implementing a strcpy() function without using <string.h> in C - c

Question

My task is like this: I should implement the strcpy function under the following constraints:
The function should use pointer expression (*(d+i))
I should implement it without using <string.h>
I'm programming in Visual Studio 2019.
I searched some source code in google and run them, but my program has a logical error. The program ends right away, each time. I don't know what I'm doing wrong.
Here's my code in Visual Studio 2019 on Windows. Please tell me what's wrong.
#include <stdio.h>
void strcpy(char*, char*);
int main()
{
char* sen1 = "Hello";
char* sen2 = "Friends";
strcpy(sen1, sen2);
printf("The result: %s\n", sen1);
return 0;
}
void strcpy(char* str1, char* str2)
{
int i = 0;
while (*(str2 + i) != '\0')
{
*(str1 + i) = *(str2 + i);
i++;
}
*(str1 + i) = '\0';
}

Answer 1

In addition to needing to provide writable storage for sen1, you should also check to ensure str2 != NULL in your function before dereferencing str2 (otherwise, even if you fix all other errors -- a segfault will likely result)
For example, in your code you can define a constant to use in setting the size of a sen1 array (or you can allocate storage with malloc(), calloc(), or realloc() -- save that for later). Using an array you can do, e.g.
#include <stdio.h>
#include <stdlib.h>
#define MAXC 64 /* if you need a constant, #define one (or more) */
...
int main (void)
{
char sen1[MAXC] = "Hello";
char *sen2 = "Friends";
mystrcpy (sen1, sen2);
printf ("The result: %s\n", sen1);
}
In your strcpy function, check that str2 isn't NULL before using str2 in your function, e.g.
char *mystrcpy (char *dest, const char *src)
{
char *p = dest;
if (!src || !dest) { /* ensure src or dest is not NULL */
fputs ("error: src or dest parameters NULL in mystrcpy().\n", stderr);
exit (EXIT_FAILURE);
}
do /* loop */
*p++ = *src; /* copy each char in src to dest */
while (*src++); /* (including the nul-termianting char) */
return dest; /* return pointer to dest */
}
Now you will copy your source string to your destination string in your (renamed) mystrcpy() function, receiving the results you expect:
Example Use/Output
$ ./bin/mystrcpy
The result: Friends
Look things over and let me know if you have further questions.

Answer 2

Two problems, at least:
String literals are not writable in C. Often the symptom is a crash (SIGSEGV).
You are not allowed to use the identifier strcpy for your own function. Use another name.
Three clean code issues, at least:
Turn int main() into int main(void) to make it properly typed.
str1 and str2 are too generic names. They don't indicate which is the source and which is the destination pointer. What about my_strcpy(char *dest, char *src)?
I'd use size_t i for the index counter instead of int, because that's the type all the string length functions and the sizeof operator return. It's also an unsigned type and can copy really long strings :-) The size_t is available after #include <stddef.h>.

Answer 3

You want this:
...
char* source = "Hello";
// or char source[] = "Hello";
char destination[1000]; // destination buffer long enough for playing around
my_strcpy(destination, source);
printf("%s\n", destination); // should print "Hello" if my_strcpy is corect
...
For the rest read Jens's answer.

Answer 4

Among the other good answers, just regarding the implementation of your strcpy function and not a detailed issue analyze of your actual code, another approach is this:
char * n_strcpy(char * dest, char const * src)
{
if (dest == NULL || src == NULL)
{
return NULL;
}
char *ptr = dest;
while ((*dest++ = *src++));
return ptr;
}