How can i copy the contents of one variable to another using pointers?

Question

char *foo(char *dest, const char *src)
{
    unsigned i;
    for (i=0; src[i] != '\0'; ++i)
        dest[i] = src[i];
    dest[i] = '\0';
    return dest;

}

I know that the code is moving through both variable's arrays, and while it's doing that, it makes dest[] have the same character as src[]

How can i do this with pointers instead?

You are already using pointers. There is no operator for accessing array elements (or at least directly). More info on the subject [here](http://stackoverflow.com/questions/381542/with-c-arrays-why-is-it-the-case-that-a5-5a). — AnArrayOfFunctions, Feb 04 '16 at 21:45

score 3 · Accepted Answer · answered Feb 04 '16 at 21:43

3

Simply

while (*src != '\0')
    *dst++ = *src++;
*dst = '\0'; // Perhaps you will also need this!

Remember at the end of the loop that dst is pointing to the end of the array not to the begining. And also copy the '\0' terminator, or you don't want to?

answered Feb 04 '16 at 21:43

Iharob Al Asimi

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That can be compressed to `while ((*dst++ = *src++));`, although that's probably a little terser than most people are comfortable with. – John Bode Feb 04 '16 at 21:48
i like the way this looks XD. Thank you guys, this solution worked for me. – tisaconundrum Feb 04 '16 at 21:51
`dst[1], *(dst+1), *(1+dst), 1[dst]`. as i remember all these are same and ugly a bit to me :). – meth Feb 04 '16 at 21:55
This only works if `src` and `dst` are already pointers. From the given context this is not clear. – too honest for this site Feb 04 '16 at 22:20
@Olaf edited my question such that it is obvious that we are in fact working with pointers – tisaconundrum Feb 05 '16 at 02:54

How can i copy the contents of one variable to another using pointers?

1 Answers1