Dynamic strings

Learn to create dynamic strings with malloc.

Dynamic strings

A dynamic string is a char array allocated at runtime with malloc. Unlike fixed-size char arrays, its size is decided at runtime.

Example: char *s = malloc(20);. This allocates 20 bytes and assigns the start address to s. To fill the string use strcpy(s, "Merhaba");. malloc needs <stdlib.h>, strcpy needs <string.h>. When done, free it with free(s);; otherwise a leak occurs.

Using a dynamic string

Once allocated, a dynamic string is used like a normal string. It's filled with strcpy, printed with printf. The key is not to exceed the allocated size.

Example: char *s = malloc(20); strcpy(s, "Merhaba"); printf("%s\n", s); free(s);. Here 20 bytes are allocated, 8 are used (Merhaba + null). The remaining 12 are unused. malloc doesn't initialize; so filling with strcpy is important. The free call prevents a leak.

Exercise Create and print a dynamic string

Allocate a twenty-byte dynamic string with char *s = malloc(20);. Fill it with strcpy(s, "Merhaba");. Print it with printf("%s\n", s);. Finally free it with free(s);. Include <stdlib.h> and <string.h>.

Show example solution

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void) {
    char *s = malloc(20);
    strcpy(s, "Merhaba");
    printf("%s\n", s);
    free(s);
    return 0;
}

Test yourself

Which function is used for a dynamic string?
- malloc
- printf
- strcmp
- strlen
Show answer

malloc — malloc allocates heap memory for a dynamic string.

Allocate a dynamic string

char *s = ____(20);
strcpy(s, "Merhaba");

Show answer

char *s = malloc(20);
strcpy(s, "Merhaba");

Match the dynamic string parts

malloc(20)

Allocation
strcpy(s, ...)

Filling
printf("%s", s)

Printing
free(s)

Release

Next lesson: String concatenation

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