Output of the following C program
c
Solution
This is a result of short-circuit evaluation.
The expression `++x` evaluates to `2`, and the compiler knows that `2 || anything` always evaluates to `1` ("true") no matter what `anything` is. Therefore it does not proceed to evaluate `anything` and the values of `y` and `z` do not change.
If you try with
x=-1;
y=z=1;
You will see that `y` and `z` will be incremented, because the compiler has to evaluate the right hand side of the OR to determine the result of the expression.
Edit: asaerl answered your follow-up question in the comments first so I 'll just expand on his correct answer a little.
Operator precedence determines how the parts that make up an expression bind together. Because AND has higher precedence than OR, the compiler knows that you wrote
++x || (++y && ++z)
instead of
(++x || ++y) && ++z
This leaves it tasked to do an OR between `++x` and `++y && ++z`. At this point it would normally be free to select if it would "prefer" to evaluate one or the other expression first -- as per the standard -- and you would not normally be able to depend on the specific order. This order has nothing to do with operator precedence.
However, specifically for `||` and `&&` the standard demands that evaluation will always proceed from left to right so that short-circuiting can work and developers can depend on the rhs expression not being evaluated if the result of evaluating the lhs tells.
Problem
What should be the output of this C program? ``` #include<stdio.h> int main(){ int x,y,z; x=y=z=1; z = ++x || ++y && ++z; printf("x=%d y=%d z=%d\n",x,y,z); return 0; } ``` The given output is : x=2 y=1 z=1 I understand the output for x, but fail to see how y and z values don't get incremented.