functions default arguments

Question

#include <iostream>
using namespace std;

int volume(int l=1, int w=1, int h=1) {
return l*w*h;
}

int main() {
cout << volume() << endl;
cout << volume(5) << endl;
cout << volume(2, 3) << endl;
cout << volume(3, 7, 6) << endl;
}

help me understand this code
like when we have volume(5)
then how the formula works???

Answer 1

Default arguments work like this: if the argument isn't specified, it will be the default parameter. So volume() is 1 * 1 * 1, volume(5) is 5 * 1 * 1, and so on.

Please, if you are trying to get a good answer, include relevant tags, I don't think "leaners" is that important.
Also, this question isn't that important to ask it twice in a row, mind deleting the first one? ¯\_(ツ)_/¯

Answer 2

Imagine the parameter of a function as a box.

def f(x):
# function does stuff using x

When you call the function, you fill a value into that box.

f(5)
# so function does
# the stuff it does using 5

A default argument just puts something in the box to begin with.

def f(x=42):
# stuff with x

Now you can call the function with or without argument.

f(5)
# function uses 5

f()
# function uses what's in
# the box per default (so 42)

Sorry, I explained it with python, but the idea is the same in C++.