Using strcpy() function in C

Question

I am presently new to programming and am following the C language which is being taught to us in our College. I have doubt in the actual functioning of the strcpy() function under the header-file #include<strings.h>

The general use of this function is given below -

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main()
{
     char a[] = "google";
     char *b = (char*)malloc((strlen(a) +1)*sizeof(char));
     strcpy(b,a);
     printf("%s, %s\n", a, b); //prints google, google
     free(b);   
}

We are given the following format and prototype of the same:

char *strcpy( char *to, const char *from)
{
    char *temp = to;
    while (*to++ = *from++);
    return temp;
}

I have a couple of doubts about this-

1. What is an extra pointer ( temp ) requirement in the function? Can't we just return to; at the end of the function?

2. I convinced myself that the temp variable is used because during each while loop, the pointer to is changing itself. That is first *to will be assigned *from and then to = to +1 , and from = from +1 , will take place which will result in changing the address stored in the pointer to . But if this is true, then what about the pointer from ? Will, it also not change at the end of the strcpy() function?

3. Where is the function returning the address stored in temp? ( the function should only work if b = strcpy(b,a); be mentioned not only strcpy(b,a); )

To check the given function I also made my own function "strcpy_me" and used it with the same format and prototype, and it gave the same results.

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

char *strcpy_me( char *to, const char *from)
{
    char *temp = to;
    while (*to++ = *from++);
    return temp;
}

int main()
{
     char a[] = "google";
     char *b = (char*)malloc((strlen(a) +1)*sizeof(char));
     strcpy_me(b,a);
     printf("%s, %s\n", a, b); // prints google, google
     free(b);   
}

Please help me clarify my doubt.

Answer 1

1. see 2

2. It's a requirement from the implementation of strcpy to return a pointer to the beginning of the destination string. So if you want to implement your function in the same way as strcpy , you'll have to save a copy of the original pointer passed as parameter, in case you intend to change that parameter inside the function. This isn't a particularly useful or rational feature, but strcpy has been like that since the dawn of time.

3. Not sure what you mean; here obviously: return temp;

4. return to; is wrong. You have to realize that the function returns a value whether that value is used by the caller or not. Strict compilers might warn "value from function isn't used" in which case you'd have to write (void) strcpy(a,b); to silence it. Since your example doesn't use the returned value, it doesn't matter that it is wrong.

Answer 2

1. What you think in no. 2 is correct.

2. from does get changed but we don't need the previous value of from . to was stored in temp as to 's original value should be returned, so we need it but no need for from .

3. No, you don't need to do that. You can ignore return values in C(in most cases without warnings as well). No returning works as doing *to = *from changes the parameter's memory as well, in this case, b. So, *to = *from changes the array b directly. It is because to and b store the same memory.

Note: In your while loop, it's *to++ = *from++ . It uses post addition, which for simplification, I didn't include here but all it does is increment both to and from after the operation.

4. As I said in 3, you don't need the return value. In fact, because original to and temp point to the same memory as b , the function returns b itself. doing, b = strcpy_me(b, a) is the same as:

    strcpy(b, a); b = b; //b's value stays the same

And yes, return temp; is absolutely garbage in my opinion. It only causes confusion.