C language ...output

Question

Main ()

{

Int x,y , z;

X=y=z=1;

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

Printf ("x=%d y=%d z=%d",x,y,z);

}

Comments

&& has higher priority than || so,

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

now as we know OR logical operator (||) has left to right associativity and therefore ++x will get executed first.

https://www.geeksforgeeks.org/c-operator-precedence-associativity/

thus after execution of ++x, short circuit occurs. and (++y && ++z) expression will not get evaluated.  

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yes.just read the same thing.The main thing is to put parenthesis properly:)

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How c reads instructions from left to right or right to left ?

Answer 1

Whenever you have multiple precedence in equation,put parenthesis and then solve:-

1. $a+b*x$ => $( (a) + (b*c) )$ , so here firstly first parentheses (a) is evaluated and then next parenthesis.(b*c).If we dont put parenthesis then we can say a+b*c,multiply has higher precedence and it will be done first.Although it will be right but we can get some situation as you mentioned in question where without parenthesis we may get wrong answer.

Coming to question:-

$ (  (++x) | | (++y&&++z) ) $

 //because && and has high precedence so y is bounded to &&

++x will be done and ig gives 2,which means true and if LHS of or is true we dont evaluate RHS(shortcircuiting).

Soa answer =$2,1,1$

Comments

@hira @rahul Sharma

Please explain what is short circuit rule and what if instead of || we have && in the question.

How it output would have changed?

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@Mayankprakash

Please explain what is short circuit rule

they just named it, don't worry about it !

 

in C, the operators &&, || are work like

a && b ===> if a is false then it doesn't evaluate b due to " False && something " lead always false.

a || b ===> if a is true then it doesn't evaluate b due to " True && something " lead always true.

this phenomena called as Short Circuit Rule.

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so when x becomes 2, won't z will become 2 as Z=++x, so z = 2 so why 1 should be printed for z