Programming Forums > Application Development > C++

aloksave · Feb 5th, 2006, 8:55 AM

why does this program give me a runtime error

#include<iostream.h>

union deriv1
{
	int *ptr;
	int arr[4];
}deriv1;


struct deriv2
{
	int inte;
}deriv2;


struct base
{
	union deriv1 *basep;
	struct deriv2 base2;
	
}basicBase;




int main()
{
	int a = 5;
	struct base *intPtr = new struct base;
	(intPtr->base2.inte) = 5;
	*(intPtr->basep->ptr) = a;
	cout << "data"<<*(intPtr->basep->ptr);
	cout << "\ndata"<<(intPtr->base2.inte);

	delete intPtr;

	return 1;
}

can anyone help me out??

DaWei · Feb 5th, 2006, 9:03 AM

Quote:

*(intPtr->basep->ptr) = a;

I'm getting old and blind, but I don't see where you're ever establishing a value for ptr.

aloksave · Feb 5th, 2006, 9:10 AM

isnt it that
(intPtr->basep->ptr) = (address of ptr)
and *(intPtr->basep->ptr) will write the value at tht location ??

nnxion · Feb 5th, 2006, 9:40 AM

No it will write to somewhere outside of your program, which will cause a run time error.
You need to either allocate storage for it, so assign it.

aloksave · Feb 5th, 2006, 9:48 AM

sorry but i didnt get it....why will it write beyond my program....??
and how do i avoid it??

(Toggle Plain Text)

struct deriv2
{
	int inte;
	int *ptr;
}deriv2;


int main()
{
        struct deriv2 *obj = new struct deriv2;

	(obj->ptr) = &a;

	cout << "data"<<*(obj->ptr);
}

works fine so why not tht???

DaWei · Feb 5th, 2006, 9:55 AM

*(intPtr->basep->ptr) = a; assigns the value of a to the location pointed to by intPtr->basep->ptr; you have not established valid contents for that location yet. The contents there are either random trash or some value established by your compiler that is guaranteed to point to no useful area. Either will generate a run-time error on systems that incorporate memory protection (most desktop OSes, these days). You might want to refer to the "Pointer Basics" material referenced in my signature.

aloksave · Feb 5th, 2006, 10:00 AM

why doesnt this work then..

(Toggle Plain Text)

(intPtr->basep->ptr) = &a;
   cout << "data"<<*(intPtr->basep->ptr);

now i am making the location point to adress of a isnt it?
still it wont work??

nnxion · Feb 5th, 2006, 10:20 AM

You _really_ need to read the Pointer Basics that's in DaWei's signature first.

DaWei · Feb 5th, 2006, 10:27 AM

(intPtr->basep->ptr) = &a; You're missing the point. You've changed nothing regarding the pointer, which is unitialized still, you're merely changing the value contained in a to the address of the variable a. THIS IS WHAT'S BEING ASSIGNED; IT IS NOT THE POINTER, BUT WHAT YOU ARE TRYING TO STORE WHERE THE POINTER POINTS. The pointer, intPtr->basep->ptr, still contains an invalid address. If you don't want to read the material, or find some other material to read, then you need to pay attention and engage your thinking process, at the very least.

aloksave · Feb 5th, 2006, 10:32 AM

Ok fine sorry for the trouble, i would read it and will come back again if i dont get it..
anyways thanks for ur help.

Feb 5th, 2006, 8:55 AM	#1
aloksave Programmer Join Date: Sep 2005 Posts: 33 Rep Power: 0	A run time error why does this program give me a runtime error (Toggle Plain Text) #include<iostream.h> union deriv1 { int ptr; int arr[4]; }deriv1; struct deriv2 { int inte; }deriv2; struct base { union deriv1 basep; struct deriv2 base2; }basicBase; int main() { int a = 5; struct base intPtr = new struct base; (intPtr->base2.inte) = 5; (intPtr->basep->ptr) = a; cout << "data"<<(intPtr->basep->ptr); cout << "\ndata"<<(intPtr->base2.inte); delete intPtr; return 1; } #include<iostream.h> union deriv1 { int ptr; int arr[4]; }deriv1; struct deriv2 { int inte; }deriv2; struct base { union deriv1 basep; struct deriv2 base2; }basicBase; int main() { int a = 5; struct base intPtr = new struct base; (intPtr->base2.inte) = 5; (intPtr->basep->ptr) = a; cout << "data"<<(intPtr->basep->ptr); cout << "\ndata"<<(intPtr->base2.inte); delete intPtr; return 1; } can anyone help me out??

