return vs cout in C++

Question

int func(int x, int y) {
x = 10;
return (x+y);
}
int main () {
int i = 0; int j = 5;
i = func (j, i);
cout << i << j;
}

I'm confused about how the return statement is used in the code above.
I first assigned i = func(5,0) ----> have no idea what to do next... do I add 5 and 0 and re-assign i = 5?

Also, where would I plug in the x = 10 in the second line? How does x and y relate to i and j?

Is the final output provided by cout, return (x+y) or int func(int x, int y)? Totally lost.

Answer 1

Solus Is this something you wrote?

Here are the answers to your questions:

1. The return statement is, well, returning the integer value 10 from func(5, 0).

2. func(5, 0) ignores the first argument as x is always reassigned the value 10. Therefore, func(5, 0) will return (10 + 0) instead of (5 + 0) which is then assigned to int i in the main function.

3. You asked "where do you plug in x = 10 in the second line?" Did you mean "why" instead of "where"? Assuming you meant "why", I would say you would know better than me if you wrote this code. 😉 However, if you grabbed this code from code challenges, then it's likely the author was trying to make this a tricky question. 🤷‍♂️

4. The values j and i are passed as the input arguments x and y for func(x, y). Notice the reverse order of j and i, which map to x and y as illustrated below:

x = j
y = i

5. i is assigned 10 as the return value from func(5, 0).

j is still 5.

cout prints 10, then 5 without spaces which appears as: 105

Answer 2

@David Carroll As a beginner, how would I know that the return statement return (x+y) will return the integer value 10 from i = func(5, 0)?

If i = 10, then cout will indeed output 105. But I still don't really understand how i = 10 came to be...
This was a challenge Q that I got wrong. I'm trying to learn from my mistakes. Thank you for the explanation.

Answer 3

Return type returns the value It doesn't print value means you can assign returned value to other variable.Whereas Cout Simply prints text.