Unusual if construct in C/C++

Question:

Today I saw a strange if construct in the code:

#define ERROR_OK 0L

int foo();

// ...
if (ERROR_OK == foo() != ERROR_OK) 
{
    // ...
}
else
{
    // ...
}

So I didn’t fully understand why this code works and at what values ​​the function is executed. I realized that if foo() returns 0, then the branch inside the if will be executed. Please help me understand completely.

Answer:

Well, it's pretty simple. Let's take it apart piece by piece.

1) ERROR_OK == foo() is compared. It will return true or false , depending on the value of ERROR_OK and what the function returns.

2) The result of the previous calculation (i.e. true or fals e) is compared with ERROR_OK . In this case, the same true or false will be reduced to int (true -> 1, false -> 0) . Well, the usual comparison.

---

Accordingly, if we assume that ERROR_OK = 0 (and it is written that way in the condition), then the first part of the condition is equivalent to 0 == foo() , i.e. !foo() . The second part of the condition is equivalent to !foo() != 0 , i.e. foo() == 0 , i.e. just !foo() . This means that if foo returns 0 , then the condition will be met, otherwise it won't.