Changing a char array without even calling the array?

Question

I have studied c++ for about 2 weeks on cave of programming (while working full time :( ) and I just got an understanding for pointers. Point *pThis to &this to get the value out of this address and so on. Anyway, I just got to a video that flipped everything I knew.

I had to copy this to paper first because I am not at home I hope it is correct!

char text[] = "hello";

int nChars = sizeof(text) - 1;

char *pStart = text;
char *pEnd = text + nChars - 1;

while(pStart < pEnd)
{
    char save = *pStart;
    *pStart = *pEnd;
    *pEnd = save;

    pStart++;
    pEnd--;
}

cout >> text >> endl;

I hope I wrote it correct from this video with my phone

https://youtu.be/g9oMn4mEx14

I do understand that we are creating two pointers that points to the start and end of text and then switching them and so on. What I don't understand is how can it change text without us even assigning anything to that char array anywhere? This boggles my mind.

I hope someone can help me get this :)

`int i = 5; int *pi = &i; *pi = 6; printf("%d\n", i);` <- prints 6. what????? how can `i` be 6 when we didn't even call that int anywhere? — user253751, Nov 25 '20 at 15:49
The code in the loop does a s swap of values and moves the start and end each 1 step closer to each other. — drescherjm, Nov 25 '20 at 15:50
Please note that Stack Overflow expects a certain amount of effort on your part. Please take the time to sit down at a computer when possible, and write the code, and run the program. Posting code with pointers that you wrote by hand is not going to be well received. — cigien, Nov 25 '20 at 15:51
The changing is done here `*pStart = *pEnd;` and here `*pEnd = save;`. This is normal, you have a pointer, you dereference it and assign. That changes whatever the pointer is pointing to, in this case elements of the `text` array. — john, Nov 25 '20 at 15:51

score 4 · Answer 1 · answered Nov 25 '20 at 16:06

Draw it on paper (still the best method to both understand and write code with pointers).
Here is an ASCII version:

The beginning:

+---+---+---+---+---+---+
| h | e | l | l | o | 0 |
+---+---+---+---+---+---+
  ^               ^
  |               |
 pStart         pEnd

The loop:

save = *pStart;

+---+---+---+---+---+---+
| h | e | l | l | o | 0 |       save: h
+---+---+---+---+---+---+
  ^               ^
  |               |
 pStart         pEnd

Then:

*pStart = *pEnd;

+---+---+---+---+---+---+
| o | e | l | l | o | 0 |       save: h
+---+---+---+---+---+---+
  ^               ^
  |               |
 pStart         pEnd

Then:

 *pEnd = save;

+---+---+---+---+---+---+
| o | e | l | l | h | 0 |       save: h
+---+---+---+---+---+---+
  ^               ^
  |               |
 pStart         pEnd

And then:

pStart++;
pEnd--;

+---+---+---+---+---+---+
| o | e | l | l | h | 0 |       save: h
+---+---+---+---+---+---+
      ^       ^
      |       |
    pStart   pEnd

and you can repeat the loop yourself.
(It's much faster and easier on paper than with a computer and a keyboard.)

score 0 · Answer 2 · edited Nov 25 '20 at 16:35

0

how can it change text without us even assigning anything to that char array anywhere? This boggles my mind.

Because the char array acts like a pointer - it's just an address to where the characters are stored. You want to be changing the characters themselves. The location in memory where the array is remains constant here.

edited Nov 25 '20 at 16:35

cigien

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answered Nov 25 '20 at 16:08

Kuba hasn't forgotten Monica

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Thank you Unslander! Now I get it! – V mike Nov 25 '20 at 16:13

score 0 · Answer 3 · answered Nov 25 '20 at 16:47

To simplify this to pseudocode, we are essentially swapping the characters at the start and end pointers using a temp variable save.

In the code snippet you just posted, this occurs in 3 steps:

store start to save
store end to start
store save to end

Changing a char array without even calling the array?

3 Answers3