Explain C++ code

Question

There is a C++ challenge with the following code:

int x=1,y=1,z=1;

cout << (++x || ++y && ++z);

cout << x << y << z;

Thre result is 1211.

First '1' is from the first cout and therefore:
x = 2
y = 1
z = 1

I know that for an OR expression such as (++x || ++y), if the first term is true then there is no need to evaluate the second.

This leads to
x=2 and
y=1.

But how about the AND expression?
Why is z not set to 1?
Does z not getting evaluated?

Answer 1

z is not getting evaluated because it looks like this:
++x || (++y && ++z)
since && has higher precedence than ||

Answer 2

this reminds me of some groovy code where I was evaluating the Boolean return of a function before the simpler variable check and it slowed things down significantly... to illustrate: (someBoolFunction()||x==5) vs (x==5||someBoolFunction())

Answer 3

Aaah got it. Assumed same precedence. So it boils down to an OR expression..

Thank you Max && Udi Finkelstein !

Answer 4

Saurav Priyadarshi it is not x that is printed but the result (true, I.e. 1) of the logical expression within parentheses.

Answer 5

You got it 99% correct. The missing 1% you missed is that due to operator precedence, the && operator has a higher precedence than || and therefore,
The right hand side of the || expression, which as you said yourself, is not evaluated (because the 1st part is already true) includes the entire '++y && ++z' expression.

Answer 6

Petest Sam || stands for the logical OR operation

Answer 7

The operator precedence is not necessarily the same thing as order of execution (evaluation). As a result, the above line can be explicitely reform as follow for clarifying the evaluation order

( ++x || (++y && ++z))

Once left side of || being evaluted to true, the right side (as you aware) is discarded.

Answer 8

Vulture With Loveness☺️☺️ are you asking about my previous comment about || ? It was just in response to a question from Petest Sam about meaning of the double pipe symbol.

Answer 9

Petest Sam it is the logical or.
expression || expression is true if at least one of them is true

Answer 10

in this question what does || stand for?

Answer 11

AK-47 seems like i missed a "."

Answer 12

Saurav Priyadarshi because the result lf the logical operation is true and true gets cast to 1 for printing

Answer 13

Petest Sam

As Max mentioned, the expected result from a logical OR operation is pretty much straight forward. Note that any non-zero value is considered true or 1 when doing evaluation.

~example~

int x = 31;
int y = 0;
int z = -1;

bool b1 = x || y; // true
bool b2 = x || z; // true
bool b3 = y || y; // false

Answer 14

Max
I guess you meant logical OR operator, right? ;)

Answer 15

seems to be interested in this platform

Answer 16

Why is 1 not 2 printed in the beginning?

Since we can see that prefix increment is applied on x, so x should be incremented before being printed.

Answer 17

Saurav, because of the OR/AND operators, the term inside the brackets gets evaluated to a bool, with a value of true this is output as 1.

Answer 18

There you need precidence bro as you know that the higher precidence operator is %.
As here the &&op is the higher orecidence then the ||op.