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GO Classes DPP | GATE CS | C Programming | Short-Circuit Evaluation

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13 13 votes

What is the output of the following code?

#include <stdio.h>

int main() {
    int a = 0, b = 5, c = 10;
    int ans = a++ && ++b || c--;

    printf("%d %d %d %d", ans, a, b, c);
    return 0;
}
  1. $\texttt{1\ 1\ 6\ 9}$
     
  2. $\texttt{1\ 1\ 5\ 9}$
     
  3. $\texttt{0\ 1\ 5\ 10}$
     
  4. $\texttt{1\ 0\ 5\ 9}$

3 Answers

5 5 votes

The expression is $\texttt{a++ \&\& ++b || c--}$.

The operator $\texttt{\&\&}$ has higher precedence than $\texttt{||}$.

So the expression is evaluated as $\texttt{(a++ \&\& ++b) || c--}$.

Initially, $\texttt{a = 0}$, $\texttt{b = 5}$, and $\texttt{c = 10}$.

In $\texttt{a++}$, the current value $0$ is used first, and then $\texttt{a}$ becomes $1$.

Since the first operand of $\texttt{\&\&}$ is $0$, the second operand $\texttt{++b}$ is not evaluated due to short-circuit evaluation.

So, $\texttt{b}$ remains $5$.

Now the expression becomes $\texttt{0 || c--}$.

In $\texttt{c--}$, the current value $10$ is used first, and then $\texttt{c}$ becomes $9$.

Since $10$ is non-zero, it is treated as true.

Therefore, $\texttt{ans = 1}$.

Final values are $\texttt{ans = 1}$, $\texttt{a = 1}$, $\texttt{b = 5}$, and $\texttt{c = 9}$.

Answer: B. $\texttt{1\ 1\ 5\ 9}$

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