Question about x=y==z [closed]

It's difficult to tell 开发者_如何转开发what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. Closed 12 years ago.

I have the following code:

#include <stdio.h>
int main(void)
{
    int x = 2, y = 6, z = 6;
    x = y == z;
    printf("%d", x);
}

output is = 1

Why?

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Because the assignment is right to left, and the precedence of == is greater than =.

it is x = (y == z)

y == z is 1.

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From the precedence table == is having a higher precedence from =

Hence

x = y == z;

is same as:

x = (y == z);

Since y == z is true (1), x gets 1.

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x = y == z is read as x = (y == z), and y and z both are 6 and thus they are equal. true is 1, so x is 1.

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y == z evaluates to true, which you are assigning to x...x = true casts to a value of 1 because x is of type int.

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y == z => 6 == 6 => True

True represented as integer (%d) is 1.

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x = y == z; is the same thing as x = (y == z); and as y == 6 and z == 6, (y == z) == 1 so x == 1

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It evals == operator first, so since y==z is true, and x is int,x is set to 1 (true)

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Comparison (==) has higher precedence than assignment (=), so your middle statement is processed as x = ( y == z ); and the result of a true comparison is 1, which is assigned to x.

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== have higher precedence than =. So x = y == z is actually x = (y == z). Now y and z both are 6. So the comparison is true and outcome is 1 which is set to x.