C Program to Calculate Power of a Number
C Program to Calculate Power of a Number
A classroom arithmetic challenge needs a result built from a sequence, not a formula copied without understanding its starting value.
A nonnegative integer power multiplies the base by itself exponent times; exponent zero gives one.
Before this exercise, review repeating work with loops. The worked program below applies those ideas to this specific problem.
Explain
Explain why integer power uses this sequence of operations.
Trace
Trace the sample values and predict the printed result.
Apply
Use 0 and explain how the initial accumulator affects the answer.
Integer Power: follow the data
Given data
A classroom arithmetic challenge needs a result built from a sequence, not a formula copied without understanding its starting value.
Rule
A nonnegative integer power multiplies the base by itself exponent times; exponent zero gives one.
Observed result
Integer Power: 8
Understand the operation
For base two and exponent three, the product changes 1 → 2 → 4 → 8.
An accumulator stores the result after each completed iteration. Choose its initial value to match the operation: zero for addition, one for multiplication. The loop must eventually stop before the counter or result exceeds its representation.
Read the complete C program
The input values are written into this complete C17 example so the calculation can be reproduced without an unavailable service. Change the named data, rebuild and compare the actual output with your prediction.
C17 · complete program
#include <stdio.h>
int main(void) {
int n = 3;
unsigned long long result = 1;
for (int i = 0; i < n; i++) result *= 2;
printf("Integer Power: %llu\n", result);
return 0;
}
Expected output
Integer Power: 8
Follow the values through the code
C17 · code used in this explanation
#include <stdio.h>
int main(void) {
int n = 3;
unsigned long long result = 1;
for (int i = 0; i < n; i++) result *= 2;
printf("Integer Power: %llu\n", result);
return 0;
}
The chosen n is 3.
For base two and exponent three, the product changes 1 → 2 → 4 → 8.
Follow the worked calculation
Start with the sample
The chosen n is 3.
Apply the rule
For base two and exponent three, the product changes 1 → 2 → 4 → 8.
Check the result
Integer Power: 8
Integer Power: 8
Try a small variation
C17 · code used in this explanation
#include <stdio.h>
int main(void) {
int n = 3;
unsigned long long result = 1;
for (int i = 0; i < n; i++) result *= 2;
printf("Integer Power: %llu\n", result);
return 0;
}
Use 0 and explain how the initial accumulator affects the answer.
Worked answer
The changed loop produces 1. With zero iterations, the initialized accumulator is the result; this is why multiplying examples must not start from zero.
C17 · complete program
#include <stdio.h>
int main(void) {
int n = 0;
unsigned long long result = 1;
for (int i = 0; i < n; i++) result *= 2;
printf("Integer Power: %llu\n", result);
return 0;
}
Expected output
Integer Power: 1
Recognize the important boundaries
Restrict the base and exponent so every product fits; unsigned wraparound is not the intended mathematical power.
The example uses a deliberately small nonnegative input. A negative input needs an explicit rejection or a separately defined interpretation.
Check your understanding
Choose the answer, then compare the reasoning for every option.
Use the idea independently
A nonnegative integer power multiplies the base by itself exponent times; exponent zero gives one.
Continue with loop initialization and termination.
Continue with recursive alternatives.
Language reference: WG14 C11 committee draft. These examples use the C17-compatible subset of the explained language features.
Reaching the recap and completing the quizzes are separate milestones.