Unit 3 · Lesson 720 minAcademic review pending

Copying, joining and comparing strings

You cannot assign or compare strings with = and ==. The string library does that work, and an array of pointers holds a list of strings.

Choose explanation

After this lesson

You should be able to

Use strcpy, strncpy, strcat and strlen correctly.
Compare two strings with strcmp and interpret its return value.
Declare an array of pointers to strings and loop over it.

Why = and == do not work

A string is an array, and C will not assign one array to another. name2 = name1; does not compile. And if both were pointers, == would compare addresses — asking whether they are the same string in memory, not whether they spell the same thing.

So the library gives you strcpy to copy and strcmp to compare. Both need #include <string.h>.

Copy, then join

#include <string.h>

char a[20] = "Warangal";
char b[20];

strcpy(b, a);              /* b now holds "Warangal"     */
strcat(b, " district");    /* b now "Warangal district"  */

printf("%zu\n", strlen(b));  /* 18 — excludes the \0    */

strcmp returns an order, not a yes or no

strcmp(a, b) returns 0 when the strings are equal, a negative number when a comes before b alphabetically, and a positive number when it comes after. So equality is strcmp(a, b) == 0 — reading it as 'if strcmp is true' gets the logic exactly backwards, because 0 means equal and 0 is false.

Because it returns an ordering, strcmp is exactly what a sort needs. Swap the integer comparison in Unit III's selection sort for strcmp(...) < 0 and it sorts names.

Zero means equal; the sign gives the order

if (strcmp(town, "Warangal") == 0)
    printf("Same town\n");

/* Sorting names uses the sign, not just zero */
if (strcmp(names[j], names[smallest]) < 0)
    smallest = j;

An array of pointers is a list of strings

const char *days[] = {"Mon", "Tue", "Wed"}; stores three addresses, each pointing at text. The strings may have different lengths, and no space is wasted padding them to a common width — unlike a 2D char array, where every row occupies the same number of bytes.

Mark them const. String literals must not be modified, and const turns an attempt to do so into a compile error rather than a crash at run time.

Three strings of different lengths, no wasted space

const char *days[] = {"Monday", "Tuesday", "Wednesday"};

for (int i = 0; i < 3; i++)
    printf("%d: %s (%zu letters)\n", i, days[i], strlen(days[i]));

Try it yourself

Store five town names, sort them alphabetically, and print the sorted list.

Need a hint?

Use an array of pointers and swap the pointers, not the text.

Check the worked solution

Swapping pointers rather than copying characters is the point of this exercise: each swap moves 8 bytes regardless of how long the town names are, and no destination-size worry arises because nothing is copied. The comparison is strcmp(...) < 0, reusing the selection sort from earlier in this unit with only the comparison changed.

#include <stdio.h>
#include <string.h>
#define SIZE 5

int main(void)
{
    const char *towns[SIZE] = {"Warangal", "Adilabad", "Nizamabad",
                               "Khammam", "Karimnagar"};

    for (int i = 0; i < SIZE - 1; i++) {
        int smallest = i;

        for (int j = i + 1; j < SIZE; j++)
            if (strcmp(towns[j], towns[smallest]) < 0)
                smallest = j;

        if (smallest != i) {
            const char *temp = towns[i];
            towns[i] = towns[smallest];
            towns[smallest] = temp;
        }
    }

    for (int i = 0; i < SIZE; i++)
        printf("%s\n", towns[i]);

    return 0;
}

Quick check

How do you test whether two strings a and b are equal?

Select an answer to check your thinking.

Why this lesson exists

Syllabus mapping

String Library Functions: Assignment and Substrings · String Comparison · Arrays of Pointers

Maps to course outcome CO5.