c++ - Explain the following output? -

c++ - Explain the following output? -

please explain output:

#include<iostream.h> int main() { int i= -3, j=2, k=0, m; m = ++i || ++j && ++k; cout<< <<" " << j << " " << k <<" "<<m; homecoming 0; }

output : -2 2 0 1

here's thought: (++i || ++j) && (++k) //considering precedence order ++i becomes -2 first part of or true, won't check 2nd part. (thanks joachim pileborg telling me short circuit evaluation)

so overall, first part of , true. not plenty statement true, 2nd part must true to. ++k makes k = 1 here's wrong. why k not increasing?

whereas, in case:

#include<iostream.h> int main() { int i= -1, j=2, k=0, m; m = ++i || ++j && ++k; cout<< <<" " << j << " " << k <<" "<<m; homecoming 0; }

output: 0 3 1 1

i got 1 considering short circuit evaluation.

let's start code snippet

#include<iostream.h> int main() { int i= -3, j=2, k=0, m; m = ++i || ++j && ++k; cout<< <<" " << j << " " << k <<" "<<m; homecoming 0; }

it obvious m have boolean value converted int. ++i equal -2 unequal 0 other expressions not evaluated because known whole look equal true. after statement

m = ++i || ++j && ++k;

m equal 1 , i equal -2 other variables not changed.

in code snippet

#include<iostream.h> int main() { int i= -1, j=2, k=0, m; m = ++i || ++j && ++k; cout<< <<" " << j << " " << k <<" "<<m; homecoming 0; }

++i equal 0. right operand of operator || evaluated. operand is

++j && ++k

as ++j equal 3 , not equal 0 ++k evaluated , equal 1. both operands of operator && not equal 0 result equal true

thus == 0, j == 3, k == 1, m == 1.

from c++ standard

5.14 logical , operator

1 && operator groups left-to-right. operands both contextually converted bool (clause 4). result true if both operands true , false otherwise. unlike &, && guarantees left-to-right evaluation: sec operand not evaluated if first operand false.

5.15 logical or operator

1 || operator groups left-to-right. operands both contextually converted bool (clause 4). returns true if either of operands true, , false otherwise. unlike |, || guarantees left-to-right evaluation; moreover, sec operand not evaluated if first operand evaluates true.

c++ operator-precedence