void ** ?? sizeof(void*)?? what is this?

nacho4d · 10-02-2009, 06:04 AM

Hi,
I am trying to use a program that has the a variable of type void** but I don't understand well what is meaning of this.
Does any knows what is the meaning of the following code?
I specifically would like to know how many bytes are allocated in pdata and data.(and the reason)
Al least one of them is data of pixels (since imgtmp is an image structure)
imgtmp->pdata is void** type and
imgtmp->data is void* type,
I replaced some vars for constants as you can see.

This a fragment of code that creates an image structure.

Code:

        void		**pdata;
	if ( NULL == (pdata = (void **)malloc(sizeof(void *) * 4)) ){ //4 is the image height
		//free data, function failed
		return();
	}
	imgtmp->pdata  = pdata;
	if ( NULL == (imgtmp->data = (char *)malloc(12)) ){ //12 is image hole buffer size
		//free data, function failed
		return();
	}
	for (int i=0 ; i<4 ; i++ ) //4 is the image height
		pdata[i] = (void *)(imgtmp->data + i*3); //3 is bytes per line in the image

As you can see I am trying to allocate a 3*4 image.

BTW: I tried sizeof(void *) and it gives me a long unsigned 8, so pdata should be 32, but is always 8 , why?

anon · 10-02-2009, 06:13 AM

sizeof tells you the size of a type (void* or void**), it couldn't possibly tell you how much memory you allocated with malloc (sizeof is evaluated by the compiler, not at runtime).

Your only way to know how much memory has been allocated for something, is to store that in a variable.

Epy · 10-02-2009, 06:14 AM

All I know is, you're not supposed to do this:
(void **)malloc(sizeof(void *) * 4)
or this:
(char *)malloc(12)

Epy · 10-02-2009, 06:21 AM

Why would void return a nonzero size anyway?

C_ntua · 10-02-2009, 06:28 AM

sizeof(void*), sizeof(void**), sizeof(any_pointer) is most likely the same size. Pointers are pointers they just store addresses. So no suprise they are both 8 bytes

imgtmp->data seems to be char* not void*

void** is a pointer that points to a pointer of type void*.

Code explanation. Lets say a pointer is 8bytes
You allocate 4 x void* pointers of space. So 32 bytes of space.
You assign that to pdata. Pdata is still 8 bytes but it points to a memory that is 32 bytes.

Now imgtmp->pdata = pdata means that imgtmp->pdata points to that memory as well.
imgtmp->data points to a memory worth of 12 bytes.

Now, since imgtmp->pdata is type of void** it is can be seen as an array of void* pointers. You have 4 such pointers. The first will point on the first byte of the 12 bytes that imgtmp->data points, the second on the 4th, the third on the 8th and the fourth on the 12th byte. That way you can use imgtmp->pdata with indexes to get whichever byte you mean. Like imgtmp->pdata[1][2] will give you 2nd row, 3rd column byte.

nacho4d · 10-02-2009, 06:31 AM

Quote:

Originally Posted by Epy

All I know is, you're not supposed to do this:
(void **)malloc(sizeof(void *) * 4)
or this:
(char *)malloc(12)

even though i got normal resulta when running sample data with this

Epy · 10-02-2009, 06:32 AM

gotcha, I was thinking about plain void instead of a pointer, my mistake.

Epy · 10-02-2009, 06:33 AM

Quote:

Originally Posted by nacho4d

even though i got normal resulta when running sample data with this

I'm sure.

FAQ: Casting malloc?

grumpy · 10-02-2009, 06:45 AM

Quote:

Originally Posted by nacho4d

even though i got normal resulta when running sample data with this

You will. The conversions you're doing are legal C, but people generally consider them to be a bad idea as they can obscure some programming errors.

