_int64位字段 [英] _int64 bit field

问题描述

我需要在结构中使用6字节(48位)的位域，该结构可以用作无符号整数来进行比较等.

I need to use a 6-byte (48-bit) bitfield in a structure that I can use as unsigned integer for comparison etc. Something along the following:

pack (1)
struct my_struct {
  _int64 var1:48;
} s;

if (s.var >= 0xaabbccddee) { // do something }

但是以某种方式在64位Windows上，sizeof此结构始终返回8个字节，而不是6个字节.任何指针都值得赞赏吗?

But somehow on 64-bit Windows, sizeof this struct always returns 8 bytes instead of 6 bytes. Any pointers are appreciated?

推荐答案

您已使用_int64，因此sizeof返回8.这就像您已决定在可用的64位中使用48位一样.即使我们声明这样的东西-

You have used _int64 and hence sizeof returns 8. Its like you've decided to use up 48 bits out of the available 64 bits. Even if we declare something like this-

struct my_struct {
  _int64 var1:1;
} s;

仍然sizeof表示8.简而言之，将根据位域的类型分配位域.在这种情况下，它的_int64分配为8个字节.

Still sizeof would say 8. In short, allocation of bitfields would take place according to the type of the bitfield. In this case its _int64 and hence 8 bytes allocation.

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