Pointers and operations with pointers

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Hi, everybody;
I found the code above. But I don't understand the out where x=3 y=4. Please help me.
Why didn't work the operations ++ when case 0 twice? And why y=4, becase I think it should be 5.

What I have tried:

C++

#include<stdio.h>

int main(){
int x=2,y=3,*ptr,i=0;
int a[9]={1,2,3,4,5,6,7,8,9};

ptr=a;
//     printf("X= %d\n",x);
//     printf("*ptr= %d\n\n",*ptr);
while(i<5)
{
    //printf("i=%d\n",i);
    switch(i%3)
    {
    case 0:
        x=*(ptr)++;
//        printf("X= %d\n",x);
//        printf("*ptr= %d\n\n",*ptr);
    break;
    case 1:
       y=++*(ptr);
//       printf("Y= %d\n",y);
//       printf("*ptr= %d\n\n",*ptr);
    break;
    default:
        break;
    }
    i++;
    printf("x = %d\ty = %d\n",x,y);
}

return 0;
}

Posted 30-Jun-20 3:26am

goksurahsan

Updated 30-Jun-20 4:39am

KarstenK

v2

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OriginalGriff · Accepted Answer · 2020-06-30T04:31:00

Solution 1

You are playing with two things here, neither of which I am sure you understand well.

Pre- and post-increment operations can be difficult to understand, because they are "side effect" operators: they change the value of variables without an obvious assignment.
In your case,

C++

x = *(ptr)++;

is the equivelant of saying this:

x = *ptr;
ptr = ptr + 1;

And

C++

y = ++*(ptr);

is the equivelant of this:

C++

*ptr = *ptr + 1;
y = *ptr;

The second version changes the value in the array a instead of altering the pointer itself.

Run the code through the debugger and look closely at what is happening to the values of ptr and the array and you'll see what I mean.

It might be worth a quick look here: Why does x = ++x + x++ give me the wrong answer?[^] before you start diving into complicated pre- and post- increment statements!

Posted 30-Jun-20 4:31am

OriginalGriff

Updated 30-Jun-20 4:32am

v2

Comments

goksurahsan 30-Jun-20 10:37am

I understand pre and post-increment before. But I actually ask for this action why when i=3 post-increment didn't work for ptr. I used debugger but debeugger showed me post-increment passed

OriginalGriff 30-Jun-20 11:02am

It worked fine. It did exactly what you told it to.

goksurahsan 30-Jun-20 10:42am

For example I think these codes are same the before codes:

case 0:
x=*(ptr);
*ptr=*ptr+1;
//x=*ptr++;
// printf("X= %d\n",x);
// printf("*ptr= %d\n\n",*ptr);
break;
case 1:
*(ptr)=*(ptr)+1;
y=*(ptr);
//y=++(*ptr);

Richard MacCutchan 30-Jun-20 10:43am

No, see my comment below.

Richard MacCutchan 30-Jun-20 10:43am

Not quite. The second case is equivalent to:

y = *ptr;
y = y + 1;

ptr does not get incremented in this case.

Took me half an hour to figure it out.

goksurahsan 30-Jun-20 10:48am

But ++ near ptr. I should it change ptr. Is this true?

Richard MacCutchan 30-Jun-20 10:50am

~~No, the * is closest to ptr so it does the indirection operation first. i.e. it takes the value (*ptr) and increments that, it does not increment the pointer.~~

OriginalGriff 30-Jun-20 11:18am

Are you sure?
It's a pre-increment, and if I test it in VS:
int main()
{
int arr[5] = { 10, 21, 32, 43, 54 };
int* p = arr;
int x = ++ * p;
return 0;
}I get arr[0] == 11, x == 11, p == arr[0] as I'd expect.

Richard MacCutchan 30-Jun-20 11:28am

You are right, I looked at this three times and still managed to misread my own answers.

OriginalGriff 30-Jun-20 11:36am

They do my head in from time to time as well ... :laugh:

Richard MacCutchan 30-Jun-20 11:43am

And I used to write that sort of thing all the time back in my pure C days. Mind you, I doubt that I ever understood it properly.

OriginalGriff 30-Jun-20 12:03pm

Ditto.
And the worst bit is yet to come ... if the OP hasn't followed the article link I gave him, he's going to get really confused with his next try - they are implementation dependent side-effect operators as well ... :EvilLaughSmiley:

goksurahsan 30-Jun-20 10:44am

And while these codes work as I thought, why does the increment process don't work in the previous code when i = 3 in * (ptr) ++ section. Unfortunately this is the part I do not understand.

OriginalGriff 30-Jun-20 11:39am

It does. Read what we said above. ++*p changes the value that p points to, not the value of p.
But because your numbers in the array also increment by one, it's difficult to tell unless you look closely.
The debugger will show you that if you look at ptr and the array content.

Richard MacCutchan 30-Jun-20 11:41am

Because after the first case 0 a = { 1,2,3,4 ...} and ptr points at 2
We than go to case 1: ptr points at 2, which gets changed to 3. y gets 3.
so a now equals {1,3,3,4 ...};
Next time we get to case 0, x gets 3 (the first 3) and ptr gets changed to point to the second 3.
Next case 1, the second 3 gets changed to 4, so a = { 1,3,4,4, ...} and y gets the first 4.

Change the numbers in the array to be non-contiguous (e.g. { 3,7,15,22 } and it becomes clearer.

Note: Apologies for my previous 'solution' it was incorrect.

goksurahsan 30-Jun-20 12:23pm

Really thank you so much. I couldn't understand because I was new to C. I get it now. While the address indicated by the pointer changes in x = * ptr ++ section, the value of y changes according to the address indicated by ptr in y = ++ * ptr section.

OriginalGriff 30-Jun-20 12:29pm

Close - the value of what ptr is pointing to gets changed, and that affects what gets written into y.

goksurahsan 30-Jun-20 12:45pm

Yes, I understand now. Thank you again and again. I learned new thing today :)

OriginalGriff 30-Jun-20 14:36pm

You're welcome!