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.