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Question

Exponentiation operation in C is carried out as

This question was previously asked in
UGC NET 2014 Paper 1 Question Paper (28-Dec-2014)
The correct answer is

a = pow (3, 2)

 C has no exponentiation operator at all — the job is done by a library function, pow(), option 3.

#include <math.h>
double pow(double base, double exponent);

so pow(3, 2) returns 9.0.

OptionWhat it actually is
3 ** 2Parsed as 3 * (*2) — a dereference of 2. Syntax error
3 ^ 2Bitwise XOR — gives 1, silently
pow(3, 2)✓ Correct
3 exp(2)Not valid syntax; exp(x) computes ex

Option 2 is the dangerous one and the real subject of the question. In many languages ^ means "raised to the power", but in C it is the bitwise exclusive-OR. The expression compiles without complaint and produces a wrong answer:

\(3=011_{2},\quad 2=010_{2},\quad 3\oplus2=001_{2}=1\)

A silent wrong result is far worse than a compile error, which is why this particular confusion is worth remembering.

Option 1 fails for a related reason. C has no ** token; the parser sees two separate * operators, and the second is read as a pointer dereference applied to the integer 2 — meaningless, so it is rejected at compile time.

Two practical cautions about pow().

It works in floating point. Both arguments and the return value are double, so pow(10, 2) may come back as 99.999999 and truncate to 99 when assigned to an int. Round explicitly, or use integer multiplication where the exponent is a small constant — x * x is both exact and much faster than pow(x, 2).

It needs its header and its library. Omitting <math.h> lets older compilers assume a return type of int, giving nonsense; and on Unix-like systems the maths library must be linked with -lm.

Hence, exponentiation is carried out as pow(3, 2).

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