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 1
x=y=z=1;
z=++x||++y&&++z;
This evaluates to
x=2 y=1 z=1
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 1
x=y=z=1;
z=++x||++y&&++z;
This evaluates to
x=2 y=1 z=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
.
++
has higher priority than||
, so the whole RHS of the assignment boils down to an increment ofx
and an evaluation to a truth value (1).This is because
++x
evaluates to true and the second branch is not executed.++x
is2
which, in a boolean context, evaluates to true or1
.z
takes the value of1
, giving you the observed final state.