Writing replacement for strcpy

Question

I have to write a replacement for strcpy without using pointers or the function having a return value... How could this be done?!

this is what i have so far but it uses return values.

void str_copy(char destination [], char source []) {
    int i;
    for (i = 0; source[i] != '\0'; ++i)
        destination[i] = source[i];
    destination[i] = '\0';
    return destination;
}

without using pointers? Not possible. In your example, `destination` and `source` are adjusted to pointers. — ouah, Nov 20 '15 at 00:24
In C, arrays decays into pointers. See: http://c-faq.com/aryptr/aryptrparam.html — ViniciusArruda, Nov 20 '15 at 04:25
If is useful for you, you can use `memcpy(destination, source, strlen(source) * sizeof(char))` — ViniciusArruda, Nov 20 '15 at 04:29
One might note that `strcpy()` returns a `char *` with the target address. If you want a proper replacement, `void` is probably not the proper return type, while `return destination;` is actually fine. — dhke, Nov 20 '15 at 10:07

Vlad from Moscow · Answer 1 · 2015-11-20T06:48:18.730

4

Just remove the return statement.

The function can look the following way

void str_copy( char destination [], const char source [] ) {
    size_t i = 0;
    while ( destination[i] = source[i] ) i++;
}

edited Nov 20 '15 at 06:48

answered Nov 20 '15 at 00:23

Vlad from Moscow

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@iharob I think this requirement is understood incorrectly. – Vlad from Moscow Nov 20 '15 at 00:24
Minor: `int i` --> `size_t i`. – chux - Reinstate Monica Nov 20 '15 at 02:58
an assignment is always 'true', so this will result in a 'forever' loop – user3629249 Nov 20 '15 at 09:57
@user3629249 Are you sure that the assignment is always true? – Vlad from Moscow Nov 20 '15 at 10:04
@user3629249 Where did you get that? In C, an assignment [has the value of the left operand after the assignment](https://stackoverflow.com/questions/9514569/what-does-an-assignment-return). The loop terminates *after* copying the final `\0`. – dhke Nov 20 '15 at 10:04

score 1 · Answer 2 · answered Nov 20 '15 at 00:22

1

Why do you have to return destination? Since you passed the address of the first element by using the empty brackets, the destination array should be changed without the return.

answered Nov 20 '15 at 00:22

SteveO

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user3629249 · Accepted Answer · 2015-11-20T10:01:54.337

1

suggest writing the function this way:

void str_copy( char destination [], const char source [] ) 
{
    for(size_t i=0; source[i]; i++ )  destination[i] = source[i];
    destination[i] = '\0';
}

edited Nov 20 '15 at 10:01

answered Nov 20 '15 at 09:56

user3629249

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Writing replacement for strcpy

3 Answers3