C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  size of class          C++Talk.NET Forum Index -> C++ Language (Moderated) View previous topic :: View next topic   Author Message Keshav Guest Posted: Wed Nov 09, 2005 2:22 pm    Post subject: size of class Hi, code snnipet is class A { char a; }; class B { char b; int a; }; class C { char c; int d; char e; }; int main() { cout << sizeof(A) < return 0; } Output is 1 8 12 My question is why size of class A is 1. And if it is 1 then why size of class C is 12. Compiler does padding but why there is different behaviour for class A and C. either it should be 4 8 12 or 1 8 9? M i wrong somewhere or it is compiler dependent? I m using Visual C++ ..net Thanks keshav [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Krakers Guest Posted: Thu Nov 10, 2005 1:18 pm    Post subject: Re: size of class Keshav wrote... Quote: class A { char a; }; class B { char b; int a; }; class C { char c; int d; char e; }; int main() { cout << sizeof(A) < Output is 1 8 12. My question is why If the hardware architecture requires 32-bit integers to be aligned at addresses divisible by 4 (which often is the case), then, that is the only way to guarantee that arrays of these classes can still be accessed. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top John Potter Guest Posted: Thu Nov 10, 2005 1:19 pm    Post subject: Re: size of class On 9 Nov 2005 09:22:55 -0500, "Keshav" <gupta.keshav (AT) gmail (DOT) com> wrote: Quote: class A { char a; }; class B { char b; int a; }; class C { char c; int d; char e; }; int main() { cout << sizeof(A) < return 0; } Output is 1 8 12 It has to do with alignment. sizeof(X) % align(X.m) == 0 always. The reason is arrays. Your implementation has align(int) == sizeof(int) == 4 and align(char) == 1. Sizeof(char) == 1 by definition. Padding between array elements is not allowed. To align the int in B, there must be padding between it and the preceeding char. The same is true for C. To align the second C in an array, the size must be a multiple of the alignment forcing padding either between the int and the following char or after the char or both. Alignment requirements are implementation defined. John [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Greg Herlihy Guest Posted: Thu Nov 10, 2005 1:22 pm    Post subject: Re: size of class Keshav wrote: Quote: Hi, code snnipet is class A { char a; }; class B { char b; int a; }; class C { char c; int d; char e; }; int main() { cout << sizeof(A) < return 0; } Output is 1 8 12 My question is why size of class A is 1. And if it is 1 then why size of class C is 12. Compiler does padding but why there is different behaviour for class A and C. either it should be 4 8 12 or 1 8 9? M i wrong somewhere or it is compiler dependent? I m using Visual C++ .net The sizes of A, B and C would be the same when compiled with any other C++ compiler whose ints had the same size and alignment requirements. The size of a POD struct (or POD class) depends only on the size and the alignment requirements of its member types. To illustrate this point, let's take the case of a compiler whose ints are both four bytes in size and four-byte aligned. We can then calculate the sizes of A, B, and C for such a compiler as follows: Calculating the size of class A is easy, since it has only one member, a char. A char is always one byte in size and has no alignment requirements. Class A therefore will contain no padding. The total size of an A class object, therefore, is one byte. Calculating the sizes of classes B and C is a lttle trickier. First, we know that the compiler will always align a struct or class to meet the most stringent alignment requirements of any of its members. In the case of B and C, the only member type with an alignment requirement of any kind is an int, so objects of class B or C will always be 4-byte aligned in our implementation. Now we just add up the members of the class, observing proper alignment along the way. B and C's first member, c, is a char aligned to a four byte address. So the member that follows c, d, needs three bytes of padding so that it can properly aligned to the next four byte address. Adding up what we have so far: 1 (char) + 3 (padding) + 4 (int) = 8 bytes The size of a class B object is eight bytes. Now, adding in C's one additional member, e, a char brings the total to 9 bytes, not the 12 bytes of its actual size. How do we account for C's additional three bytes? The answer is that the three extra bytes are needed in order to make C large enough so that multiple C objects can be stored in an array contiguously (or more precisely, without any padding between any two elements). If class C were nine bytes in size, the next C object in an array could not be both four-byte aligned (to satisfy the aligment requirements of an int) and also be immediately adjacent to the preceding C object in memory. So in order