''warning C4715 not all control paths return a value'' What does it mean?

[Mohammadtaqi]

warning C4715: 'Pow' : not all control paths return a value
warning C4715: 'Sery' : not all control paths return a value

#include "stdafx.h"
#include <iostream>
 
double Pow(double x, int n)
{
    if (n == 0)
        return 1;
    if (n > 0)
        return x * Pow(x, n - 1);
    if (n < 0)
        return Pow(1 / x, -n);
}
 
double Sery(int n, int t)
{
    if (n > 0)
        return Sery(n - 1, t) + Pow(n, t);
}
 
int _tmain(int argc, _TCHAR* argv[])
{
    std::cout << Sery(1, 1) << std::endl;
        return 0;
}

[hype261]

Basically what this is telling you is that your code could potentially return garbage. Take a look at the Sery function.

double Sery(int n, int t) 
{ 
    if (n > 0) 
        return Sery(n - 1, t) + Pow(n, t); 
 
}

What is going to happen if n equals -1. So -1 > 0 will evaluate to false and the statement return Sery(n - 1, t) + Pow(n, t); won't be executed. This is why it is giving you the warning.

[donbock]

hype261 is right about 'Sery', but the situation is a little more confusing in 'Pow'.

double Pow(double x, int n) 
{ 
    if (n == 0) 
        return 1; 
    if (n > 0) 
        return x * Pow(x, n - 1); 
    if (n < 0) 
        return Pow(1 / x, -n); 
}

The compiler is worried that the last if test might be false, causing execution to fall out of the function without hitting a return statement. You might argue that the previous two if statements guarantee that the last test can't be false, but your compiler is unable to recognize this dependency between these statements.

The following alternative will satisfy the compiler:

double Pow(double x, int n) 
{ 
    if (n == 0) 
        return 1; 
    else if (n > 0) 
        return x * Pow(x, n - 1); 
    else   /* n must be < 0. */
        return Pow(1 / x, -n); 
}

By the way, you might want to change the name of your 'Pow' function so that it won't be confused with the Standard Library 'pow' function.