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passing pointers into functions

Started by Psycho_68 Jul 17, 2003 at 8:33 AM 8 replies 400+ views

Original Post

Jul 17, 2003 at 8:33 AM

hi, ive not been using c++ for a quite long time now and i feel quite stupid about this, ... ,but here goes: im trying to pass a pointer to a function so when i change it in the function it changes also the value in the rest of the program, when i try this: #include <iostream.h> void func(int* pass) { int bla = 15; int* ho = &bla cout << *pass << "\n"; pass = ho; cout << *pass << "\n"; } int main(int argc, char* argv[]) { int aaa = 888; int* ptr = &aaa cout << *ptr << "\n"; func(ptr); cout << *ptr << "\n"; return 0; } this dont work, a new pointer is created with a new adress within the function and it doesnt change the value outside the function, i need someway to pass the pointer to the function, so a new one is not created, or is created with the same adress as the one outsuide the function, so i can change the value outside the function within the function ive also come up with this code, which works, but uses references, i would like to get it to work without them void func(int& pass) { int bla = 15; int* ho = &bla cout << pass << "\n"; pass = *ho; cout << pass << "\n"; } int main(int argc, char* argv[]) { int aaa = 888; int* ptr = &aaa cout << *ptr << "\n"; func(*ptr); cout << *ptr << "\n"; return 0; }

Koen

119

Koen

Jul 17, 2003 at 8:39 AM

Try this:


void func(int* pass)
{
int bla = 15;
cout << *pass << "\n";
*pass = bla;
cout << *pass << "\n";
}

You were assigning the adress-value, while you wanted to assign the actual integer value. Therefore, you need to dereference the pointer (*pass, which indicates the value pass points to).

Yanroy

Jul 17, 2003 at 8:42 AM

If you''re trying to change the value that is being pointed to, then you need to difference the pointer before assigning to it:

void func(int* pass)
{
int bla = 15;
//int* ho = &bla
cout << *pass << "\n";
*pass = bla;
cout << *pass << "\n";
}

But that''s probably not what you''re trying to do... my guess is you want to change the address being pointed to. In which case, you need a pointer to a pointer

void func(int** pass)
{
int bla = 15;
int* ho = &bla
cout << *pass << "\n";
*pass = ho;
cout << *pass << "\n";
}

Now when the function returns to main(), ptr will be equal to the address of bla. You''ll have to call func() like this in main(): func(&ptr);

NOTE: this will probably crash if you try to use the pointer, because bla is created locally in func() and thus goes out of scope when func() returns.

--------------------

You are not a real programmer until you end all your sentences with semicolons; (c) 2000 ROAD Programming

You are unique. Just like everybody else.

"Mechanical engineers design weapons; civil engineers design targets."
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Psycho_68

Author

Psycho_68

Jul 17, 2003 at 9:05 AM

thx for the help all
this think doesnt work:

#include <iostream.h>
void func(int* pass)
{
int bla = 15;
int* ho = &bla
cout << *pass << "\n";
pass = ho;
cout << *pass << "\n";
}

int main(int argc, char* argv[])
{
int aaa = 888;
int* ptr = &aaa
cout << *ptr << "\n";
func(&ptr);
cout << *ptr << "\n";

return 0;
}

it gives this error:

C:\Windows\Plocha\lam\lam.cpp(19) : error C2664: ''func'' : cannot convert parameter 1 from ''int ** '' to ''int *''
Types pointed to are unrelated; conversion requires reinterpret_cast, C-style cast or function-style cast
Error executing cl.exe.

lam.exe - 1 error(s), 0 warning(s)

I really wanna change the adress so this dont help>
void func(int* pass){int bla = 15;cout << *pass << "\n";*pass = bla;cout << *pass << "\n";}

thx for the help anyway.

so i tryed this from Yanroy>

void func(int** pass)
{
int bla = 15;
int* ho = &bla
cout << *pass << "\n";
*pass = ho;
cout << *pass << "\n";
}

it works, somehow, the first cout prits the number, but the other 2 print a adress and the last one prints a weird number

think i need to work on this a little more, btw i never heard of pointers to pointers, i think i may stick with references, heh

Psycho_68

Author

Psycho_68

Jul 17, 2003 at 9:07 AM

uhh anybody got some more ideas, other than those references, but those pointers to pointers look even more evil heh

Trienco

146

Trienco

Jul 17, 2003 at 10:16 AM

i get the feeling you might need to get a better understanding of pointers and how they work.

what is this ho all about? it is completely pointless, never used and passing something over 20 useless variables wont change a thing. so throw it out.

void func(int** pass)
{
int bla = 15;
cout << *pass << "\n";
*pass = &bla
cout << *pass << "\n";
}

now you output *pass but seem to expect to get a number. how? pass is a pointer to a pointer to a int and dereferencing it just once will still be a pointer, not a number.

cout << **pass << "\n";

then you would still have a pointer to an address thats invalid as soon as you leave the function. so:
a) bla wont exist anymore
b) your pointer to the address of bla wont make any sense anymore
c) trying to dereference *pass will most likely crash

if this is just for testing inside func its ok, but you should _never_ set a pointer from "outside" to a temporary inside the function without expecting access violations if you ever try to use the pointer outside the function.

Yanroy

Jul 17, 2003 at 11:23 AM

Heh... I screwed up... in the code I posted, the cout << *pass should have been cout << **pass

--------------------

You are not a real programmer until you end all your sentences with semicolons; (c) 2000 ROAD Programming

You are unique. Just like everybody else.

"Mechanical engineers design weapons; civil engineers design targets."
"Sensitivity is adjustable, so you can set it to detect elephants and other small creatures." -- Product Description for a vibration sensor

Yanroy@usa.com

Psycho_68

Author

Psycho_68

Jul 17, 2003 at 2:24 PM

quote:
Original post by Trienco
i get the feeling you might need to get a better understanding of pointers and how they work.

we were dooing Java at school and, I kinda forgotten what c++ is like, i think i get the picture now, only the pointer to poiter thing surprized me

quote:
Original post by Trienco
if this is just for testing inside func its ok, but you should _never_ set a pointer from "outside" to a temporary inside the function without expecting access violations if you ever try to use the pointer outside the function.

I get the picture so this is wrong, cause the pointer gets asigned an adress which goes out of scope as soon as the function quits: ??

void func(int** pass)
{
int bla = 15;
cout << *pass << "\n";
*pass = &bla
cout << *pass << "\n";
}

I think Ill use references then, im kinda getting to like them lately.

[edited by - Psycho_68 on July 17, 2003 3:27:51 PM]

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