to meet both its alignment requirements and the array storage requirement, class C must be of a size that will accomodate both. A 12 byte-sized C does so by ensuring both four-byte alignment and contiguous storage for an array of C objects. Lastly, I would also point out that the order that members are declared within a class can affect its size. For example, rearranging C's members like so: class D { char c; char e; int d; }; declares a class that stores the same amount of data as C, but in only two thirds of the space (that is, 8 instead of 12 bytes). Greg [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top kanze Guest Posted: Thu Nov 10, 2005 1:27 pm    Post subject: Re: size of class Keshav wrote: Quote: code snnipet is class A { char a; }; class B { char b; int a; }; class C { char c; int d; char e; }; int main() { cout << sizeof(A) < return 0; } Output is 1 8 12 My question is why size of class A is 1. And if it is 1 then why size of class C is 12. One obvious answer is: why not? The compiler is free to add padding where ever it wants (except at the beginning of a POD struct). In practice, it seems pretty obvious that alignment considerations are coming into play, and that your compiler enforces an alignment of 4 for int (and uses 4 byte ints). So in class B, the offset of a must be 4. Quote: Compiler does padding but why there is different behaviour for class A and C. Because class A doesn't contain an int, and class C does. There's no point in maintaining alignment within a structure if you don't ensure that the size of the structure itself is a multiple of that alignment. Think of what happens with the alignment of the second element if you write C[ 2 ]. Quote: either it should be 4 8 12 or 1 8 9? M i wrong somewhere or it is compiler dependent? I m using Visual C++ .net Well, it's very compiler dependant, but compiler implementors don't generally just make random decisions -- I'd be very surprised if any compiler for an Intel platform (or any platform using 4 byte int's which require an alignment of 4) did otherwise. About the only variation I've seen is that a compiler will insist that all struct's respect a minimum alignment, even when the contents of the struct doesn't require it. But this is really only common on word addressed machines, where the alignment affects pointer size as well. (And as far as I know, there is only one word addressed machine still being sold today.) -- James Kanze GABI Software Conseils en informatique orientée objet/ Beratung in objektorientierter Datenverarbeitung 9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34 [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Antoun Kanawati Guest Posted: Thu Nov 10, 2005 1:32 pm    Post subject: Re: size of class Keshav wrote: Quote: Hi, code snnipet is class A { char a; }; class B { char b; int a; }; class C { char c; int d; char e; }; int main() { cout << sizeof(A) < Output is 1 8 12 My question is why size of class A is 1. And if it is 1 then why size of class C is 12. Compiler does padding but why there is different behaviour for class A and C. either it should be 4 8 12 or 1 8 9? M i wrong somewhere or it is compiler dependent? I m using Visual C++ .net Padding has to do with alignment. To do the arithmetic, assume an array of instances; if you align the first instance correctly, then all the following instances should also be aligned correctly. So, if you have an int in the middle, it may be necessary to pad before AND after it, to make the struct line up nicely. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Francis Glassborow Guest Posted: Thu Nov 10, 2005 1:39 pm    Post subject: Re: size of class In article <1131521503.655390.120080 (AT) f14g2000cwb (DOT) googlegroups.com>, Keshav <gupta.keshav (AT) gmail (DOT) com> writes Quote: class A { char a; }; class B { char b; int a; }; class C { char c; int d; char e; }; int main() { cout << sizeof(A) < return 0; } Output is 1 8 12 My question is why size of class A is 1. And if it is 1 then why size of class C is 12. Compiler does padding but why there is different behaviour for class A and C. either it should be 4 8 12 or 1 8 9? M i wrong somewhere or it is compiler dependent? I m using Visual C++ .net In general the alignment requirements of char types lead to no requirement for padding. However any user defined type must always have a size that is at least a multiple of the member with the strongest alignment requirement. Hence on systems where int is aligned to a multiple of four bytes, any type with an int member must have a size that is a multiple of four. -- Francis Glassborow ACCU Author of 'You Can Do It!' see http://www.spellen.org/youcandoit For project ideas and contributions: http://www.spellen.org/youcandoit/projects [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Victor Bogado Guest Posted: Thu Nov 10, 2005 1:41 pm    Post subject: Re: size of class This is not really a C++ question, since as far as I know the C++ spec don't say how the variables should be layouted in memory nor does he says how much memory should each type is(*). My guess is that the compiler is aligning the int value. The layout of the structure in memory is like that : class a { [a] } = 1 class b { [b] [ ] [ ] [ ] [a] [a] [a] [a] } = 8 class c { [b] [ ] [ ] [ ] [a] [a] [a] [a] [e] } = 9 In my test using the gcc in linux Ubuntu I got the same result, but this is not necessary, for instance if I had the amd 64 processor the resul would be certainly different. This is why is not a good idea to write your objects to the disk as memory dump. