类型转换 && 符号扩展 && 算术右移

K&R Exercise 2-1

char a = (unsigned char) ~0;

char b = (unsigned char) ~0 >> 1;

char c = (unsigned char)(~0 >> 1);

unsigned char d = (unsigned char) ~0;

unsigned char e = (unsigned char) ~0 >> 1;

unsigned char f = (unsigned char)(~0 >> 1);

printf("%u\n", x); // x is a, b, c, d, e or f.

a. 2^32-1 截断11111111, printf中char首先转换为int, 符号扩展2^32-1, int再转换为unsigned, 输出2^32-1;

b. 2^32-1 截断11111111, 无符号数逻辑右移01111111, printf中char先转换为int, 再转换为unsigned, 符号位为0, 扩展不变，输出2^7-1;

c. 2^32-1 有符号数算术右移2^32-1, 截断11111111, 余下同a;

d. 2^32-1 截断11111111, printf中unsigned char直接转换为unsigned, 输出2^8-1;

e. 2^32-1 截断11111111, 无符号数逻辑右移01111111, printf中unsigned char直接转换unsigned, 输出2^7-1;

f. 2^32-1 有符号数算术右移2^32-1, 截断11111111, 余下同d.

总结：

1. 算术右移因为截断在输出看不出差异，但是有本质不同;

2. char, short, int, long的signed和unsigned在内存中没有区别;

3. char->short->int->long类型转换时, 如果被转换数为signed型则存在符号扩展问题;

4. signed char首先转换为signed int, 符号扩展, signed int再转换为int;

5. 符号扩展的根本目的在于保证以补码形式保存的signed数在位数增加后其值不变,因此也必然以被转换数的类型作为是否进行符号扩展的标准.