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What is the output of fun1()?

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int fun(char b, char c)
{
    b = c = 1;
    printf("%d\n", b >> -1);
    printf("%d\n", c >> -1);
}

int fun1(char b, char c)
{
    b = b << 1;
    c = b = 1 << 1;
    printf("%d\n", c);
}

What is the output of fun1()?

Show answer & explanation

Answer: A. 2

In fun1(), the expression c = b = 1 << 1 is evaluated right-to-left: 1 << 1 equals 2, then b = 2, then c = 2. The printf("%d\n", c) outputs the single integer value 2. The first statement b = b << 1 has no effect on the final output since b is reassigned later. Note: fun() itself contains undefined behavior due to right-shifting by a negative count (b >> -1), but the question asks only about fun1()'s output.

Step-by-step Derivation:
Step-by-step execution of fun1(char b, char c):

b = b << 1; — Shifts b left by 1 bit (but b is later overwritten, so this has no lasting effect)
c = b = 1 << 1; — Evaluated right-to-left:
- 1 << 1 = 2 (shift 1 left by 1 position: binary 01 → 10)
- b = 2 (assign 2 to b)
- c = 2 (assign 2 to c)
printf("%d\n", c); — Prints the value of c, which is 2

Output: 2