Void function returns value of 1000

[Deanm007]

It says all over that void f() means that it returns no value. But this function below does return a value to the main function which is 1000. Am I correct?

Basically void is taking place of a return type. Maybe it returns a value but no return type(ie. int, char, bool)?

void SetTo1000(int iaArray[], int iIndex) 
{
    iaArray[iIndex] = 1000;
}
 
int main() 
{
    int iMyArray[5] = {-3,2,0,-1,5};
    SetTo1000(iMyArray,3);
    for (int iIndex=0; iIndex<5; iIndex++)
    {
        cout << iMyArray[iIndex] << endl;
    }
    return 0;

[Dheeraj Joshi]

Your code(function) do not return any value to main function.

Regards
Dheeraj Joshi

[Deanm007]

Hi Dheeraj

i see 1000 as part of the output. That 1000 gets returned from the function to main, no?

[sridhard2406]

Hi Deanm007,
Your funtion will store the 1000 into iaArray[iIndex]. please find the below funtion for returning the int.



int funtion(int a)
{
a++;
return a ;
}

[Dheeraj Joshi]

Please read pass by reference and pass by value to understand why you get the value 1000 in your output.

Regards
Dheeraj Joshi

[Banfa]

No 1000 is not returned to main in the strictest sense of the word.

The return mechanism for any function is quite different to writing data to a memory location.

Your function does not return a value it returns void and so it does not return a value to main. It writes a value to a memory location that is accessible by main. The net result is that the value stored by the function is available in main but it has not been returned to main, it has been communicated to main by an alternate mechanism.

The actual return process often involves data being transferred on the stack as part of the stack unwinding that happens when a function returns.

[Dheeraj Joshi]

Nice answer banfa, In fact i wanted to ans in same way, could not articulate it. Good explanation.

Regards
Dheeraj Joshi

[Deanm007]

@Banfa
Thanks a lot. that was very clear.

So you wouldn't need to pass in by reference to store a function's result in the memory, would you? Because in my code, the function would need ampersands like this:

# void SetTo1000(int& iaArray[], int& iIndex) 
# {
#     iaArray[iIndex] = 1000;
# }

[Deanm007]

sridhar, dheeraj, thanks for responding

[Banfa]

No you would need to pass a pointer to an array by reference because it is already a pointer, you would be unnecessarily dereferencing twice assuming you where just changing the contents of the array. Same for the size.

This should work just as well

 
void SetTo1000(int iaArray[], int iIndex) 
{
    iaArray[iIndex] = 1000;
}
 

Note aiArray is a pointer to int (int *) with this syntax.

[Deanm007]

@Banfa
I got it, thank you.