in C addresses do not work in functions , why is that?

Question

When trying to use a C++ style in C:

void square(int &x){
    x = x * x;
};

This gets an error. error: expected ';', ',' or ')' before '&' token

i'm most comfortable with c++, but i'm learning C, is there a way to to have adressess in void functions

Tried switching from void -> int, double, char. It only works when i take away the & symbol, but i would like to have an address there. Is there a way to do that? Should i just use * instead of &, like this :

void square(int *x){
    x = (*x) * (*x);
};

Hint: `int*` and `int` are two *very* different things, that's why. — tadman, Jan 16 '23 at 19:43
"i would like to have an address there" First question: Why? If you want an address, you're going to have to deal with the consequences. — tadman, Jan 16 '23 at 19:44
`void square(int &x)` declares `x` as a *reference*, and that's totally something that C++ has that C does not. Yes, the way to do it in C is with a pointer `int *x` and explicit dereferences `x = *x * *x` (note you don't need the parentheses). — Steve Summit, Jan 16 '23 at 19:47

score 1 · Answer 1 · answered Jan 16 '23 at 19:47

1

C language does not have C++ references. You need to use pointers

void square(int *x)
{
    *x = *x * *x;
}

Your second code is invalid as you assign a local pointer with integer converted to pointer.

answered Jan 16 '23 at 19:47

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Vlad from Moscow · Answer 2 · 2023-01-16T20:15:11.493

Even in C++ the sign & does not denote an address in a declaration. It denotes a reference.

This declaration in C++

void square(int &x){
//...
}

means that the function accepts its argiment by reference.

In C passing by reference means passing an object indirectly through a pointer to it. So dereferencing a pointer within a function you get a direct access to the original object passed to the function through the pointer.

So this function in C++

void square(int &x){
    x = x * x;
}

accepts an object of the type int by reference.

In C this function

void square(int *x){
    *x = *x * *x;
}

accepts an object of the type int by reference indirectly through a pointer to it.

Pay attention to that the assignment statement in the second function

void square(int *x){
    x = (*x) * (*x);
}

is incorrect. You are trying to assign an integer expression to a pointer. You have to write

    *x = *x * *x;

as already shown above. That is you need to change the passed object of the type int to the function indirectly through a pointer to it.

in C addresses do not work in functions , why is that?

2 Answers2