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铺垫在结构上如何运作?
原标题:how does padding work in structure?
  • 时间:2012-05-24 18:54:13
  •  标签:
  • c

这个程序的输出是28 我不懂如何? 根据我, 这应该是 32( 4+4+4+4+4+4+12)+4( 维持对齐) = 32。 请解释显示输出 28 的原因? < precode> struct test{c; int d; int x; int y; long dowy p;} t1; pintf ("%d""%d", sizeof( t1) ;

最佳回答

也许你的"长双倍"其实和双倍(8字节)一样, 如果你在32位处理器上,对齐是4字节。

4+4+4+4+4+8=24

size of( long double) 是什么?

<强度 > EDIT:

我用海合会 s 来调查。 解释结构大小的实际答案是 :

4+4+4+4+4+12=28

size of( 长双) 是 12 。

无需挂贴, 因为 is 4 and. 有趣的是, 是 8 。

问题回答

根据 this ,在gcc 32-bit模式(使用 gcc-m32 或为产生32位输出而建造的gcc)中,长的双倍(使用 gcc>-gcc-m32 或为产生32位输出而建造的gcc,不管你的平台实际上是什么),只有4位比特的对齐。也许最好参考gcc手册来验证这一点。

On a 64 bit system sizes are char - 1 byte(its not 4 bytes) int - 4 bytes long double - 12 bytes

1+4+4+4+12+平铺=28字节!

在我看来这是正确的。 紧接于p 之后的字段将会是 4 字节对齐, 所以不需要在结构的结尾处划线 。

4+4+4+4+4+4+12=28

在我的系统上,输出量是32, 但在我的系统上,尺寸(长两倍)是16(x86_64,LLVM3)。

28 输出的 < code> size of 输出当然正确 = 1 字节用于字符、3 字节贴布、3 x 4 英寸和12 字节用于长的双倍(96 字节大小,但80 字节精确度), 共28 字节。

记住,即使你用x86-64机器编集,你也可能用 mingw32 来编集x86-32机器,这有区别。

Mingw32 使用 4 字节对应 < code > loong 双 。



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