C Program to Concatenate Two Strings
C Program to Concatenate Two Strings
A school text exercise needs to concatenate strings while preserving the buffer’s valid characters and final terminator.
String concatenation appends one string’s characters after another, placing a single terminator after the combined content.
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 concatenate strings uses this sequence of operations.
Trace
Trace the sample values and predict the printed result.
Apply
Use Hill as the prefix and append through an explicit copy loop including the suffix terminator.
Concatenate Strings: follow the data
Given data
A school text exercise needs to concatenate strings while preserving the buffer’s valid characters and final terminator.
Rule
String concatenation appends one string’s characters after another, placing a single terminator after the combined content.
Observed result
River School
Understand the operation
Check the known destination capacity before strcat. The source’s terminator replaces the destination’s old terminator after the appended characters.
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) {
char name[32] = "River";
const char suffix[] = " School";
size_t used = strlen(name), added = strlen(suffix);
if (added >= sizeof name - used) { puts("Text is too long"); return 0; }
strcat(name, suffix);
puts(name);
return 0;
}
Expected output
River School
Follow the values through the code
C17 · code used in this explanation
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main(void) {
char name[32] = "River";
const char suffix[] = " School";
size_t used = strlen(name), added = strlen(suffix);
if (added >= sizeof name - used) { puts("Text is too long"); return 0; }
strcat(name, suffix);
puts(name);
return 0;
}
River occupies five characters and the suffix seven. The result has twelve visible characters plus one terminator.
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
River occupies five characters and the suffix seven. The result has twelve visible characters plus one terminator.
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.
Check the result
River School
River School
Try a small variation
C17 · code used in this explanation
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main(void) {
char name[32] = "River";
const char suffix[] = " School";
size_t used = strlen(name), added = strlen(suffix);
if (added >= sizeof name - used) { puts("Text is too long"); return 0; }
strcat(name, suffix);
puts(name);
return 0;
}
Use Hill as the prefix and append through an explicit copy loop including the suffix terminator.
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) {
char name[32] = "Hill";
const char suffix[] = " School";
size_t used = strlen(name), added = strlen(suffix);
if (added >= sizeof name - used) { puts("Text is too long"); return 0; }
for (size_t i = 0; i <= added; i++) name[used + i] = suffix[i];
puts(name);
return 0;
}
Expected output
Hill School
Recognize the important boundaries
strcat does not accept a destination size. The explicit remaining-capacity check is what establishes sufficient storage.
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.
Check your understanding
Choose the answer, then compare the reasoning for every option.
Use the idea independently
String concatenation appends one string’s characters after another, placing a single terminator after the combined content.
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.