Programming Forums > Application Development > C++

BaroN NighT · Apr 25th, 2005, 2:06 PM

Reference, address. Same thing.

Eggbert · Apr 25th, 2005, 3:38 PM

>Reference, address. Same thing.
It depends on what you are talking about. In theory, pointers, addresses, and references are all basically the same concept. In C++, a pointer is an address, but a reference has different semantics. Saying that they are the same thing would only confuse the issue more.

>But the & sign in addition to standing for an address, can also be a reference.
Or even bitwise AND.

brokenhope · Apr 25th, 2005, 7:13 PM

Hmm I think im still having trouble understanding

Is & and * the same thing?

So when you use * it defines it to the address of a variable?

So why do you use it when you declare a variable? Wouldnt you only do it when you define a variables name... like whats the point of doing this char* bloh; ? What does it do diffrently than char bloh; ?

And this is the last question I have, I think it doesnt work but I dont know...

char one = "One";
char two = *one;
char one = "Two";
cout << two;

Would that ouput two? or nothing? or how could I make it output two without drasticaly changing the code? I really dont get this... I thought I did but now I dont...

bl00dninja · Apr 25th, 2005, 7:31 PM

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int * px;  <--declares a pointer of type int (points to an int)

x=1;  <--  just a regular variable

px = &x;  <--assigns the address of x to px's value

now px is assigned the address of x. if you were to

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cout<<px<<endl

it would print the same thing as if you were to

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cout<<&x<<endl;

it would print the address of x.

if you

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cout<<*px<<endl;

it print's the VALUE that px POINTS to, which is x which in this case is 1.

you use the * operator to declare a pointer. that's why a lot of people like to put it next to the type like this:

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int* px;

px is the pointer, it is the address of x.

*px is the value pointed to.

&x is the address of x.

the * is called the dereference operator because when you use it on a pointer you "dereference" the pointer which means to get the value of what it is referencing. the & and * symbols are unfortunately used in some different ways which can lead to confusion. as to your pseudo-code

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char one = "One";
char two = *one;
char one = "Two";
cout << two;

here is what you want

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char one = 'o';
char * two = &one;
one = 't';  //  <--  or *two = 't';  either way
cout<<*two<<endl;

i didn't declare the whole words because you can't do it like you did you have to use an array like

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char one[] = "One";

and this will severely complicate this issue for you. i suggest you play with ints or something else right now for messing around with pointers.

brokenhope · Apr 25th, 2005, 7:38 PM

hmm I think im starting to understand, so what happens in an example where y points to x, and the x variable changes several times, does the y variable change with the x variable? Like

int* y;
int x = 4;
cout << y << endl;
y = &x;
x = x*2;
cout << y << endl;

Would that output:
4
8

or what?

Ooble · Apr 25th, 2005, 7:49 PM

You need to output *y instead of y. Plain ol' vanilla "y" is the address of the variable x; "*y" points to the value of it.

brokenhope · Apr 25th, 2005, 7:53 PM

Ah, now I understand.. when I compile it and try to run it theres nothing on the screen, and a warning error coming up saying that the program encountered a problem and needs to close comes up...

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#include <iostream.h>
#include <stdlib.h>

using namespace std;

int main()
{
      int* y;
      int x = 4;
      cout << *y << endl;
      y = &x;
      x = x*2;
      cout << *y << endl;
      system("PAUSE");
      return 0;
}

bl00dninja · Apr 25th, 2005, 7:58 PM

(Toggle Plain Text)

#include <iostream.h>
#include <stdlib.h>

using namespace std;

int main()
{
      int* y;
      int x = 4;
      cout << *y << endl; //<--y isn't pointing to anything yet...stray pointer!
      y = &x;                  //<--  move this line up one and you should be good
      x = x*2;
      cout << *y << endl;
      system("PAUSE");
      return 0;
}

now corrected

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#include <iostream.h>
#include <stdlib.h>

using namespace std;

int main()
{
      int* y;
      int x = 4;
      y = &x;
      cout << *y << endl;
      x = x*2;
      cout << *y << endl;
      system("PAUSE");
      return 0;
}

tada!!! wasn't that easy?

brokenhope · Apr 25th, 2005, 8:03 PM

Oh that was just a mistake of not paying attention to where I added the *, thanks a lot I think I finally understand this.

bl00dninja · Apr 29th, 2005, 2:37 AM

feel free to ask more questions in the future.

good luck!

Apr 25th, 2005, 7:13 PM	#13
brokenhope Hobbyist Programmer Join Date: Apr 2005 Posts: 126 Rep Power: 4	Hmm I think im still having trouble understanding Is & and * the same thing? So when you use * it defines it to the address of a variable? So why do you use it when you declare a variable? Wouldnt you only do it when you define a variables name... like whats the point of doing this char* bloh; ? What does it do diffrently than char bloh; ? And this is the last question I have, I think it doesnt work but I dont know... char one = "One"; char two = one; char one = "Two"; cout << two; Would that ouput two? or nothing? or how could I make it output two without drasticaly changing the code? I really dont get this... I thought I did but now I dont... Last edited by brokenhope; Apr 25th, 2005 at 7:15 PM.*

