Thread: Output of program

Hello,

When I executed this program I got output -2 3 0 1. Can anyone explain what how this calculation is reached in program below. Thanks

Code:

int main(){
int i = -3, j = 2, k = 0, m;
m = ++i && ++j || ++k;
printf("%d%d%d%d", i,j,k,m);
}

I have put this question for help. If you know answer tell it to me.

Code:

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

the logical operators && and || are short-circuit operators with left to right associativity. so in above expression, evaluation starts from the left most variable, i.e. i, and goes left till the truth value of the expr is not determined. as soon as the truth value is found, the furthur evaluation of expression stops.

so in above expression, we have

m = (++i) && (++j) || (++k)

starting from left we first evaluate the && along with it's operands. in C, any non-zero value is logically true. therefore (++i) && (++j) evaluates true, as their values during comparison can never be zero due to their initial values, and the truth value is also determined for the whole expression ( true || 0 == true). so no further evaluation takes place and the statement ++k is not executed as control never reaches that statement.

note that during the comparison the exact values of i and j cannot be determined as in an expression, only precedence and associativity rules are guaranteed, not the order of evaluation of sub-expressions like ++i. however, after the above expr, the values of i and j have been incremented as they were evaluated, even if their order was not known.

finally, m is set to the logical result of the expression, which is equal to 1.

Thanks a lot xabhi. Actually where I made mistake was that I was treating negative value as false.

It is generally true that the order of evaluation of subexpressions is unspecified, but in this case we can determine the order of evaluation by inspection because of the sequence points that are introduced.

clearly missed these points... thanks