Returns in Recursive Function in C++ - c++

Question

I'm a little confused here. Let's look at the following code:
bool testing(int i) {
if((i%2)==0) {
return true;
} else {
--i;
testing(i);
}
return false;
}
When I do testing(5), I was expecting the function to return true because at some point, 5 will become 4, so 4 % 2 == 0, so the function will return true but it just wasn't the case. What's wrong?

Answer 1

You should return testing(i); instead of just testing(i);

Answer 2

The idea of recursion is when a function calls itself, directly or indirectly.
The function in your code will become recursive if it is modified to:
bool testing(int i){
// test if even, if so return true
if((i % 2) == 0){
return true;
// otherwise decrement and test again
}else{
// at this point the function calls itself with decremented argument
return testing(--i);
}
// I doubt that this case will be ever returned
// more likely your function will return "true" or run "forever" decrementing
return false;
}
To avoid infinite cycles you need a base case, termination condition that produces result without recursion. For example if i becomes very small or negative you return false.
bool testing(int i){
// base case
if(i < 0) return false;
// rest of the function
if((i % 2) == 0){
return true;
}else{
return testing(--i);
}
}
Making it a bit more concise, you finally have three cases:
bool testing(int i){
// base case
if(i < 0) return false;
// test if even
if((i % 2) == 0) return true;
// recursion step
return testing(--i);
}
For further reading, check this

Answer 3

You don't bubble up the final return value; you need to use return on the recursive call. Additionally, you can simplify the pre-decrement:
return testing(--i);

Answer 4

Because of you only call testing(i) function. That's why it's not call recursively.
you should write return testing(i)

Answer 5

it returns false because its return value is overridden by the last statement of "return false".