Apr 25th, 2005, 7:31 PM	#14
bl00dninja Programming Guru Join Date: Oct 2004 Location: namespace std Posts: 1,246 Rep Power: 6	(Toggle Plain Text) int * px; <--declares a pointer of type int (points to an int) x=1; <-- just a regular variable px = &x; <--assigns the address of x to px's value int * px; <--declares a pointer of type int (points to an int) x=1; <-- just a regular variable px = &x; <--assigns the address of x to px's value now px is assigned the address of x. if you were to (Toggle Plain Text) cout<<px<<endl cout<<px<<endl it would print the same thing as if you were to (Toggle Plain Text) cout<<&x<<endl; cout<<&x<<endl; it would print the address of x. if you (Toggle Plain Text) cout<<px<<endl; cout<<px<<endl; it print's the VALUE that px POINTS to, which is x which in this case is 1. you use the * operator to declare a pointer. that's why a lot of people like to put it next to the type like this: (Toggle Plain Text) int* px; int* px; px is the pointer, it is the address of x. px is the value pointed to. &x is the address of x. the is called the dereference operator because when you use it on a pointer you "dereference" the pointer which means to get the value of what it is referencing. the & and * symbols are unfortunately used in some different ways which can lead to confusion. as to your pseudo-code (Toggle Plain Text) char one = "One"; char two = one; char one = "Two"; cout << two; char one = "One"; char two = one; char one = "Two"; cout << two; here is what you want (Toggle Plain Text) char one = 'o'; char * two = &one; one = 't'; // <-- or two = 't'; either way cout<<two<<endl; char one = 'o'; char * two = &one; one = 't'; // <-- or two = 't'; either way cout<<two<<endl; i didn't declare the whole words because you can't do it like you did you have to use an array like (Toggle Plain Text) char one[] = "One"; char one[] = "One"; and this will severely complicate this issue for you. i suggest you play with ints or something else right now for messing around with pointers. __________________ i put on my robe and wizard hat... Have you ever heard of Plato, Aristotle, Socrates?...Morons. Last edited by bl00dninja; Apr 25th, 2005 at 7:39 PM.

Apr 25th, 2005, 7:49 PM	#16
Ooble I eat cake for breakfast. Join Date: Jul 2004 Location: In my box. Posts: 4,434 Rep Power: 9	You need to output y instead of y. Plain ol' vanilla "y" is the address of the variable x; "y" points to the value of it. __________________ Me :: You :: Them

Apr 25th, 2005, 7:53 PM	#17
brokenhope Hobbyist Programmer Join Date: Apr 2005 Posts: 126 Rep Power: 4	Ah, now I understand.. when I compile it and try to run it theres nothing on the screen, and a warning error coming up saying that the program encountered a problem and needs to close comes up... (Toggle Plain Text) #include <iostream.h> #include <stdlib.h> using namespace std; int main() { int* y; int x = 4; cout << y << endl; y = &x; x = x2; cout << y << endl; system("PAUSE"); return 0; } #include <iostream.h> #include <stdlib.h> using namespace std; int main() { int y; int x = 4; cout << y << endl; y = &x; x = x2; cout << *y << endl; system("PAUSE"); return 0; }

Apr 25th, 2005, 7:58 PM	#18
bl00dninja Programming Guru Join Date: Oct 2004 Location: namespace std Posts: 1,246 Rep Power: 6	(Toggle Plain Text) #include <iostream.h> #include <stdlib.h> using namespace std; int main() { int* y; int x = 4; cout << y << endl; //<--y isn't pointing to anything yet...stray pointer! y = &x; //<-- move this line up one and you should be good x = x2; cout << y << endl; system("PAUSE"); return 0; } #include <iostream.h> #include <stdlib.h> using namespace std; int main() { int y; int x = 4; cout << y << endl; //<--y isn't pointing to anything yet...stray pointer! y = &x; //<-- move this line up one and you should be good x = x2; cout << y << endl; system("PAUSE"); return 0; } now corrected (Toggle Plain Text) #include <iostream.h> #include <stdlib.h> using namespace std; int main() { int y; int x = 4; y = &x; cout << y << endl; x = x2; cout << y << endl; system("PAUSE"); return 0; } #include <iostream.h> #include <stdlib.h> using namespace std; int main() { int y; int x = 4; y = &x; cout << y << endl; x = x2; cout << *y << endl; system("PAUSE"); return 0; } tada!!! wasn't that easy? __________________ i put on my robe and wizard hat... Have you ever heard of Plato, Aristotle, Socrates?...Morons.

Apr 25th, 2005, 2:06 PM	#11
BaroN NighT Programmer Join Date: Mar 2005 Location: USA Posts: 60 Rep Power: 4	Reference, address. Same thing.

Apr 25th, 2005, 3:38 PM	#12
Eggbert Professional Programmer Join Date: Nov 2004 Posts: 250 Rep Power: 5	>Reference, address. Same thing. It depends on what you are talking about. In theory, pointers, addresses, and references are all basically the same concept. In C++, a pointer is an address, but a reference has different semantics. Saying that they are the same thing would only confuse the issue more. >But the & sign in addition to standing for an address, can also be a reference. Or even bitwise AND.

Apr 25th, 2005, 7:38 PM	#15
brokenhope Hobbyist Programmer Join Date: Apr 2005 Posts: 126 Rep Power: 4	hmm I think im starting to understand, so what happens in an example where y points to x, and the x variable changes several times, does the y variable change with the x variable? Like int* y; int x = 4; cout << y << endl; y = &x; x = x*2; cout << y << endl; Would that output: 4 8 or what?

Apr 25th, 2005, 8:03 PM	#19
brokenhope Hobbyist Programmer Join Date: Apr 2005 Posts: 126 Rep Power: 4	Oh that was just a mistake of not paying attention to where I added the *, thanks a lot I think I finally understand this.

Apr 29th, 2005, 2:37 AM	#20
bl00dninja Programming Guru Join Date: Oct 2004 Location: namespace std Posts: 1,246 Rep Power: 6	feel free to ask more questions in the future. good luck! __________________ i put on my robe and wizard hat... Have you ever heard of Plato, Aristotle, Socrates?...Morons.

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Programming Forums > Application Development > C++

Question about Pointers