This question already has answers here:
What is bit masking?
(2 answers)
Closed 1 year ago.
A struct I'm dealing with has a field defined as uint8_t. The spec also states that all multiple-byte fields are represented in host-endian format.
Bits 0:3 contain the information that I need (also an unsigned integer). Using plain C, how do I extract those 3 bits and convert it to a number type?
unit8_t a;
unit8_t b;
a = input data;
b = a & 0x0F; // b contains a number from 0 to 15
Related
This question already has answers here:
Most efficient way to set n consecutive bits to 1?
(3 answers)
Closed 2 years ago.
I want to set an N amount of bits in a byte (byte always starts as 0) and store it using a pointer. Imagine:
void SetBits(uint8_t bytesToSet, uint8_t* var) {}
How would I go about implementing this (using C)?
Where should that N bits be set? To the left, to the right?
If you want to have those N bits to the right, your function should look like this:
void SetBits(uint8_t bitsToSet, uint8_t* var)
{
if (bitsToSet < 8)
*var = (1 << bitsToSet) - 1;
else {*var = 0; *var = ~(*var);}
}
For an example, for the call SetBits(5,&a), variable a will hold the value 0b00011111.
This question already has answers here:
":" (colon) in C struct - what does it mean? [duplicate]
(3 answers)
Closed 9 years ago.
While looking through the source package for QEMU, I found in the exec.c file:
struct PhysPageEntry {
/* How many bits skip to next level (in units of L2_SIZE). 0 for a leaf. */
uint32_t skip : 6;
/* index into phys_sections (!skip) or phys_map_nodes (skip) */
uint32_t ptr : 26;
};
I was wondering what the : operator means. I could not find it in a list of syntax definitions for C.
This is a structure declared with bit fields and the structure members are called bit fields: A
bit field is set up with a structure declaration that labels each field and determines its width. The above definition causes PhysPageEntry to contain one 6-bit field and one 26 bit field members namely skip and ptr respectively. Its signature is
struct
{
type [member_name] : width ;
};
Here width is the number of bits in the bit-field. The width must be less than or equal to the bit width of the specified type.
struct PhysPageEntry declares two bit fields skip and ptr. This is basically to allow the struct to pack these odd lengths (in terms of bits) efficiently. If the author didn't do this, the struct would likely be 8 bytes long (on a 32-bit architecture).
It represents number of bits. It's mostly used for unions:
struct byte {
unsigned a : 1;
unsigned b : 1;
unsigned c : 1;
unsigned d : 1;
unsigned e : 1;
unsigned f : 1;
unsigned g : 1;
unsigned h : 1;
};
You can read this too for better understanding.
Its called a bitfield. Within a structure or union
declaration, this declares 'skip' to be a "bit field" of 6 bits width. They are to be used inside of structures. If it helped please vote as right!
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
How do you set, clear and toggle a single bit in C?
I want to create an assembler , so I need to assign 32 bits bit by bit or field by field to create 32 bit opcode from assembly ... how can I do this in C ? how can I assign bits in integer ? can this be done ?
You can declare these two macros to help you:
#define Set_Bit(IntValue, BitNumber) IntValue = IntValue | (1<<BitNumber)
#define Clr_Bit(IntValue, BitNumber) IntValue = IntValue & (~((1) << (BitNumber))))
Some questions have discussed these before:
Macros to set and clear bits
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
How to get the value of individual bytes of a variable?
How can I get single bytes from a word (for example an unsigned int in C) without using bit-wise operations (that is using arithmetic operations?)
I don't know why, but this formula (C-Like) for an x number doesn't seem to work:
floor(x / pow(R, i)) % R
where R is the radix with which the number is represented, and i is used to indicate the i-th byte to obtain.
If you really need to avoid bitwise operations, you can cheat alternatively (beware if you're using a little or big endian machine!):
char *int_16_storage;
uint16_t the_word = 0xabcd;
int_16_storage = &the_word;
uint8_t low_byte = int_16_storage[0];
uint8_t high_byte = int_16_storage[1];
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
how to get bit by bit data from a integer value in c?
I have a 8-bit byte and I want to get a bit from this byte, like
getByte(0b01001100, 3) = 1
Firstoff, 0b prefix is not C but a GCC extension of C.
To get the value of the bit 3 of an uint8_t a, you can use this expression:
((a >> 3) & 0x01)
which would be evaluated to 1 if bit 3 is set and 0 if bit 3 is not set.
First of all C 0b01... doesn't have binary constants, try using hexadecimal ones. Second:
uint8_t byte;
printf("%d\n", byte & (1 << 2);
Use the & operator to mask to the bit you want and then shift it using >> as you like.