Why does this if-statement combining assignment and an equality check return true?

This has to do with operator precedence.

if (i = 1 && i == 0)

is not

if ((i = 1) && (i == 0))

because both && and == have a higher precedence than =. What it really works out to is

if (i = (1 && (i == 0)))

which assigns the result of 1 && (i == 0) to i. So, if i starts at 0 then i == 0 is true, so 1 && true is true (or 1), and then i gets set to 1. Then since 1 is true, you enter the if block and print the value you assigned to i.

Assuming your code actually looks like this:

#include <iostream>
using namespace std;

int main()  {
    int i = 0;
    if (i = 1 && i == 0) {
        cout << i;
    }
}

Then this:

if (i = 1 && i == 0) {

evaluates as

 if (i = (1 && i == 0)) {

and so i is set to 1.