Printing different values for ++i||j++&&++k with printf vs cout

Question

This code prints different values for ++i||j++&&++k depending on whether the printf function or cout is used. Why is that?

#include <stdio.h>
#include <iostream>

using namespace std;

int main()
{

    int i = 1,j=1,k=1;
    cout<<++i||j++&&++k; 
    printf("%d", ++i||j++ && ++k);

}

Since you'll never, *ever* be writing code like this in a production scenario, you don't need to know the answer to this question. — Robert Harvey, Jun 23 '21 at 16:00
This doesn't look undefined behavior like `i++ + ++i` because each variables are used only once per statements. — MikeCAT, Jun 23 '21 at 16:01
duplicates: [Why doesn't ++i || ++j && ++k give the expected value](https://stackoverflow.com/q/55517958/995714), https://stackoverflow.com/q/28145152/995714, [How to solve this logical expression?](https://stackoverflow.com/q/33924522/995714), [How to solve this logical expression? (closed)](https://stackoverflow.com/q/33924522/995714) — phuclv, Jun 23 '21 at 16:07

score 3 · Accepted Answer · answered Jun 23 '21 at 16:03

According to C++ Operator Precedence - cppreference.com, the << operator has higher precedence than || operator. Therefore, the statement

cout<<++i||j++&&++k;

means

(cout<<++i) || j++&&++k;

Therefore, the value of i after the increment is printed.

In the other hand, the value of ++i||j++ && ++k will be printed by

printf("%d", ++i||j++ && ++k);

Also note that the execution of cout<<++i||j++&&++k; may affect the result of printf("%d", ++i||j++ && ++k); because the ++ operator has side effects and the values of the variables aren't reset between the statements.

Printing different values for ++i||j++&&++k with printf vs cout

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