C Programming - Structures, Unions, Enums - Discussion

Bit field can not be used for floating values
it can be used with => signed int ,unsigned int and char
the meaning of "char scheme:4;"
is only 4-bit are reserved for "scheme" not 4-byte(people always make that confusion) here is one example
[1][1][1][1][1][1][1][1] -->8-bit field for char
but for char scheme:4;
the first 4 are not reserved and the last 4 will be reserved for "scheme".
like this (assume 0 for unreserved and 1 for reserved)
[0][0][0][0][1][1][1][1]

When we run this program it will show an error that "Bit Field must be an Integral Type". This means that bit fields must be of type signed int or unsigned int and char. We cannot define float data type. Therefore,

int a:3; /* valid */

float b:3; /* Invalid */

One more thing, If we do not mention the type then also it will show an error.

struct a
{ /*
category: 5; Invalid code
scheme: 4; */
}

@Prashant O_O: Friend I ran this code on gcc compiler after making few changes, here is my code:

#include<stdio.h>

int main()
{
struct a
{
int category:5;
char scheme:4;
};

printf("size=%d", sizeof(struct a));
return 0;
}

And according to your suggestion output must be 9 but it is 8 even take any values in the bit field still the output will be 8. Can you please explain me this.

Float and double will fallow the IEEE-754 standard to store the values into the memory.

If we use that as a bit-field member in a structure. It won't able to follow the method so in big field concept float I avoided to use and also for this reason if you store the same value in float and double, they won't be equal. Because the storage method is different for float and double
Float x=1.1f,
Double y=1.1,
X n't equal to Y.

#Monisha

The bitfield method is to avoid structure padding.
Eg:

Struct monisha
{
Int a: 4
Char b: 3
};

Here the size of the structure is 4 byte as it takes max size data type as int is having 4 bytes which is more than char hence structure takes 4 bytes and in that 4 bytes int a will take 4 bits and char b will take 3 bits.

This is known as but fled to avoid memory leakage.

1. The amount of memory required to store a structure variable is the sum of the size of all the members.

On the other hand, in the case of unions, the amount of memory required is always equal to that required by its largest member.

2. In the case of structure, each member has their own memory space but In union, one block is used by all the member of the union.

@Professor_X.

As you said, The minimum size of a structure variable will be the size of the biggest data type in the structure.

This sentence is valid only for union not structure because the structure will allocate different memory for different data types. But union allocates memory for the largest data type.

Please correct if I'm wrong. Thank you.

You can test this.

The minimum size of a structure variable will be the size of the biggest data type in the structure.

i.e.,

In this example (*replace float with int).

sizeof (struct a) will be a minimum of 4 in 32-bit turboc.

Please correct me if I'm wrong.

@Prema.

The main purpose of reserving bits in declaring them is the memory allocation for the variables can be used for what we want.

So by allocating required number of bits, we can save the memory.

What is the purpose of reserving the bits for a variable?

As the size of integer is 2 bytes or 4 bytes depending on the compiler why we should reserve 5 bits in this code?