Implementing square() without using the multiplication operator - c++

Question

Task:
Implement square() without using the multiplication operator; that is, do the
x*x by repeated addition (start a variable result at 0 and add x to it x times).
Then run some version of "the first program" using that square().
Solution #1:
#include <iostream>
int square(int x){
int result = 0; // same output of we declare result without initializing
for (int i = 0; i < x; i++) {
result += x;
}
}
int main() {
std::cout << square(19);
}
Output: 19.
Solution #2:
#include <iostream>
int square(int x){
int result = 0;
for (int i = 0; i < x; i++) {
return result += x;
}
}
int main() {
std::cout << square(19);
}
Output: 19.
Solution #3:
#include <iostream>
int square(int x){
int result;
for (int i = 0; i < x; i++) {
return result += x;
}
}
int main() {
std::cout << square(19);
}
Output: 22006.
Why does only the third one work?

Answer 1

Solution #1
You don't return result, thus you calculate it but never return it
i.e.
#include <iostream>
int square(int x){
int result = 0; // same output of we declare result without initializing
for (int i = 0; i < x; i++) {
result += x;
}
return result;
}
int main() {
std::cout << square(19);
}
Solution #2
You return the result after the first iteration, this will always return x's value.
#include <iostream>
int square(int x){
int result = 0;
for (int i = 0; i < x; i++) {
result += x; // remove "return" here
}
return result;
}
int main() {
std::cout << square(19);
}
Solution #3
This is an interesting one. With languages like C++, you need to initialize your variables such as integers, see this for more info. Since you don't initialize it, your the integer gets a memory allocation and what ever is in memory at that point is what your int value is now. Since you only add to it, it will add the calculation to that value.
Thus, you need:
#include <iostream>
int square(int x){
int result = 0; // initialize values
for (int i = 0; i < x; i++) {
result += x; // remove return here
}
return result; // return result here
}
int main() {
std::cout << square(19);
}

Answer 2

This is invalid code: the function square is declared as int, but doesn't return anything.
You return prematurely, during the first iteration of the loop.
Same as #2, but with unitialized variable result - garbage.

Answer 3

You seem to have understood the task you have to accomplish. And you're very close to the result. But there's several major issues, which is why none of the provided solutions work.
Solution 1
You seem to be doing the task correctly, but never return the result. The fact that the output returns 19 in your case is surprising, but it won't always be like that. See here
Solution 2
You might to read up again on return. It returns the value immediately. So in essence, you're only doing one pass inside the loop, so you're only returning the parameter of the function, which in this case is 19.
Solution 3
This one has the same issue with return than the previous one. But you're also adding to an uninitialised value of result, and because of that can have litterally any possible value inside of it. Which is why you're getting a weird number as a result.
For Solution 1, never returning a value inside a function that expects a return value, and in Solution 3, accessing the value of an uninitialised value, are called undefined behaviours. In short, never have undefined behaviour in your code, it might work some times on some machines/compilers, but most of the time it will simply not work, or worse, "work" but in unpredictable ways, causing major issues and bugs.