Note that, if your compiler is a C++ compiler, the conversions are necessary - even if your code is C - because C++ does not allow an implicit conversion from (void *) to other pointer types.

nacho4d · 10-02-2009, 07:22 AM

Thank you very much C_nua for your kind explanation
i think I got the idea, thanks again.

...

This board is great since I register like 3 years ago!. Is the best for c questions

10-02-2009, 06:04 AM	#1
nacho4d Registered User Join Date: Nov 2006 Location: japan Posts: 104	void ?? sizeof(void)?? what is this?* Hi, I am trying to use a program that has the a variable of type void but I don't understand well what is meaning of this. Does any knows what is the meaning of the following code? I specifically would like to know how many bytes are allocated in pdata and data.(and the reason) Al least one of them is data of pixels (since imgtmp is an image structure) imgtmp->pdata is void type and imgtmp->data is void* type, I replaced some vars for constants as you can see. This a fragment of code that creates an image structure. Code: void pdata; if ( NULL == (pdata = (void )malloc(sizeof(void ) 4)) ){ //4 is the image height //free data, function failed return(); } imgtmp->pdata = pdata; if ( NULL == (imgtmp->data = (char )malloc(12)) ){ //12 is image hole buffer size //free data, function failed return(); } for (int i=0 ; i<4 ; i++ ) //4 is the image height pdata[i] = (void )(imgtmp->data + i3); //3 is bytes per line in the image As you can see I am trying to allocate a 34 image. BTW: I tried sizeof(void *) and it gives me a long unsigned 8, so pdata should be 32, but is always 8 , why? __________________ Mac OS 10.6 Snow Leopard : Darwin
nacho4d is offline

10-02-2009, 06:14 AM	#3
Epy I like turtles Join Date: Sep 2009 Location: Ohio Posts: 179	All I know is, you're not supposed to do this: (void )*malloc(sizeof(void ) * 4) or this: *(char )**malloc(12) __________________ Jake, that CAD guy Hazudra Fodder
Epy is offline

10-02-2009, 06:21 AM	#4
Epy I like turtles Join Date: Sep 2009 Location: Ohio Posts: 179	Why would void return a nonzero size anyway? __________________ Jake, that CAD guy Hazudra Fodder
Epy is offline

10-02-2009, 06:28 AM	#5
C_ntua Registered User Join Date: Jun 2008 Posts: 1,134	sizeof(void), sizeof(void), sizeof(any_pointer) is most likely the same size. Pointers are pointers they just store addresses. So no suprise they are both 8 bytes imgtmp->data seems to be char not void* void** is a pointer that points to a pointer of type void. Code explanation. Lets say a pointer is 8bytes You allocate 4 x void pointers of space. So 32 bytes of space. You assign that to pdata. Pdata is still 8 bytes but it points to a memory that is 32 bytes. Now imgtmp->pdata = pdata means that imgtmp->pdata points to that memory as well. imgtmp->data points to a memory worth of 12 bytes. Now, since imgtmp->pdata is type of void** it is can be seen as an array of void* pointers. You have 4 such pointers. The first will point on the first byte of the 12 bytes that imgtmp->data points, the second on the 4th, the third on the 8th and the fourth on the 12th byte. That way you can use imgtmp->pdata with indexes to get whichever byte you mean. Like imgtmp->pdata[1][2] will give you 2nd row, 3rd column byte.
C_ntua is offline

10-02-2009, 06:32 AM	#7
Epy I like turtles Join Date: Sep 2009 Location: Ohio Posts: 179	gotcha, I was thinking about plain void instead of a pointer, my mistake. __________________ Jake, that CAD guy Hazudra Fodder
Epy is offline

10-02-2009, 07:22 AM	#10
nacho4d Registered User Join Date: Nov 2006 Location: japan Posts: 104	Thanks C_nua Thank you very much C_nua for your kind explanation i think I got the idea, thanks again. ... This board is great since I register like 3 years ago!. Is the best for c questions __________________ Mac OS 10.6 Snow Leopard : Darwin
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