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Martin Bonner Guest Posted: Thu Nov 10, 2005 5:01 pm    Post subject: Re: size of class John Potter wrote: Quote: On 9 Nov 2005 09:22:55 -0500, "Keshav" <gupta.keshav (AT) gmail (DOT) com> wrote: class A { char a; }; class B { char b; int a; }; class C { char c; int d; char e; } int main() { cout << sizeof(A) < return 0; } Output is 1 8 12 It has to do with alignment. sizeof(X) % align(X.m) == 0 always. The reason is arrays. Your implementation has align(int) == sizeof(int) == 4 and align(char) == 1. Sizeof(char) == 1 by definition. Padding between array elements is not allowed. To align the int in B, there must be padding between it and the preceeding char. The same is true for C. To align the second C in an array, the size must be a multiple of the alignment forcing padding either between the int and the following char or after the char or both. Alignment requirements are implementation defined. This raises a question that has been discussed in the boost list. Boost has some utility templates that calculate (a multiple of) the alignment of a type, and a type corresponding to a particular alignment. On some platforms this latter template is unable to find a type which corresponds to the result of the former template. The situation could be improved if it could assume that the alignment requirement is a power of two. Clearly the standard does not guarantee that the alignment requirement is a power of two. However, in practise it usually is. My question is then: does anyone know of any architecture (in use, proposed, or defunct) where the natural alignment requirement for a type was not a power of two? [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top James Kanze Guest Posted: Sun Nov 13, 2005 3:42 pm    Post subject: Re: size of class Martin Bonner wrote: [...] Quote: Clearly the standard does not guarantee that the alignment requirement is a power of two. However, in practise it usually is. My question is then: does anyone know of any architecture (in use, proposed, or defunct) where the natural alignment requirement for a type was not a power of two? Unisys (ex-Burroughs) Series A. It's not been made for a couple of years now, but it had 48 bit word size, and I would expect 8 bit bytes. Which would mean an alignment of 6. On the other hand, I think it was a word addressed machine. Does the Boost stuff work for word addressed machines. (If not, there's been a lot of them, and at least one is still being made.) -- James Kanze mailto: [email]james.kanze (AT) free (DOT) fr[/email] Conseils en informatique orientée objet/ Beratung in objektorientierter Datenverarbeitung 9 pl. Pierre Sémard, 78210 St.-Cyr-l'École, France +33 (0)1 30 23 00 34 [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Martin Bonner Guest Posted: Mon Nov 14, 2005 3:35 pm    Post subject: Re: size of class James Kanze wrote: Quote: Martin Bonner wrote: [...] Clearly the standard does not guarantee that the alignment requirement is a power of two. However, in practise it usually is. My question is then: does anyone know of any architecture (in use, proposed, or defunct) where the natural alignment requirement for a type was not a power of two? Unisys (ex-Burroughs) Series A. It's not been made for a couple of years now, but it had 48 bit word size, and I would expect 8 bit bytes. Which would mean an alignment of 6. BOTHER! Quote: On the other hand, I think it was a word addressed machine. Does the Boost stuff work for word addressed machines. (If not, there's been a lot of them, and at least one is still being made.) I think it does. It uses (roughly): alignment = sizeof( struct { char; T } ) - sizeof(T) I would expect that to produce 6 for anything other than T=char on the Series A. