how is x&&y||z evaluated?

Question

Given

int x=1,y=2,z;

Could you explain why the result for:

x && y || z 

is 1?

x && y = 1
x && y || z = 1

Answer 1

x && y || z 

is equivalent to

(x && y) || z

if x=1 and y=2 then x&&y is 1 && 2 which is true && true which is true.

true || z 

is always true. z isn't even evaluated

Answer 2

x && y || z => (x && y) || z => 1 || z => 1

Answer 3

(bool)1 = true
(bool)2 = true

Uninitialized int refers to data that was saved in memory, where it is placed on stack... and it rarely is 0x00000000, and even if it was, true || false = true.

Answer 4

The && operator has higher precedence than the || operator. See, e.g., this operators precedence table, numbers 13 and 14.

Your example evaluates as (x && y) || z. Thanks to the short circuiting rule, z is never evaluated because the result of x && y is already true.