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Question

Predict the output of following source code in C language :

int main ( )

{

    char x = 65;

    printf(“%c |%d|%f”, x, x, x);

    return 0;

}

This question was previously asked in
MH SET 2026 Paper 1 Question Paper (25-Oct-2026)
The correct answer is

A | 65 | 0.000000

printf interprets each argument according to its conversion specifier. Arguments of a variadic function undergo default argument promotion, so the char x (value 65) is passed as an int.

Step 1 — %c prints the int as a character; 65 is the ASCII code of A, so A is printed.

Step 2 — %d prints the same value as a decimal integer, 65.

Step 3 — %f expects a double, but an int was passed. No conversion to 65.0 takes place; this is undefined behaviour in the C standard. On the common x86-64 calling convention a double is read from a floating-point register that was never loaded, which typically holds zero, so 0.000000 is printed. The output is therefore A, 65 and 0.000000.

The value 65.000000 would appear only if x were cast to double, and printing 65 for %c ignores that %c prints the character whose code is 65.

A | 65 | 0.000000 option is correct.

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