How do you calculate the size of a bit field?

In above student structure size of the structure without bit field is size of (StdId) + size of (Age) = 8 bytes + 8 Bytes = 16 bytes. After using bit fields to its members, it is 8 bits + 4 bits = 12 bits = 1.5 bytes which is very much less.

How do you determine the size of a struct?

Size of struct can be calculated as the total sizes of all datatypes. char – 1 Byte(In the struct 26 characters are creating so Size is 26*1 = 26).

How are bit fields used in structures?

In C, we can specify size (in bits) of structure and union members. The idea is to use memory efficiently when we know that the value of a field or group of fields will never exceed a limit or is withing a small range.

How do you determine the size of a struct in C++?

6 Answers. The size is at least sizeof(int) + sizeof(struct node *) + sizeof(struct node *) . But it may be more as the compiler is allowed to add padding bytes to your structure if it wishes.

What is the purpose of bit fields give an example for bit fields?

Bit fields can be used to reduce memory consumption when a program requires a number of integer variables which always will have low values. For example, in many systems storing an integer value requires two bytes (16-bits) of memory; sometimes the values to be stored actually need only one or two bits.

What are bit fields what is the use of bit fields in a structure declaration?

These space-saving structure members are called bit fields, and their width in bits can be explicitly declared. Bit fields are used in programs that must force a data structure to correspond to a fixed hardware representation and are unlikely to be portable.

How do you determine the size of a structure without using sizeof?

First, create the structure. create an array of structures, Here aiData[2]. Create pointers to the structure and assign the address of the first and second element of the array. Subtract the pointers to get the size of the structure in c.

What is the size of the struct?

Above is the alignment of the structure A, and that’s why the size of the struct is 32 Bytes. Also, the object a of type struct A is 32 Bytes.

Which equation is used to calculate the range for bit field?

Remember the formula for finding out the range of signed integers: 2(n-1) to 2(n-1)-1 where N is the number of bits. For example, if N is 8 (number of bits in a byte), i.e., the range of a signed integer of size 8 is -2(8-1) to 2(8-1)-1, which is -128 to +127.

How are bit fields stored in memory?

Some bit field members are stored left to right others are stored right to left in memory. If bit fields too large, next bit field may be stored consecutively in memory (overlapping the boundary between memory locations) or in the next word of memory.

How do you find the size of a variable without using sizeof?

The idea is to use pointer arithmetic ( (&(var)+1) ) to determine the offset of the variable, and then subtract the original address of the variable, yielding its size. For example, if you have an int16_t i variable located at 0x0002 , you would be subtracting 0x0002 from 0x0006 , thereby obtaining 0x4 or 4 bytes.

How do you find the size of an array without using sizeof operator?

*(a+1) => Dereferencing to *(&a + 1) gives the address after the end of the last element. *(a+1)-a => Subtract the pointer to the first element to get the length of the array. Print the size.

How big is a struct in C++?

If you are talking about a 32-bit application, integers, and pointers are 4 bytes. Thus, your struct is 12 bytes in size.

Why is the size of a struct 32 bytes?

Of course, the compiler adds padding and tries to align the data members. So for the above structure, the data alignment looks like below, Above is the alignment of the structure A, and that’s why the size of the struct is 32 Bytes. Also, the object a of type struct A is 32 Bytes.

How many bytes do I need for a bitfield?

Clearly 3 bitfields totalling 33 bits don’t fit into a single unsigned int, hence the need for 8 bytes in the first example. 3 bitfields totalling 3 bits certainly do fit into an unsigned int, hence only 4 bytes in the second example. Furthermore, a bitfield cannot span multiple integers.

How big are sizeof(int) and sizeof(struct node*)?

How big sizeof(int)and sizeof(struct node*)are depends on your system. They are likely to be 4 bytes each if you have a 32-bit system, or they could be 8 bytes if you have a 64-bit system. But really the only way to know for sure is to use the compiler to print out the size. Share Improve this answer Follow answered Oct 20 ’12 at 8:31