在 C++ 中给出错误返回类型的错误

Question

#include<iostream>
using namespace std;

void factor(int n)
{
    if(n<=1)
    return;
    for(int i=2;i*i<=n;i++){
        while(n%i==0){
            printf(i);
            n=n/i;
        }
    }
    if(n>1)
    {
    return n;
    }
}

int main(){
    factor(10);
}

Error: return-statement with a value, in function returning 'void' [-fpermissive] 16 | return n; | ^

Answer 1

The return type shoud not be void but int.

Answer 2

The problem is that your function factor has a return type of void but it actually returns an int object.

To solve this, you should match the return type of the function factor with what you're actually returning from inside the function as shown below.

Secondly you should also take care of the situation when none of the conditions(like if ) are satisfied by adding appropriate return statements as shown below:

int factor(int n) //VOID CHANGED TO INT HERE
{
    if(n<=1)
    return 0; //CHANGED RETURN TO RETURN 0 HERE
    for(int i=2;i*i<=n;i++){
        while(n%i==0){
            //printf(i);
            n=n/i;
        }
    }
    if(n>1)
    {
    return n;
    }
    return 0;//ADDED RETURN HERE for the case if none of the if are satisfied and control flow reaches here
}