CareerCup

int strcpy(char *dest, char *src)
{
    char *from = src;
    char *to = dest;

    while(*from != \0)
    {
        *to = *from;
        to++;
        from++;
    }
    *to = \0;
}

No need to keep the "char* from", and it should return dest.

char* strcpy(char* dest, const char* src)
{
char* first = dest;
while(*dest++ = *src++);
return first;
}

The function needs to

1. Make sure that dest points to a valid memory address (for example, not null or 0) and

2. It has enough memory to hold all characters + null character from the source string.

3. There is no need to return anything. The function can be void if it is copying to the memory location passed via the argument.

xstrcpy(string str1, string str2)
{
char *s1 = str1; //target string
char *s2 = str2; //source string
while(*s2)
{
*s1 = *s2;
s1++; s2++;
}

cout << str1 ;
}

xstrcpy(string str1, string str2)
{
char *s1 = str1; //target string
char *s2 = str2; //source string
while(*s2)
{
*s1 = *s2;
s1++; s2++;
}

cout << str1 ;
}

<pre lang="" line="1" title="CodeMonkey99059" class="run-this">

void strcpy(char *dest, char *src)
{
while(*dest++ = *src++);
}</pre><pre title="CodeMonkey99059" input="yes">
</pre>

static string ImplementStrCpy(string src, string dest)
{
while(src != null)
{
dest = src;
break;
}
return dest;
}

dont we need to worry about memory allocation??

void myStrcpy(char *src,char *dest)
{
    int i=0;
    
    *dest=*src;
    while(*src){
        *dest=*src;
        dest++;
        src++;
       // i++;
    }
    dest[i]='\0';
    
   // Complete this function