C Program to Check a Palindrome String

A school text exercise needs to palindrome string while preserving the buffer’s valid characters and final terminator.

A bytewise palindrome reads the same from each end under an explicitly stated comparison policy.

Before this exercise, review strings and their terminator and array bounds. The worked program below applies those ideas to this specific problem.

Explain
Explain why palindrome string uses this sequence of operations.
Trace
Trace the sample values and predict the printed result.
Apply
Try School without silently ignoring case or reordering its letters.

Palindrome String: follow the data

Given data
A school text exercise needs to palindrome string while preserving the buffer’s valid characters and final terminator.
Rule
A bytewise palindrome reads the same from each end under an explicitly stated comparison policy.
Observed result
Palindrome

A guided model of this example. The illustration explains the process; run the complete C program to check its output.

Understand the operation

Compare symmetric positions only through half the length. No reversal buffer is needed to test equality, and the loop naturally accepts empty or one-character text under this exact-byte definition.

A C string ends at its first null character. Read only within its accessible storage and reserve a position for that terminator when producing a changed string. These examples use basic text bytes, not a Unicode grapheme-processing library.

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>
#include <string.h>
#include <ctype.h>
int main(void) {
  const char text[] = "level";
  size_t length = strlen(text);
  int palindrome = 1;
  for (size_t i = 0; i < length / 2; i++) if (text[i] != text[length - 1 - i]) palindrome = 0;
  puts(palindrome ? "Palindrome" : "Not palindrome");
  return 0;
}

Expected output

Palindrome

Run in C compiler →

Follow the values through the code

C17 · code used in this explanation

#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main(void) {
  const char text[] = "level";
  size_t length = strlen(text);
  int palindrome = 1;
  for (size_t i = 0; i < length / 2; i++) if (text[i] != text[length - 1 - i]) palindrome = 0;
  puts(palindrome ? "Palindrome" : "Not palindrome");
  return 0;
}

The pairs in level are l/l and e/e; the middle v needs no comparison.

Compare the printed answer with the original text and the stated ordering or filtering rule. The presence of a terminator makes %s usable; it does not by itself make every earlier write fit.

Follow the worked calculation

Start with the sample
The pairs in level are l/l and e/e; the middle v needs no comparison.
Apply the rule
Compare the printed answer with the original text and the stated ordering or filtering rule. The presence of a terminator makes %s usable; it does not by itself make every earlier write fit.
3
Check the result
Palindrome
```
Palindrome
```

A guided model of this example. The illustration explains the process; run the complete C program to check its output.

Try a small variation

C17 · code used in this explanation

#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main(void) {
  const char text[] = "level";
  size_t length = strlen(text);
  int palindrome = 1;
  for (size_t i = 0; i < length / 2; i++) if (text[i] != text[length - 1 - i]) palindrome = 0;
  puts(palindrome ? "Palindrome" : "Not palindrome");
  return 0;
}

Try School without silently ignoring case or reordering its letters.

Worked answer

The complete worked answer applies the same explicitly bounded operation to a small variation. Read its stored bytes and terminator separately from the visible output.

C17 · complete program

#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main(void) {
  const char text[] = "School";
  size_t length = strlen(text);
  int palindrome = 1;
  for (size_t i = 0; i < length / 2; i++) if (text[i] != text[length - 1 - i]) palindrome = 0;
  puts(palindrome ? "Palindrome" : "Not palindrome");
  return 0;
}

Expected output

Not palindrome

Run in C compiler →

Recognize the important boundaries

This definition is case-sensitive and includes spaces and punctuation. A phrase-palindrome task needs a separate normalization policy.

An untrusted input string needs a verified buffer bound and a valid terminator. strlen is not a safe way to prove arbitrary raw storage has a terminator.

Choose the answer, then compare the reasoning for every option.

Use the idea independently

A bytewise palindrome reads the same from each end under an explicitly stated comparison policy.

Continue with character-array storage.

Continue with bounded line reading.

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.

Next lesson: C Program to Reverse a String Using Recursion →

C Program to Check a Palindrome String