C 运算符中的优先级 == 和 ( = )

Question

I have to analyze what some code in C does, and I have a doubt about what happens in a certain line. The code is

#define PRINTX printf("%d\n", x)

void problem() { 
    int x = 2, y, z;
    x *= 3 + 2; PRINTX;
    x *= y = z = 4; PRINTX;
    x = y == z; PRINTX;
    x == (y = z); PRINTX; // This line is the problem
} 

This code snippet prints the resulting numbers:

10
40
1
1  // This result **

the problem is that I'm still trying to figure out why does the last line prints out x = 1 , when the operation is x == (y = z) . I'm having trouble finding out what that 1 means and the precedence of the operations. Hope someone can help me: :)

Answer 1

Nothing in the last statement changes the value of x , so its value remains unchanged.

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Parens were used to override precedence, forcing the = to be the operand of the == .

An operator's operands must necessarily be evaluated before the operator itself, so we know the following:

• y is evaluated at some point before the = .
• z is evaluated at some point before the = .
• x is evaluated at some point before the == .
• = is evaluated at some point before == .

That's it. All of these are valid orders:

• z y = x ==
• y z = x ==
• x y z = ==
• etc.

But whenever x , y and z are evaluated, we can count on the following happening:

1. = assigns the value of z (currently 4 ) to y and returns it.
2. == compares the value of x (currently 1 ) with the value returned by = ( 4 ). Since they're different, == returns 0 (which isn't used by anything).

As you see, nothing changed x , so it still has the value it previously had ( 1 ).

Answer 2

In the last statement, nothing is changing the value of x. We are testing if x equals something, but we aren't changing it's value.

So it continues having the same value as it had in the previous statement, in particular, a value of 1.

Answer 3

the reason is because the == operator checks if the 2 numbers are equal, and returns 1 if equal and 0 if not equal that is why it returns one you can check by making x= 1 and y=2 and using the == operator between them

Answer 4

The comparison result of x and assignment of y with (y = z) is discarded. Last line could have dropped the compare: y = z; PRINTX; y = z; PRINTX; .

The assignment is not subsequently used either, so the line could have been PRINTX; .