为什么打印出来的长度不一样呢？？ - V2EX

为什么打印出来的长度不一样呢？？ - V2EX

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import struct

s = struct.Struct ('Is')
s1 = struct.Struct ('sI')

print ("s.size:",s.size )
print ("s1.size:",s1.size )

################
运行结果：
s.size: 5
s1.size: 8

4 replies 2015-08-27 23:20:04 +08:00

1

wartime

Aug 27, 2015

int 类型在内存中存放地址 4 字节对齐, 导致 s1 占了额外空间.

s.pack (1, 'a')
'\x01\x00\x00\x00a'

s1.pack ('a', 1 )
'a\x00\x00\x00\x01\x00\x00\x00'

可以看到内存中实际存放方式

2

hnsxyhh

OP

Aug 27, 2015

@wartime 为什么 s1 有对齐， s 没有对齐啊

3

exch4nge

Aug 27, 2015

@hnsxyhh 可以去了解了解 C/C++的结构体对齐，就大概知道为啥 python 这个也要对齐了

4

wartime

Aug 27, 2015

1

@hnsxyhh s 的起始地址是四字节对齐的，首先存放 int 类型值，就不需要再偏移， s1 里起始地址存放的是一个 char 类型，占用一个字节，假设是 s1 起始地址是 0x1000, int 存放地址就是从 0x1001 开始，没有对齐，只有再偏移三个字节, 到 0x1004, 才能四字节对齐。

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