Thread: What is a output?. How will Come?.

the o/p of this program is 00131
in given expression (m=i++&&j++&&k++||l++) operater play an imp role
i.e. we will first evaluted incre. op. that has higher predency after that we will evaluted logical AND op(&&) and after that logical OR(||) op.and at the end as.op(=) will be evaluted . note that it is a logical exp. so m have value either 1 or 0. while in this exp it is true so has 1 as o/p
and the value of i,j,k and l is incremented when write exp , so it print these incremented value by 1 so these value become 0,0,1,and 3 and the value of m is evaluted in exp ,1
dilip agrawal mca indore mp

00131 is the answer.
m=i++ && j++ && k++ || l++;
let us leave all the postfix increments because they will change the values only in the next statement of their respective variables. so m=i && j&& k|| l
-1 && -1=0
0 && 0=0
0 || 2=1
so m is 1. this statement is over then all the variables will be incremented by 1 because of that postfix increments. so output is 00131

out put vil b
-1,-1,0,2,1

Originally Posted by Golda

main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}

i=0,j=0,k=1,l=3,m=0

0 0 1 2 -1

0,0,1,3,0
is this correct

the output will be 00131
All the effect will be occur at next statement because of post increment
the || operator is the last element and the value of l is 2 so that any value of the previous generation will be ommited and the value of m will be true i.e. 1
m=0||1 is 1
i=-1 --> 0
j=-1 --> 0
k=0 --> 1
l=2 --> 3
m = 1

Originally Posted by Golda

main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}

Output : 0 0 1 3 1
EXP : As the expression have 3 && operators at the begin it will check for all the case s ,& will get 0 as resultant wch is || with 1 (i.e.,2) So that m=0||1=1.

i=0,j=0,k=1,l=3,m=0