What happened to my strings in both strcpy() and custom function? - c

Question

I have been doing well with the string functions I've been making recently. So then, I made my 3rd string function that is basically like strcpy():
#include <stdio.h>
#include <string.h>
const char *copy_str(const char *str);
int main() {
char name[256];
printf("What's your name? "); scanf("%255s", name);
char result[256] = copy_str(name);
printf("Hello, %s.\n", result);
}
const char *copy_str(const char *str) {
return str;
}
I didn't try using the strcpy() function itself in the function I made, because I'm not really familiar with it. Then I got an error:
copy_str.c:9:10: error: array initializer must be an initializer list or string literal
char result[256] = copy_str(name);
^
I fixed it this way:
int main() {
char name[256];
printf("What's your name? "); scanf("%255s", name);
char result[256];
strcpy(name, result);
printf("Hello, %s.\n", result);
}
But then the output went like this:
What's your name? builderman
Hello, .
No string for some reason. Even if you typed in a different string, it would be the same result.
Q: Why did strcpy() ruin my string? What ways can I improve my function?

Answer 1

I will try to answer your questions in parts.
copy_str.c:9:10: error: array initializer must be an initializer list or string literal
char result[256] = copy_str(name);
the meaning of this error is that in C syntax you cannot initialize an array the way that you did.
you can only initialize an array of chars with either a literal like so:
char name[] = "Hello";
or an initialization list like so:
char name[] = {'H', 'e', 'l', 'l', 'o'};
you can't initialize an array with a pointer to a char (which is the return value of your function)
strcpy doesn't magically copy an entire string to another one, it iterates over the string until it finds a NULL terminator '\0'
so a simple implementation of strcpy can be:
char *StrCpy(char *dest, const char *src)
{
char *returned_str = dest;
while (*src != '\0')
{
*dest = *src;
src++;
dest++;
}
*dest = '\0';
return returned_str;
}
so strcpy iterates over the strings and copies each character from src to dest.
a problem may arise when sending an array of characters that doesn't have a null terminator at the end of the array, which is how a string should be terminated in C.

Answer 2

The function strcpy has the following declaration
char *strcpy(char * restrict s1, const char * restrict s2);
That is the first parameter defines the array where the string pointed to by the second parameter is copied.
However in your program
int main() {
char name[256];
printf("What's your name? "); scanf("%255s", name);
char result[256];
strcpy(name, result);
printf("Hello, %s.\n", result);
}
you are copying the non-initialized character array result into the array name.
You have to write
strcpy( result, name );
instead of
strcpy(name, result);
As for the first program then the function copy_str is redundant. You could just write
char result[256] = name;
However the initializer has the type char * due to the implicit conversion of an array to pointer to its first element. So such initialization of an array is incorrect. You may initialize a character array with a string literal like
char s[] = "Hello";
But you may not use a character pointer to initialize a character array like
char t[] = "Hello";
char s[] = t;
Your function copy_str could look the following way
char * copy_str( char *dsn, const char *src )
{
for ( char *p = dsn; ( *p++ = *src++ ); );
return dsn;
}