Reproduce the behaviour of strcpy

Question

I'm trying to reproduce the behaviour of strcpy in c, my problem is that the function works but it append extra stuff at the end.

char    *ft_strcpy(char * dst, const char * src)
{
int i;
i = 0;
while (src[i] != '\0') {
    dst[i] = src[i];
    i++;
}
return dst;
}

When i run it I get the following.

int main()
{
char p[] = {};
char z[] = "Hello World";
ft_strcpy(p,z);
printf("%s\n", p);
return 0;
}

function results

John3136 · Accepted Answer · 2016-02-09T02:40:48.790

2

You aren't copying the nul terminator so the printf doesn't know to stop.

Just add dst[i] = 0; after the loop.

Also, you haven't allocated any space for p, so you're going to get undefined behavior. For a first test just do something like char p[100]; and make sure z is never > 99 chars long. Eventually you need a better fix, but that will get you started.

edited Feb 09 '16 at 02:40

answered Feb 09 '16 at 02:35

John3136

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Yes, thanks but I still get `Segmentation fault (core dumped)` text. – Feb 09 '16 at 02:38
Did you address the fact that `p` doesn't have "space"? – John3136 Feb 09 '16 at 02:39
Ok, I used your suggestion and its working. I just need a way to make it better. – Feb 09 '16 at 02:47

score 0 · Answer 2 · answered Feb 09 '16 at 05:05

TenTen,

Set dest[i] to '\0' after ending the loop. Here is the code:

int main()
    {
    char src[100], dest[100], i;
    printf("Enter string Src: ");
    scanf("%s",src);
    for(i=0; src[i]!='\0'; ++i)
        {
        dest[i]=src[i];
        }
    dest[i]='\0';
    printf("String Dest: %s",dest);
    return 0;
    }

Reproduce the behaviour of strcpy

2 Answers2