struct的代码如何工作，请在下面解释代码

问题描述

struct abc
{
  int a:3;
  int b:3;
  int c:2;
}
void main()
{
   struct abc a=(2,3,4)
   printf(" %d %d %d",a.a,a.b,a.c);
   gethc();
}

推荐答案

struct abc
{
    int a:3;
    int b:3;
    int c:2;
};

int _tmain(int argc, _TCHAR* argv[])
{
    struct abc a = {2,3,4};
    printf(" %d %d %d, %d",a.a,a.b,a.c,sizeof(struct abc));
    return 0;
}

因此，在您的结构中，这里a被分配了3位，b被分配了3位，c被分配了2位.现在，当您将值赋予结构变量时.
位中提供的值

a = 10(二进制为2)(从三分之二中占用2位空间)
b = 11(二进制为3)(在三分之二内占用2位空间)
c = 100(二进制为4)(需要3位空间，但只有2位)

因此a和b的值将正确打印，但是c的值将为0，因为不会存储(100)中的第三位，即1.

So here a is assigned 3 bits, b is assigned 3 bits and c is assigned 2 bits in your structure. Now when you give value to a structure variable.
Values provided in bit

a = 10 (2 in binary) (take 2 bit space out of three)
b = 11 (3 in binary) (take 2 bit space out of three)
c = 100 (4 in binary) (needs 3 bit space but have 2 bit only)

so value of a and b will be printed correctly, but value of c will be 0, as third bit from (100) that is 1 is not stored.

struct abc
{
int a:3;
int b:3;
int c:2;
}

在结构abc中定义(请注意:缺少的分号是语法错误)三个位字段.全部共享相同的类型，即有符号整数，a和b的长​​度为三位(它们可以保存值{-4,-3,-2,-1,0,1,2,3})，而c的长度为两位(可以保留值)值{-2,-1,0,1}).

Defines (note: the missing semicolon is a syntax error) three bit fields in the struct abc. All share the same type, namely signed integer, a and b are three bits long (they may hold the values {-4,-3,-2,-1,0,1,2,3}) while c is two bits long (it may hold the values {-2,-1,0,1}).

struct abc a=(2,3,4)

上一行(再次，缺少的分号是语法错误)通过以下方式初始化结构abc的实例a:

The above line (again, the missing semicolon is a syntax error) initializes the instance a of the struct abc this way:

a.a = 2;
a.b = 3;
a.c = 4; // NOTE: 4 is an out of range value for c

printf(" %d %d %d",a.a,a.b,a.c);

上面的行显示了位字段的值.

The above line prints the values of the bit fields.

gethc();

我不知道那是什么您是说getch();吗?

I don''t know what it is. Do you mean getch(); instead?

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