & symbol prefixing the variable.

Ed Swangren Over a year ago

@SevenBits: No, there most certainly is not. & is the address of operator. Passing by reference is not the same as taking the address of soem variable. I was wondering how long it would take for someone to come along and incorrectly equate adding another level of indirection to passing by reference.

AisIceEyes Over a year ago

I'm still trying to internalize this but is there a way where stringB can also be mutated in another function if it was declared like how I declared it?

Ed Swangren Over a year ago

@AisIceEyes: What it refers to can be mutated, but stringB cannot be unless you pass a pointer to it. It's easier if you think of everything being passed by value (i.e., a copy, which it is). You can't mutate the copy, but in this case, what you are passing (the value) is an address. So, you can still muck around with what lives at that address, but if you want to modify the inout itself, you will need to get its address. Note that, in this case, you have an array, so taking its address results in a pointer to array, not a pointer to pointer to char. You cannot modify arrays directly

Ed Swangren Over a year ago

@SevenBits: No, it is different. Take, for example, the method which receives the argument; void whatever(int *p) { p = malloc(sizeof(int)). The change to p will not be visible to the caller because the function is only mutating its local copy. That is not pass by reference; you are passing a pointer by value. Of course, you can mutate what the pointer refers to by dereferencing it, but that is not what it means to pass something by reference. C is pass by value only.

|

SevenBits · Accepted Answer · 2013-03-25 03:05:43Z

0

You are currently doing:

manipulateStrings(stringA, stringB);

And printing from within manipulateStrings. Instead, to verify you're doing it correctly, I would do:

manipulateStrings(stringA, stringB);
printf ("2 string1: %s", string1);
printf ("2 string2: %s", &string2);

And not print from within that function. That way, you can test if the variables are being copied by-value or by-reference.

answered Mar 25, 2013 at 3:05

SevenBits

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