string copy function

Question

#include<stdio.h>
#include<malloc.h>
void my_strcpy(char *sour,char *dest){
 if(sour == NULL || dest == NULL){
  return;
 }
 while(*sour != '\0'){
  *dest++ = *sour++;
 } 
 *dest = '\0';
}
int main(){
 char *d = NULL;
 char *s = "Angus Declan R";
 d = malloc(sizeof(char*));
 my_strcpy(s,d);
 printf("\n %s \n",d);
 return 0;
}

This func works fine and prints the string. My doubt is as the pointer "dest" will be pointing to the '\0' how does it prints the whole string(as it didnt point to the initial address of the string).

That first check should be (sour == NULL || dest == NULL) (OR, rather than AND). Also, the malloc line should be d = malloc(strlen(s)+1); (d needs to point to a buffer that is large enough to hold every character in string s plus the NULL character) — user1354557
– user1354557, Commented Mar 7, 2013 at 19:39
d = malloc(sizeof(char*)); doesn't allocate enough memory on most systems. Usually a pointer is four or eight bytes large. — Daniel Fischer
– Daniel Fischer, Commented Mar 7, 2013 at 19:40
Just a small suggestion, you could improve your function by having it return an int. Maybe the number of characters copied, and something to indicate if the pointers are NULL (like -1). — teppic
– teppic, Commented Mar 7, 2013 at 21:01

Nik Bougalis · Accepted Answer · 2013-03-07 19:40:41Z

It's true that dest will point to the end of the string. But you are not printing the string by using dest - you are printing the string by using d which is a different variable.

Remember that in C and C++ values are passed by value by default - so when you call the function my_strcpy the value of the variable d is copied into the variable dest which is local to the function my_strcpy only and any changes to that variable will not affect d.

Also note that you are not allocating enough space for your d variable:

d = malloc(sizeof(char*));

This will allocate enough space for a pointer to character which will usually mean enough space for 4 (or maybe 8) characters. You should allocate enough space for the string you intend to copy plus one character for the terminating null byte. What is the size of the string you are trying to copy? Hint: strlen should help.

adelbertc · Accepted Answer · 2013-03-07 19:39:05Z

0

The d pointer passed in is a copy of the pointer, so the position/movement of the pointer in your function is not reflected back to where it was called in main. While d and dest both point to the same block of memory (at least initially) and any changes to that block of memory will be reflected on both ends, the dest pointer is only a copy.

answered Mar 7, 2013 at 19:39

adelbertc

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Comments

ouah · Accepted Answer · 2013-03-07 19:39:05Z

0

 my_strcpy(s,d);

C passes arguments by value. The value of d is passed to my_strcpy, which means d object inside main is not modified in your my_strcpy function.

answered Mar 7, 2013 at 19:39

ouah

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