code

#include <iostream>
using namespace std;

int main()
{
    int i = 1;
    while (i < 10)
        if (i   % 2 == 0)
            cout << i << endl;

    return 0;
}

The output is

3
5
7
9

Since i is 1, I thought that the if statement satisfies 2% 2 == 0 and 2 should be output, but I don't know why 3.

CodePudding user response：

if (i % 2 == 0) means evaluate i%2 and compare to 0, and increment i at some point after evaluating it but before cout << i

That's why you're seeing odd numbers printed. In every case, an even value of i caused the statement controlled by the if to be executed, and the increment changed the even value to an odd one.

CodePudding user response：

You are using post-increment, which increments the variable and then returns the old value before the increment. You are then printing the incremented value.

So, lets look at your loop step-by-step:

• On the 1st iteration, i is 1. i sets i to 2, but returns 1. So the if compares 1 % 2 == 0, which is false.

• On the 2nd iteration, i is 2. i sets i to 3, but returns 2. So the if compares 2 % 2 == 0, which is true, so cout << i prints 3.

• On the 3rd iteration, i is 3. i sets i to 4, but returns 3. So the if compares 3 % 2 == 0, which is false.

• On the 4th iteration, i is 4. i sets i to 5, but returns 4. So the if compares 4 % 2 == 0, which is true, so cout << i prints 5.

• And so on...

It sounds like perhaps you were expecting i to pre-increment instead, returning the new value rather than the old value. In which case, there is a slightly different syntax for that purpose:

#include <iostream>
using namespace std;

int main()
{
    int i = 1;
    while (i < 10)
        if (  i % 2 == 0) // <--   i vs i  
            cout << i << endl;

    return 0;
}

Note the position of in relation to i. Both syntaxes increment i, but positioning to the left of i returns the new value after i is incremented, while positioning to the right of i returns the old value before i was incremented.

Same with the decrement operator --, too. It has pre-decrement and post-decrement versions.

CodePudding user response：

According to the C 14 Standard (5.2.6 Increment and decrement)

1 The value of a postfix expression is the value of its operand. [ Note: the value obtained is a copy of the original value — end note ] The operand shall be a modifiable lvalue. The type of the operand shall be an arithmetic type or a pointer to a complete object type. The value of the operand object is modified by adding 1 to it, unless the object is of type bool, in which case it is set to true.

So for example in the second iteration of the while loop the variable i is indeed equal to 2 after incrementing it in the first iteration of the loop

while (i < 10)
    if (i   % 2 == 0)
        cout << i << endl;

Thus the condition of the if statement

    if (i   % 2 == 0)

evaluates to true. But the value of the variable i after the evaluation of the condition was incremented. So this statement

        cout << i << endl;

outputs the new value 3.

You can equivalently rewrite the while loop the following way

while (i < 10)
    if (i % 2 == 0)
        cout <<   i << endl;
    else
          i;