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top kanze Guest Posted: Tue Nov 15, 2005 2:32 pm    Post subject: Re: size of class Martin Bonner wrote: Quote: James Kanze wrote: Martin Bonner wrote: [...] Clearly the standard does not guarantee that the alignment requirement is a power of two. However, in practise it usually is. My question is then: does anyone know of any architecture (in use, proposed, or defunct) where the natural alignment requirement for a type was not a power of two? Unisys (ex-Burroughs) Series A. It's not been made for a couple of years now, but it had 48 bit word size, and I would expect 8 bit bytes. Which would mean an alignment of 6. BOTHER! Before getting too upset about it, I'd find out if it was relevant to whatever you're doing. I don't know if there was ever a C++ compiler for it. A priori, the only reason anyone would be writing code for this machine today is when maintaining existing code, and if there was never a C++ compiler, then there probably isn't any existing code to be maintained in C++ either. Note that this is the only one I know of which was manufactured recently (i.e. within the last ten years), that if you go back further in time, you'll find others. CDC's Cyber line (mid 1970's?) used 60 bit words with 12 bit bytes, for example. The standard convention on a PDP-10 was 7 bit bytes in a 36 bit word, with one bit not used, but of course, a conforming C/C++ implementation would have to break with that -- sizeof(int) == 5.14286 just doesn't cut it. (On a PDP-10, byte size was a field in the instruction, and could in theory be anything from 1 to 36. The obvious solution for C/C++ would be to use 9.) Note that as far as I know, all of these machines are word addressed, not byte addressed. I think the IBM 360 (mid 1960's?) was the first machine to use byte addressing, ignoring the low order bits when accessing larger types (which works best if the larger types have a size which is a power of 2); before that, Some convenient word size was chosen (and there was no real reason for it to be a power of two bits), and then several characters were stuffed into it, and special instructions created to access them, with the character selector either in the high, unused bits of the address or on another word. (On the other hand, there were also 8 bit machines which worked on arbitrarily long sequences of bytes, like the 1401. Of course, the 1401 used decimal arithmetic anyway, and not binary, which would have caused problems for a C/C++ implementation as well.) I think that the Unisys was the only one sold after about 1980, however. And there wasn't much C++ being written back then to worry about. Quote: On the other hand, I think it was a word addressed machine. Does the Boost stuff work for word addressed machines. (If not, there's been a lot of them, and at least one is still being made.) I think it does. It uses (roughly): alignment = sizeof( struct { char; T } ) - sizeof(T) I would expect that to produce 6 for anything other than T=char on the Series A. I expect so. Since you asked about power of two, I was expecting some sort of bit twiddling, perhaps supposing the byte address on the low order bits of the pointer. -- James Kanze GABI Software Conseils en informatique orientée objet/ Beratung in objektorientierter Datenverarbeitung 9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34 [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Martin Bonner Guest Posted: Wed Nov 16, 2005 12:50 pm    Post subject: Re: size of class kanze wrote: Quote: Martin Bonner wrote: James Kanze wrote: Martin Bonner wrote: [...] Clearly the standard does not guarantee that the alignment requirement is a power of two. However, in practise it usually is. My question is then: does anyone know of any architecture (in use, proposed, or defunct) where the natural alignment requirement for a type was not a power of two? Unisys (ex-Burroughs) Series A. It's not been made for a couple of years now, but it had 48 bit word size, and I would expect 8 bit bytes. Which would mean an alignment of 6. BOTHER! Before getting too upset about it, I'd find out if it was relevant to whatever you're doing. I'm not doing anything. I'm shouting from the sidelines. Other people are trying to write portable code. "Portable" can have a range of meanings, a sample (from most to least portable): - "guaranteed by the standard and implemented by every known implementation". - "guaranteed by the standard". - "not guaranteed by the standard, but implemented that way on every conceivable machine". - "all desktops" - "All Neato compilers" - "Neato 3.1 with the fimble patch and the /gurble command line option supplied" Assuming that alignment was a power of two would step (a little) below "implemented that way on every conceivable machine". Quote: I don't know if there was ever a C++ compiler for it. A priori, the only reason anyone would be writing code for this machine today is when maintaining existing code, and if there was never a C++ compiler, then there probably isn't any existing code to be maintained in C++ either. Well quite, but it does weaken the argument "this code will never break in practise" if somebody can say "what happens if someone comes out with a Unisys style architecture?" [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. 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