如何从字符数组复制到字符指针？

Question

I am trying to copy from a character array to character pointer. My code:

char str[] = "Hello World";
char *result = (char *)malloc(strlen(str)+1);
int index=0;
while(index <= strlen(str))
{
  *result = str[index];
  result++;
  index++;
}

This above code is not working and below code is working

char str[] = "Hello World";
char *result = (char *)malloc(strlen(str)+1);
int index=0;
while(index <= strlen(str))
{
  result[index] = str[index];
  index++;
}

Can anyone explain this behavior?

Answer 1

we simply use the strcpy function to copy the array into the pointer.

     char str[] = "Hello World";
     char *result = (char *)malloc(strlen(str)+1);
     strcpy(result,str);

Answer 2

In your first snippet you modify the pointer in the loop, so after the loop it will no longer point to the same location returned by the malloc call.

Look at it this way, after the call to malloc the pointer and the memory look like this:

result
|
v
+---+---+---+---+---+---+---+---+---+---+---+---+
|   |   |   |   |   |   |   |   |   |   |   |   |
+---+---+---+---+---+---+---+---+---+---+---+---+

After the first iteration of the loop it will look like this:

result
    |
    v
+---+---+---+---+---+---+---+---+---+---+---+---+
| H |   |   |   |   |   |   |   |   |   |   |   |
+---+---+---+---+---+---+---+---+---+---+---+---+

And after the loop is done it will look like this

result
                                                 |
                                                 v
+---+---+---+---+---+---+---+---+---+---+---+----+
| H | e | l | l | o |   | W | o | r | l | d | \0 |
+---+---+---+---+---+---+---+---+---+---+---+----+

The copying is made, but the pointer no longer points to the original position.

[Note: Before the copying, the memory allocated by malloc will not be "empty" or initialized in any way, I just show it like that here for simplicity's sake.]

Answer 3

Try this code.. Use strcpy() function.

char str[] = "Hello World";
char *result = (char *)malloc(strlen(str)+1);
strcpy(result,str);

Answer 4

Already you allocate a memory for that pointer, so you can simply use the strcpy() function to copy the characters

 strcpy(result,str);

str copied to result.

Answer 5

In my opinion, there is no practical reason to modify your pointers (like you do in the first snippet) when you can just add an offset (like you do in the second snippet). It just makes the code a little bit harder to get right.

For the first snippet, try saving the original result like this:

char *result = ...
char *start = result;

And after the loop, try printing start , instead of result . It should point to the newly copied string, like you expect.

Answer 6

Your first code snippet copied the string perfectly , but in this process you moved allocated character pointer. If it has to work save its initial value in some temporary pointer

char str[] = "Hello World";
char *result = (char *)malloc(strlen(str)+1);
char *tmp = result;
int index=0;
while(index <= strlen(str))
{
  *result = str[index];
  result++;
  index++;
}
printf("%s\n", tmp);