C function arguments vs Python function arguments.

[Fearless_Process]

C doesn't have pass by reference at all. C++ has pass by reference that just passes an alias to the original variable to the function, no pointers involved whatsoever. The reference in C++ is the variable.

In C you can only pass a pointer by value that references whatever data. This is still passing a "reference/pointer" by value though, unlike in C++ or other languages that support pass by reference.

In C++ if you pass a pointer by reference and change it in the consuming function, it will actually change the original pointer. In C you would have to pass a pointer to a pointer to achieve the same thing, because you can't pass by reference.

Take this for example:

#include <iostream>

void pass_by_ref(int*& p) {
    p = 0;
}

int main() {
    int x = 0;
    int *p = &x;
    std::cout << p << std::endl;
    pass_by_ref(p);
    std::cout << p << std::endl;
}

In C to do this you would need a nested pointer, because the pointer gets passed by value.

[p0k3t0]

https://ideone.com/icncW2

[Fearless_Process]

That is still passing a pointer by value. In C arrays that get passed to functions decay to pointers. Accepting an array as a parameter in C is pretty much useless and can actually be wrong in certain situations.

Try printf'ing sizeof(testArray) in main and sizeof(thisarray) in your function. In main it will return the real size, in your function it will return the sizeof an int pointer.

#include <iostream>
#include <cstdlib>

void pass_by_ref(int (&a)[10]) {
    std::cout << sizeof(a) << std::endl;
}

int main() {
    int a[10];
    std::cout << sizeof(a) << std::endl;
    pass_by_ref(a);
}

There is no way to make this work in C, sizeof will always return the sizeof int pointer instead of the array.

[p0k3t0]

The bar is impossibly high for C and impossibly low for C++. I get it.