Data manipulation inside a function in "C"

Question

Here's something I don't understand. Maybe someone can shed some light on it.

I know that the standard method for data manipulation is passing a reference to a function and changing the data in the function. Like this:

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


void function2( float *param) {
    printf("I've received value %f\n", *param);
    (*param)++;
}

int main(void) {
    float variable = 111;

    function2(&variable);
    printf("variable %f\n", variable);
    return 0;
}

When calling function2 with (&variable) I expect that the function can change the data. So far no questions.

But why does this work as well?

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


void function2( float &param) {
    printf("I've received value %f\n", param);
    (param)++;
}

int main(void) {
    float variable = 111;

    function2(variable);
    printf("variable %f\n", variable);
    return 0;
}

In my understanding, when calling function2(variable) a copy of the value of "variable" is passed to the function. But nevertheless the value of "variable" has changed after the function call.

When reading a code like this I would never expect the data of "variable" to change, no matter what happens inside the function.

Answer 1

This declaration of the parameter

void function2( float &param) {

is incorrect. In C there is no reference. Such a function declaration will be valid in C++, where references exist.

In C++ when the function is called like

function2(variable);

neither copy of the variable is created. The function refers to the original variable declared in main.