Choose the correct operators to fill in the blanks:
int i,j,k; i=1;j=2;k=3; printf("%d",i___5___j___2____k);
Output is: $2$
precedence *,/,% L->R +,- L->R
i+1-2+k
= 2-2+k
= 0+k = 3
5/j-2+k = 2-2+k = 0+k = 3
i+1+2/k = i+1+0 = 2
i*5 %2 -2/k =5%2 - 2/k = 1 - 2/3
= 1 - 0
= 1 So, answer is option C .
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