How to copy a char array in C?

Question

In C, I have two char arrays:

char array1[18] = "abcdefg";
char array2[18];

How to copy the value of array1 to array2 ? Can I just do this: array2 = array1?

Solution 1

You can't directly do array2 = array1, because in this case you manipulate the addresses of the arrays (char *) and not of their inner values (char).

What you, conceptually, want is to do is iterate through all the chars of your source (array1) and copy them to the destination (array2). There are several ways to do this. For example you could write a simple for loop, or use memcpy.

That being said, the recommended way for strings is to use strncpy. It prevents common errors resulting in, for example, buffer overflows (which is especially dangerous if array1 is filled from user input: keyboard, network, etc). Like so:

// Will copy 18 characters from array1 to array2
strncpy(array2, array1, 18);

As @Prof. Falken mentioned in a comment, strncpy can be evil. Make sure your target buffer is big enough to contain the source buffer (including the \0 at the end of the string).

Solution 2

If your arrays are not string arrays, use: memcpy(array2, array1, sizeof(array2));

Solution 3

If you want to guard against non-terminated strings, which can cause all sorts of problems, copy your string like this:

char array1[18] = {"abcdefg"};
char array2[18];

size_t destination_size = sizeof (array2);

strncpy(array2, array1, destination_size);
array2[destination_size - 1] = '\0';

That last line is actually important, because strncpy() does not always null terminate strings. (If the destination buffer is too small to contain the whole source string, sntrcpy() will not null terminate the destination string.)

The manpage for strncpy() even states "Warning: If there is no null byte among the first n bytes of src, the string placed in dest will not be null-terminated."

The reason strncpy() behaves this somewhat odd way, is because it was not actually originally intended as a safe way to copy strings.

Another way is to use snprintf() as a safe replacement for strcpy():

snprintf(array2, destination_size, "%s", array1);

_{(Thanks jxh for the tip.)}

Solution 4

As others have noted, strings are copied with strcpy() or its variants. In certain cases, you could use snprintf() as well.

You can only assign arrays the way you want as part of a structure assignment:

typedef struct { char a[18]; } array;
array array1 = { "abcdefg" };
array array2;

array2 = array1;

If your arrays are passed to a function, it will appear that you are allowed to assign them, but this is just an accident of the semantics. In C, an array will decay to a pointer type with the value of the address of the first member of the array, and this pointer is what gets passed. So, your array parameter in your function is really just a pointer. The assignment is just a pointer assignment:

void foo (char x[10], char y[10]) {
    x = y;    /* pointer assignment! */
    puts(x);
}

The array itself remains unchanged after returning from the function.

This "decay to pointer value" semantic for arrays is the reason that the assignment doesn't work. The l-value has the array type, but the r-value is the decayed pointer type, so the assignment is between incompatible types.

char array1[18] = "abcdefg";
char array2[18];
array2 = array1; /* fails because array1 becomes a pointer type,
                    but array2 is still an array type */

As to why the "decay to pointer value" semantic was introduced, this was to achieve a source code compatibility with the predecessor of C. You can read The Development of the C Language for details.

Solution 5

You cannot assign arrays, the names are constants that cannot be changed.

You can copy the contents, with:

strcpy(array2, array1);

assuming the source is a valid string and that the destination is large enough, as in your example.

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