bit-fields

问题 I do have a structure having bit-fields in it.Its comes out to be 2 bytes according to me but its coming out to be 4 .I have read some question related to this here on stackoverflow but not able to relate to my problem.This is structure i do have struct s{ char b; int c:8; }; int main() { printf("sizeof struct s = %d bytes \n",sizeof(struct s)); return 0; } if int type has to be on its memory boundary,then output should be 8 bytes but its showing 4 bytes?? 回答1: Source: http://geeksforgeeks

问题 I am going through a windows device driver and I saw struct code like this: struct driver1 { UINT64 Readable : 1; UINT64 Writable : 1; UINT64 Executable : 1; UINT64 Control : 3; UINT64 Status : 1; UINT64 Reserved : 51; UINT64 Available1 : 5; UINT64 IsMapped : 1; }; Does each UINT64 represent a single bit ? Does a colon represent bits? 回答1: This means that Readable , Writable and Executable each take up a bit, Control takes 3, Reserved 51 and so on. Refer to http://publications.gbdirect.co.uk

问题 I want to read the 2nd, 5th and the 6th bit from a 32-bit register. I have decided to use struct bit fields to store them. Is the following data structure correct? struct readData { int unwanted:1; int reqbit1:1; int unwanted1:2; int reqbit2:2; int unwanted2:26; }; I'm not sure about how the bit fields are created. I'm going to use an API that will copy bytes from the h/w register to this structure directly. In that case, will reqbit1 contain the 2nd bit? As per my understanding, the compiler

问题 I have to work with the InputDevice.getSources() method to determine the type (source) of InputDevice. But instead of returning a predetermined integer, it returns a combined bitfield, for example: 16786707 (this is an actual value from my gamepad). As you can see 16786707 is not listed in the InputDevice documentation page because it is generated on the fly. How do I parse the number 16786707 to determine whether the InputDevice is: a SOURCE_CLASS_JOYSTICK (16), or a SOURCE_GAMEPAD (1025),

问题 This question already has answers here : What's the purpose of unnamed bit field at the end of structure (2 answers) Closed 3 years ago . Suppose I have a structure defined as follows: typedef struct Counters { uint8_t counterSec : 6; uint8_t : 3; uint8_t counterMin : 6; uint8_t : 3; uint8_t counterHr : 5; uint8_t : 1; }; As I want to dedicate 6 bits per first two counters and 5 bits per last counter, I end up with 17 bits, which means that a total of 24 bits will be allocated to an instance

问题 I was trying to find a reason why I should not write struct bitfield { signed foo:4; unsigned bar:2; }; instead of specifying verbosely struct bitfield { signed int foo:4; unsigned int bar:2; }; As the size of each member of the bitfield is explicitly specified after the colon, could there be any drawbacks? Does it matter if I use char , short , long , long long ? Must the number of specified bitfield bits probably always be smaller than the width of the declaration type? Found some related

问题 I'm writing a parser for a binary format. This binary format involves different tables which are again in binary format containing varying field sizes usually (somewhere between 50 - 100 of them). Most of these structures will have bitfields and will look something like these when represented in C: struct myHeader { unsigned char fieldA : 3 unsigned char fieldB : 2; unsigned char fieldC : 3; unsigned short fieldD : 14; unsigned char fieldE : 4 } I came across the struct module but realized

问题 Let consider the following program test.c : #include <stdio.h> struct test { unsigned int a:5; }; int main () { unsigned int i; struct test t = {1}; for (i = 0; i < t.a << 1; i++) printf("%u\n", i); return 0; } When compiled with gcc -Wsign-compare test.c the following warning is produced (tested with gcc 4.8.1): test.c:9:19: warning: comparison between signed and unsigned integer expressions [-Wsign-compare] for (i = 0; i < t.a << 1; i++) ^ clang -Wsign-compare test.c produces the following

问题 How to access entire structure members in C.Means I want to take all the data of variables in structure. struct data { char a:1; char b:2; char c:3; char d:1; } arr; I can access individual members by using . operator.But i need to access all members in that structure.Colud you please tell me how to do. 回答1: As suggested make an union of the whole data and of the bitfield structure. union { char Whole; struct data { char a:1; char b:2; char c:3; char d:1; } arr; } MyData; 来源： https:/

问题 I got a bit field with a bunch of flags, and I need a quick and dirty way to set everything to zero, so instead of blindly casting the struct to an integer, I decided it would be "better" to put the bit fields in a union with an actual integer. union Flags { uint _all; struct { uint status : 2; uint expanded : 1; uint draw : 1; uint drawChildren : 1; uint hidden : 1; uint disabled : 1; uint used : 1; uint deletable : 1; uint incomplete : 1; uint isStatic : 1; uint isConst : 1; uint isVolatile