Operator precedence in C for the statement z=++x||++y&&++z

Question

I was studying operator precedence and I am not able to understand how the value of x became 2 and that of y and z is

Answer 1

The and && and the or || operation is executed from left to right and moreover, in C 0 means false and any non-zero value means true. You write

x=y=z=1;
z= ++x || ++y && ++z;

As, x = 1, so the statement ++x is true. Hence the further condition ++y && ++z not executed.

So the output became:

x=2 // x incremented by 1
y=1 // as it is
z=1 // assigned to true (default true = 1)

Now try this,

z= ++y && ++z || ++x ;

You will get

x=1 // as it is because ++y && ++z are both true 
y=2 // y incremented by 1
z=1 // although z incremented by 1 but assigned to true (default true = 1)

And finally try this:

int x = 1;
int y = 0;
int z = 1;

z= y && ++z || ++x;

The output will be:

So the output became:

x=2 
y=0 
z=0 

Because, now the statement for z is look like this:

z = false (as y =0) && not executed || true
z = false || true
z = true

So, y remains same, x incremented and became 2 and finally z assigned to true.

Answer 2

z=++x||++y&&++z;

NOTE: ++ has higher priority than ||

Now after this line is executed the value of x is incremented and x=2 now ++y&&++z are never executed as the first condition is true and hence you are getting the value as x=2 y=1 z=1

Answer 3

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

is equivalent to

x=y=z=1
z=((++x)||((++y)&&(++z)));

Since ++x returns 2, which is nonzero, the ++y && ++z branch is never executed, and thus the code is equivalent to:

x=y=z=1;
z=(++x)||(anything here);

Answer 4

++ has higher priority than ||, so the whole RHS of the assignment boils down to an increment of x and an evaluation to a truth value (1).

z = ++x         ||  ++y&&++z;
    truthy (1)     never executed

This is because ++x evaluates to true and the second branch is not executed. ++x is 2 which, in a boolean context, evaluates to true or 1. z takes the value of 1, giving you the observed final state.