Understanding sizeof(char) in 32 bit C compilers

Question

(sizeof) char always returns 1 in 32 bit GCC compiler.

But since the basic block size in 32 bit compiler is 4, How does char occupy a single byte when the basic size is 4 bytes???

Considering the following :

struct st 
{
int a;
char c;
};

sizeof(st) returns as 8 as agreed with the default block size of 4 bytes (since 2 blocks are allotted)

I can never understand why sizeof(char) returns as 1 when it is allotted a block of size 4.

Can someone pls explain this???

I would be very thankful for any replies explaining it!!!

EDIT : The typo of 'bits' has been changed to 'bytes'. I ask Sorry to the person who made the first edit. I rollbacked the EDIT since I did not notice the change U made. Thanks to all those who made it a point that It must be changed especially @Mike Burton for downvoting the question and to @jalf who seemed to jump to conclusions over my understanding of concepts!!

Answer 1

sizeof(char) is always 1. Always. The 'block size' you're talking about is just the native word size of the machine - usually the size that will result in most efficient operation. Your computer can still address each byte individually - that's what the sizeof operator is telling you about. When you do sizeof(int), it returns 4 to tell you that an int is 4 bytes on your machine. Likewise, your structure is 8 bytes long. There is no information from sizeof about how many bits there are in a byte.

The reason your structure is 8 bytes long rather than 5 (as you might expect), is that the compiler is adding padding to the structure in order to keep everything nicely aligned to that native word length, again for greater efficiency. Most compilers give you the option to pack a structure, either with a #pragma directive or some other compiler extension, in which case you can force your structure to take minimum size, regardless of your machine's word length.

char is size 1, since that's the smallest access size your computer can handle - for most machines an 8-bit value. The sizeof operator gives you the size of all other quantities in units of how many char objects would be the same size as whatever you asked about. The padding (see link below) is added by the compiler to your data structure for performance reasons, so it is larger in practice than you might think from just looking at the structure definition.

There is a wikipedia article called Data structure alignment which has a good explanation and examples.

Answer 2

It is structure alignment with padding. c uses 1 byte, 3 bytes are non used. More here

Answer 3

Sample code demonstrating structure alignment:

struct st 
{
int a;
char c;
};

struct stb
{
int a;
char c;
char d;
char e;
char f;
};

struct stc
{
int a;
char c;
char d;
char e;
char f;
char g;
};

std::cout<<sizeof(st) << std::endl; //8
std::cout<<sizeof(stb)  << std::endl; //8
std::cout<<sizeof(stc)  << std::endl; //12

The size of the struct is bigger than the sum of its individual components, since it was set to be divisible by 4 bytes by the 32 bit compiler. These results may be different on different compilers, especially if they are on a 64 bit compiler.

Answer 4

First of all, sizeof returns a number of bytes, not bits. sizeof(char) == 1 tells you that a char is eight bits (one byte) long. All of the fundamental data types in C are at least one byte long.

Your structure returns a size of 8. This is a sum of three things: the size of the int, the size of the char (which we know is 1), and the size of any extra padding that the compiler added to the structure. Since many implementations use a 4-byte int, this would imply that your compiler is adding 3 bytes of padding to your structure. Most likely this is added after the char in order to make the size of the structure a multiple of 4 (a 32-bit CPU access data most efficiently in 32-bit chunks, and 32 bits is four bytes).

Edit: Just because the block size is four bytes doesn't mean that a data type can't be smaller than four bytes. When the CPU loads a one-byte char into a 32-bit register, the value will be sign-extended automatically (by the hardware) to make it fill the register. The CPU is smart enough to handle data in N-byte increments (where N is a power of 2), as long as it isn't larger than the register. When storing the data on disk or in memory, there is no reason to store every char as four bytes. The char in your structure happened to look like it was four bytes long because of the padding added after it. If you changed your structure to have two char variables instead of one, you should see that the size of the structure is the same (you added an extra byte of data, and the compiler added one fewer byte of padding).

Answer 5

All object sizes in C and C++ are defined in terms of bytes, not bits. A byte is the smallest addressable unit of memory on the computer. A bit is a single binary digit, a 0 or a 1.

On most computers, a byte is 8 bits (so a byte can store values from 0 to 256), although computers exist with other byte sizes.

A memory address identifies a byte, even on 32-bit machines. Addresses N and N+1 point to two subsequent bytes.

An int, which is typically 32 bits covers 4 bytes, meaning that 4 different memory addresses exist that each point to part of the int.

In a 32-bit machine, all the 32 actually means is that the CPU is designed to work efficiently with 32-bit values, and that an address is 32 bits long. It doesn't mean that memory can only be addressed in blocks of 32 bits.

The CPU can still address individual bytes, which is useful when dealing with chars, for example.

As for your example:

struct st 
{
int a;
char c;
};

sizeof(st) returns 8 not because all structs have a size divisible by 4, but because of alignment. For the CPU to efficiently read an integer, its must be located on an address that is divisible by the size of the integer (4 bytes). So an int can be placed on address 8, 12 or 16, but not on address 11.

A char only requires its address to be divisible by the size of a char (1), so it can be placed on any address.

So in theory, the compiler could have given your struct a size of 5 bytes... Except that this wouldn't work if you created an array of st objects.

In an array, each object is placed immediately after the previous one, with no padding. So if the first object in the array is placed at an address divisible by 4, then the next object would be placed at a 5 bytes higher address, which would not be divisible by 4, and so the second struct in the array would not be properly aligned.

To solve this, the compiler inserts padding inside the struct, so its size becomes a multiple of its alignment requirement.

Not because it is impossible to create objects that don't have a size that is a multiple of 4, but because one of the members of your st struct requires 4-byte alignment, and so every time the compiler places an int in memory, it has to make sure it is placed at an address that is divisible by 4.

If you create a struct of two chars, it won't get a size of 4. It will usually get a size of 2, because when it contains only chars, the object can be placed at any address, and so alignment is not an issue.

Answer 6

Sizeof returns the value in bytes. You were talking about bits. 32 bit architectures are word aligned and byte referenced. It is irrelevant how the architecture stores a char, but to compiler, you must reference chars 1 byte at a time, even if they use up less than 1 byte.

This is why sizeof(char) is 1.

ints are 32 bit, hence sizeof(int)= 4, doubles are 64 bit, hence sizeof(double) = 8, etc.

Answer 7

Because of optimisation padding is added so size of an object is 1, 2 or n*4 bytes (or something like that, talking about x86). That's why there is added padding to 5-byte object and to 1-byte not. Single char doesn't have to be padded, it can be allocated on 1 byte, we can store it on space allocated with malloc(1). st cannot be stored on space allocated with malloc(5) because when st struct is being copied whole 8 bytes are being copied.

Answer 8

It works the same way as using half a piece of paper. You use one part for a char and the other part for something else. The compiler will hide this from you since loading and storing a char into a 32bit processor register depends on the processor.

Some processors have instructions to load and store only parts of the 32bit others have to use binary operations to extract the value of a char.

Addressing a char works as it is AFAIR by definition the smallest addressable memory. On a 32bit system pointers to two different ints will be at least 4 address points apart, char addresses will be only 1 apart.