10 replies to this topic

[Roger]

What is the size of the C structure below on a 32-bit system? On a 64-bit?

struct foo {

        char a;

        char* b;

};

[artificial]

You can use C's/C++'s sizeof() operator to calculate the size of any variable, structure etc. in bytes.

struct foo

{

   char a;

   char *b;

};


int nSize = sizeof(foo);

printf("Size of foo is %i bytes!\n", nSize);

Greets,
artificial

[WingedPanther]

It depends somewhat on both the compiler and the OS.

[Roger]

So, it's not as straightforward as 3 bytes (1 for char and 2 for char*) for 16bit and 5 bytes (1 for char and 4 for char*) for 32bit?

[Guest]

Nope. Probably the most annoying part about C is that there are so many things that are implementation dependent. Heck, the number of bits in a byte can't even be assumed in the C language.

[Ancient Dragon]

some compilers such as Microsoft has a packing pragma that let you change how the compiler aligns the data in the structure or c++ class.

#pragma pack(1) // byte align the data

struct foo

{

   char a;

   char *b;

};

With byte alignment the compiler does not put holes in the structure, so sizeof(struct foo) will be 5

#pragma pack(2) // word align the data

struct foo

{

   char a;

   char *b;

};

With word alignment the compiler will align the pointer in an even byte boundry (or address) and sizeof(struct foo) will be 6.

[Roger]

I see.. Does that mean it can be 3 or 4 bytes in 16bit and 5 or 6 bytes in 32bit OS?

[Alexander]

Roger said:

So, it's not as straightforward as 3 bytes (1 for char and 2 for char*) for 16bit and 5 bytes (1 for char and 4 for char*) for 32bit?

More or less (pointer wise), on 16, 32 and 64-bit arches, pointer sizes are 2, 4 and 8 respectively.

It's not safe to assume padding, but natural padding of a char resulting in 4 octets will often be the case, yes. For x86, fields are usually 32-bit aligned. The reason for this is to increase the system's performance at the cost of memory usage.

Similarly, for x64, fields are usually 64-bit/8-byte aligned, so sizeof(foo) would be 16.

[Roger]

So, if x86:

4 bytes for 16-bit OS
8 bytes for 32-bit OS
16 bytes for 64-bit OS

??

[Alexander]

Roger said:

So, if x86:

x86 refers to the 8086 family of instruction set architectures, but you would imply 32-bit if you would say just "x86". x86-16 implies 16-bit and x64 implies x86-64.

As for the rest of your quote, pointers are 16 bits in x86-16, so you'd be correct to assume it would be what you wrote.

[Roger]

Okay, thanks!