Does the use of a global variable within a function cause a local copy to be created

Question

Unfortunately I can't find anything on the subject although I can't believe this question hasn't been asked before.

When I use a global variable inside a function, is a local copy of the variable created as with a passed argument or is the global variable accessed directly? Does it even make sense to define a global variable as a parameter of a function using the reference declarator & or as a pointer *?

PS: I know that global variables are a bad practice, but I'm programming for a microcontroller where it sometimes make sense to use global variables.

Answer 1

When I use a global variable inside a function, is a local copy of the variable created as with a passed argument or is the global variable accessed directly?

The global variable is used directly. A local copy is not made.

Does it even make sense to define a global variable as a parameter of a function using the reference declarator & or as a pointer *?

Yes, there are some situations in which this may make sense. For example, if you have several global variables and have a function which writes to these global variables, but not always to the same global variable(s), then it may make sense to pass a pointer or a reference to a global variable.

In the code block below, I have created a simple example program, which defines three global variables a, b and c and first swaps the values of the variables a and b and then the values of the variables b and c. For both swaps, the same function can be used, because the function takes as function arguments a pointer to each global variable that is to be swapped.

It is necessary for the global variables to be passed by pointer (or in C++ by reference) instead of by value, because otherwise, the function would swap the local function arguments instead of the global variables.

#include <stdio.h>

int a = 5;
int b = 3;
int c = 2;

void swap( int *first, int *second )
{
    int temp = *first;
    *first = *second;
    *second = temp;
}

int main( void )
{
    printf(
        "Before swapping,\n"
        "the global variables have the following values:\n"
        "a = %d\nb = %d\nc = %d\n\n",
        a, b, c
    );

    swap( &a, &b );

    printf(
        "After swapping the values of a and b,\n"
        "the global variables have the following values:\n"
        "a = %d\nb = %d\nc = %d\n\n",
        a, b, c
    );

    swap( &b, &c );

    printf(
        "After also swapping the values of b and c,\n"
        "the global variables have the following values:\n"
        "a = %d\nb = %d\nc = %d\n\n",
        a, b, c
    );
}

This program has the following output:

Before swapping,
the global variables have the following values:
a = 5
b = 3
c = 2

After swapping the values of a and b,
the global variables have the following values:
a = 3
b = 5
c = 2

After also swapping the values of b and c,
the global variables have the following values:
a = 3
b = 2
c = 5

Since you tagged the question with both C and C++ (which should generally not be done), I gave an example that is compatible with both C and C++. In C++, one would generally use a reference instead of a pointer, and one would use std::cout instead of printf.