int 函数参数工作不正常

Question

I'm trying to make something in C++ and I have a problem. I have this code:

#include <iostream>
#include <string>
//---MAIN---
using namespace std;
int af1 = 1;
int af2 = 1;

void lettersort(int cnt1) {
    cout << "RESULT:" << cnt1 << endl;
    cnt1++;
    cout << "RESULT WHEN+:" << cnt1 << endl;
    cout << "RESULT IN GLOBAL INT:" << af2 << endl;
}

int main()
{
    lettersort(af2);
    return 0;
}

So is there any way so that cnt1++ affects af2 too, to make it bigger ? I don't want to use af2++ directly because I want to sometimes use af1 .

Answer 1

At the moment you are just passing af2 to cnt1 by value , so any changes to cnt1 are strictly local to the function lettersort . In order to get the behaviour you want you need to pass your cnt1 parameter by reference . Change:

void lettersort(int cnt1)

to:

void lettersort(int &cnt1)

Answer 2

You are passing the argument by value . Ie, you are copying the value of af1 to a local variable in lettersort . This integer is then incremented, and disposed of when the function ends, without affecting the original af1 . If you want the function to be able to affect af1 , you should pass the argument by reference :

void lettersort(int& cnt1) { // Note the "&"

Answer 3

if i understood your question:

there are 2 ways you can do that.

1. make lettersort function return the new value, and put it in af2

    int lettersort(int cnt1) { cout << "RESULT:" << cnt1 << endl; cnt1++; cout << "RESULT WHEN+:" << cnt1 << endl; cout << "RESULT IN GLOBAL INT:" << af2 << endl; return cnt1; } int main() { af2 = lettersort(af2); return 0; }

2. pass the value by reference. you can read about it here , but generally its about passing a pointer to that value. meaning whatever you do on the argument you are passing, will happen on the original var.

example:

    void foo(int &y) // y is now a reference
    {
        using namespace std;
        cout << "y = " << y << endl;
        y = 6;
        cout << "y = " << y << endl;
    } // y is destroyed here

    int main()
    {
        int x = 5;
        cout << "x = " << x << endl;
        foo(x);
        cout << "x = " << x << endl;
        return 0;
    }

Answer 4

• here you have to just modified the argument pass to lettersort function as passed by reference.

• for example if you declare and initialize any variable like:

   int a=10; int &b = a;

• now a and b refer to the same value.if you change a then the changes also reflect in b also.

• so,

   cout << a; cout << b;

• both statement produce the same result across the program. so using this concept i modified the function argument and made it as by reference.

• your correct code is :

   #include <iostream> #include <string> using namespace std; int af1 = 1; int af2 = 1; void lettersort(int &cnt1) { cout << "RESULT:" << cnt1 << endl; cnt1++; cout << "RESULT WHEN+:" << cnt1 << endl; cout << "RESULT IN GLOBAL INT:" << af2 << endl; } int main() { lettersort(af2); return 0; }