Feb 5th, 2006, 9:40 AM	#4
nnxion Programming Guru Join Date: Jun 2005 Location: elemental plane Posts: 1,429 Rep Power: 5	No it will write to somewhere outside of your program, which will cause a run time error. You need to either allocate storage for it, so assign it. __________________ "Employ your time in improving yourself by other men's writings, so that you shall gain easily what others have labored hard for." -- Socrates

Feb 5th, 2006, 9:48 AM	#5
aloksave Programmer Join Date: Sep 2005 Posts: 33 Rep Power: 0	sorry but i didnt get it....why will it write beyond my program....?? and how do i avoid it?? (Toggle Plain Text) struct deriv2 { int inte; int ptr; }deriv2; int main() { struct deriv2 obj = new struct deriv2; (obj->ptr) = &a; cout << "data"<<(obj->ptr); } struct deriv2 { int inte; int ptr; }deriv2; int main() { struct deriv2 obj = new struct deriv2; (obj->ptr) = &a; cout << "data"<<(obj->ptr); } works fine so why not tht???

Feb 5th, 2006, 9:55 AM	#6
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	(intPtr->basep->ptr) = a; assigns the value of a to the location pointed to by intPtr->basep->ptr; you have not established valid contents for that location yet. The contents there are either random trash or some value established by your compiler that is guaranteed to point to no useful area. Either will generate a run-time error on systems that incorporate memory protection (most desktop OSes, these days). You might want to refer to the "Pointer Basics" material referenced in my signature. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers*

Feb 5th, 2006, 10:00 AM	#7
aloksave Programmer Join Date: Sep 2005 Posts: 33 Rep Power: 0	why doesnt this work then.. (Toggle Plain Text) (intPtr->basep->ptr) = &a; cout << "data"<<(intPtr->basep->ptr); (intPtr->basep->ptr) = &a; cout << "data"<<(intPtr->basep->ptr); now i am making the location point to adress of a isnt it? still it wont work??

Feb 5th, 2006, 9:10 AM	#3
aloksave Programmer Join Date: Sep 2005 Posts: 33 Rep Power: 0	isnt it that (intPtr->basep->ptr) = (address of ptr) and *(intPtr->basep->ptr) will write the value at tht location ??

Feb 5th, 2006, 10:20 AM	#8
nnxion Programming Guru Join Date: Jun 2005 Location: elemental plane Posts: 1,429 Rep Power: 5	You _really_ need to read the Pointer Basics that's in DaWei's signature first. __________________ "Employ your time in improving yourself by other men's writings, so that you shall gain easily what others have labored hard for." -- Socrates

Feb 5th, 2006, 10:27 AM	#9
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	(intPtr->basep->ptr) = &a; You're missing the point. You've changed nothing regarding the pointer, which is unitialized still, you're merely changing the value contained in a to the address of the variable a. THIS IS WHAT'S BEING ASSIGNED; IT IS NOT THE POINTER, BUT WHAT YOU ARE TRYING TO STORE WHERE THE POINTER POINTS. The pointer, intPtr->basep->ptr, still contains an invalid address. If you don't want to read the material, or find some other material to read, then you need to pay attention and engage your thinking process, at the very least. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Feb 5th, 2006, 10:32 AM	#10
aloksave Programmer Join Date: Sep 2005 Posts: 33 Rep Power: 0	Ok fine sorry for the trouble, i would read it and will come back again if i dont get it.. anyways thanks for ur help.

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Programming Forums > Application Development > C++

A run time error