Need to Rewrite strcpy() function

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November 10th, 2002, 09:49 PM

Rdesign

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Need to Rewrite strcpy() function

We have a new assignment: rewrite the strcpy() function that is usually accessable with the <cstring.h> include. I thought of a way of doing it with the use of pointers however, my instructor has outlawed them because we have not reached that lesson yet. If someone here can provide a hint or a solution I belive that I would learn alot from it. So, strcpy(dest, origin); no pointers any suggestions?

Thanks in advance.

November 18th, 2002, 02:12 AM

vpopper

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How about this-- no pointers:

Code:

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

char * mystrcpy(char to[], const char from[], size_t size)
{
    register int i;

    for (i=0; i <= size; i++)
        to[i] = from[i];

    to[size] = 0;
    return to;
}

int main (int argc, char** argv)
{
    char foo[] = {'f','o','o'};
    char bar[4];

    (void)mystrcpy(bar, foo, 3);
    printf("bar = %s\n", bar);
    return 0;
}

November 18th, 2002, 03:22 AM

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Or, without a size parameter:

Code:

void mystrcpy (char to[], char from[])
{
    int i;

    for (i = 0; from[i] != 0; i++) {
        to[i] = from[i];
    }
}

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November 18th, 2002, 02:07 PM

Rdesign

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That's not a pointer?
What is the astric after the return value in the function mean?
////////////////////////////////////////////////////////
char * mystrcpy(char to[], const char from[], size_t size)
{
register int i;

for (i=0; i <= size; i++)
to[i] = from[i];

to[size] = 0;
return to;
}
///////////////////////////////////////////////////////
I already gave up and used a pointer. The assignmentis due today....I just had a thought, is that a virtual function? If it is, were not allowed to use those either :-)
Thanks anyway though.

November 19th, 2002, 03:38 AM

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Quote:

What is the astric after the return value in the function mean?

The asterisk is part of the return type. vpopper's mystrcpy() function actually returns a pointer-to-char. With parentheses, the prototype would be

Code:

(char *) mystrcpy (char to[], const char from[], size_t size);

Quote:

That's not a pointer?

I'm not sure what exactly you're referring to, but I'll assume you're talking about the functions' parameters. Strictly speaking they are pointers, yes. The []-brackets are more a hint to the programmer that the function expects an array-of-type rather than a pointer-to-type.

Although a bit late (since you already handed in your assignment), I think the idea was that you didn't use pointer arithmetic to copy the string.

The thing is, it doesn't make sense to write a string copying routine if you're not working with the original destination array. A reference to it has to be passed to the function one way or another. C implements the "pass by reference" concept with pointers; there's no way around that.

Need to Rewrite strcpy() function

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