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This article will demonstrate various methods on how to copy a character array in C language.

memcpyCopy a character array in C using the function

charArrays are probably the most commonly used data structure in C code, and copying array contents is one of its core operations. C-style strings charare very similar to arrays, and therefore, there are multiple ways to handle copying array contents. In the following example, we declare two arrays, arrand arr2; the former is initialized with character list notation, and the latter is initialized with a string literal. Note that this results in different array structures and sizes. arrThe object is 7 characters in memory, but arr2requires 17 characters plus the terminating null byte, resulting in an 18-byte object. Therefore, we sizeof arr2 - 1pass the value of the expression as the second argument, indicating the length of the array. On the other hand, we can print the contents of the array using printfthe function and %sthe format specifier arr2.

The same details should be considered when copying charan array to a different location. memcpyThe get_by_array function is a part of the standard library string utilities and is defined in <string.h>the get_by_array header file. It takes three parameters, the first one is the destination pointer, where the contents of the array will be copied. The second parameter is a pointer to the source array and the last parameter specifies the number of bytes to be copied. Note that sizeofthe operator returns charthe size of the array object in bytes. So we call it with mallocthe sizeof arrvalue of the expression to allocate dynamic memory. The memory area returned is large enough to hold arrthe contents of and we don't need to worry about buffer overflow bugs. Note, however, that mallocthe pointer returned from should be freed.

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

void printCharArray(char *arr, size_t len)
{
    for (size_t i = 0; i < len; ++i) {
        printf("%c, ", arr[i]);
    }
    printf("\n");
}

int main(int argc, char *argv[]){
    char arr[] = {'a', 'b', 'c', 'd', 'e', 'f', 'g'};
    char arr2[] = "array initialized";

    printf("%lu\n", sizeof arr);
    printf("%lu\n", sizeof arr2);

    printCharArray(arr, sizeof arr);
    printCharArray(arr2, sizeof arr2 - 1);

    char *str = malloc(sizeof arr);
    memcpy(str, arr, sizeof arr);
    printf("str: ");
    printCharArray(str, sizeof arr);
    free(str);

    str = malloc(sizeof arr2);
    memcpy(str, arr2, sizeof arr2);
    printf("str: %s\n", str);
    free(str);

    exit(EXIT_SUCCESS);
}

Output:

7
18
a, b, c, d, e, f, g,
a, r, r, a, y,  , i, n, i, t, i, a, l, i, z, e, d,
str: a, b, c, d, e, f, g,
str: array initialized

memmoveCopy a character array in C using the function

memmoveis another memory area copy function in the standard library string utilities. It is implemented as a more powerful function to accommodate the case where the destination and source memory areas overlap. memmoveThe parameters are memcpythe same as .

When copying arr2the contents of , we sizeof arr2pass the expression as the third argument. This means that even the terminating null byte is copied to the destination pointer, but we therefore use this operation to call printfand %sto output the contents, rather than using printCharArray.

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

void printCharArray(char *arr, size_t len)
{
    for (size_t i = 0; i < len; ++i) {
        printf("%c, ", arr[i]);
    }
    printf("\n");
}

int main(int argc, char *argv[]){
    char arr[] = {'a', 'b', 'c', 'd', 'e', 'f', 'g'};
    char arr2[] = "array initialized";

    char *str = malloc(sizeof arr);
    memmove(str, arr, sizeof arr);
    printf("str: ");
    printCharArray(str, sizeof arr);
    free(str);

    str = malloc(sizeof arr2);
    memmove(str, arr2, sizeof arr2);
    printf("str: %s\n", str);
    free(str);

    exit(EXIT_SUCCESS);
}

Output:

str: a, b, c, d, e, f, g,
str: array initialized