[C++] Call-by-reference question.

#include <iostream>
using namespace std;

void Plus(int * num)
{
    ++*num;
}

void reverse(int * num)
{
    *num *= -1;
}

int main()
{
    int num = 10;
    int num1 = -9;

    Plus(&num);
    Plus(&num);
    Plus(&num);
    Plus(&num);

    reverse(&num1);

    cout << num << endl;
    cout << num1 << endl;
    return 0;
}

Here, the num value should be output as 14 Plus function content If *num++; is used, the value of num is still 10 output Use the Plus function as a reference void Plus(int &num) { num++; } If you do this, you get 14. If you use a reference to num++, it becomes 14 I wonder why it becomes 10 if you use a pointer to *num++.

c++

2022-09-21 17:05

1 Answers

It is necessary to compare the priorities of rear ++ and *.

http://en.cppreference.com/w/cpp/language/operator_precedence

If you look here, you can see that post ++ has a higher priority.

void Plus(int* num) {
    *num++;
}

If you solve the above function to make it easier to understand, it becomes a function below.

void Plus(int* num) {
    int* temp = num;
    *temp
    num = num + 1;
}

This means that no increment applies to the integer of the address you enter.

Therefore, the value of num is not modified.

void Plus(int& num) {
    num++;
}

In this case, since it is a reference type, the increment operation is applied to the variable you entered to make 14.

2022-09-21 17:05

If you have any answers or tips

Popular Tags

python x 4647

android x 1593

java x 1494

javascript x 1427

c x 927

c++ x 878

ruby-on-rails x 696

php x 692

python3 x 685

html x 656