A small problem of bit field and symbol expansion

annotation ：
Bit field ： Some information is stored , It doesn't take up a whole byte , Only a few or one binary bits are required , For example, when storing switching value , Only 0 and 1 Two kinds of state , use 1 Bit binary bit . To save storage space , And easy to handle ,C Language has provided a data structure , be called Bit field or Bit segment .

Writing method of bit field ： In the definition of structure , Definition The variable name is followed by a colon plus a limited number of digits .

typedef struct {
    
	short a;
	short b;
	short c:1; // Bit field , On behalf of the variable c Just assign 1bit Of memory space 
	short reserved : 15;
}_FORTEST;

problem ： There are the following codes

#include<stdio.h>

typedef struct {
    
	short a;
	short b;
	short c:1; // Use bit fields , Instructs the compiler c Only 1 Bit for storage 
	short reserved : 15;
}_FORTEST;

int main()
{
    
	_FORTEST f;
	f.a = 100; f.b = 200; f.c = 1;
	f.c &= 0x01;
	printf(" Decimal system ：a = %d b = %d c = %d\r\n",f.a,f.b,f.c);
	printf(" Hexadecimal ：a = %#x b = %#x c = %#x",f.a,f.b,f.c);
	return 0;
}

The running result is

Decimal system ：a = 100 b = 200 c = -1
Hexadecimal ：a = 0x64 b = 0xc8 c = 0xffffffff

among

f.a = 100; f.b = 200; f.c = 1;

because f.c Occupy 1 position , therefore f.c The complement of is 1

f.c &= 0x01;

This statement makes f.c And 0x01 Meet each other , The result is 0x1( because f.c Only occupy 1 position )

	printf(" Decimal system ：a = %d b = %d c = %d\r\n",f.a,f.b,f.c);
	printf(" Hexadecimal ：a = %#x b = %#x c = %#x",f.a,f.b,f.c);

Output f.c when ： It's all about f.c Extended to 4 byte （32bit）, And then f.c Only 1bit, The only remaining data bits replace the symbol bits for symbol expansion , So the binary complement after expansion is 32 individual 1, That is to say 10 It's binary -1, as well as 16 It's binary 0xffffffff