C++ Pointers exam revision

Question

Hi all I have an exam that involves c++ coming up and I am just struggling on pointers a little, can anyone assist,

this is an example question

What will be printed out on execution and p starts with address 0x00C0

float p = 11.0; 
p = p + 2.0;

float *q = &p; 
float *r = q; 
*r = *r + 1; 

int *s = new int(); 
*s = *q; 
*q = (*q)*10.0; 

*r = 15.0;

cout << p <<endl;
cout << *q <<endl;
cout << r <<endl;
cout << *s <<endl;
cout << *r <<endl;

now I compiled and ran this program but I cant get my head around the values of *q which = 15. Does it not get multiplied by 10?

Also r is an address in memory can anyone explain that to me please?

Help appriciated!

`r` and `q` point to the same thing, so `*r = 15.0;` changes the value of `*q` as well. — user123, Jan 19 '14 at 14:19
There is a difference between `P` and `p`. You mentioned that `P` starts with `0x00C0` but is is not used anywhere in the code. — unxnut, Jan 19 '14 at 14:21
Ah yes Mohammad thank you i completly missed that! Also unxnut that was a typo! — user2904478, Jan 19 '14 at 14:23
Do it like you do math - substitute variables - eg q with (&p)- you may gain a simplification — , Jan 19 '14 at 14:29
`Also r is an address in memory can anyone explain that to me please?` Time to read [your C++ book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) cover-to-cover. — Lightness Races in Orbit, Jan 19 '14 at 14:31

Cool_Coder · Accepted Answer · 2014-01-19T14:29:42.430

1

Always try to think in terms of variable value instead of pointer value. If multiple pointers point to the same memory location then printing the value of the pointer (*ptr) will always give same output for all pointers.

float p = 11.0; 
p = p + 2.0;//p = 13

float *q = &p; 
float *r = q; 
*r = *r + 1;//p = 14 

int *s = new int(); 
*s = *q;//*s = 14 
*q = (*q)*10.0;//p = 140 

*r = 15.0;//p = 15//somehow did not see this line :P

cout << p <<endl;//15
cout << *q <<endl;//15
cout << r <<endl;//0x00C0
cout << *s <<endl;//14
cout << *r <<endl;//15

Proof here.

edited Jan 19 '14 at 14:29

answered Jan 19 '14 at 14:21

Cool_Coder

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hint: float* r = q; Do something with it. – David Szalai Jan 19 '14 at 14:24
I mean at the end of the code *r=15 also changes q, since points to the same as q. – David Szalai Jan 19 '14 at 14:29
1

Thank you all for your help! also thank you Cool_Coder for stepping down line by line this helps alot – user2904478 Jan 19 '14 at 14:32
@unxnut Correct me, if I am wrong, but doesn't `float *r = q` initialize the pointer `r` with the pointer `q` (which in turn is initialized with `&p`, wherefore `r == &p` and not `*r == &p`, which in turn means that `*r = *r + 1` is the same as `p = p + 1` (so `*r` changes from `13` to `14`) not `*r = &p +1`? – Lmis Jan 19 '14 at 14:49
float *r = q; makes r point at the same memory location as q. Think it like this. So it doesnt make a difference if you use *r or *q because both point to the same memory location. – Cool_Coder Jan 19 '14 at 14:52
@Cool_Coder Yes ofc, my issue is with the comment by unxnut which includes (what I understand to be) the claim, that at some point `*r == &p`, whereas I claim `*r == p` and `r == &p` – Lmis Jan 19 '14 at 14:57
@Lmis, my bad. You are right that `float *r = q` assigns the address of `q` to `r`. – unxnut Jan 19 '14 at 15:08

C++ Pointers exam revision

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