210324 Little-Endian

2021-03-24 
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uint32_t a = 0;
uint16_t x = 10;
uint16_t y = 20;
a = (x << 16) | y;
uint16_t* res = (uint16_t*) &a;
printf("%u %u\n", *res, *(res + 1)); //output: 20 10

Windows和Linux都是小端
x=0x000a
y=0x0014
a=0x000a0014
在内存中：

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低地址------->高地址
x    0a 00
y    14 00
a    14 00 0a 00
      ^res    ^res+1
      20         10
#shanghaitech #programming