理解c

Question

我不明白这段代码的输出：

main() {
    int ret = ~(~0 <<5) << 2;
    printf("ret: %d, %u\n", ret, ret);
}


output:
ret: 124, 124

如果我在精神上处理，我会这样做：

• 解析〜0给出二进制1
• 解析〜0＆lt;＆lt; 5给出二进制100000
• 解析〜（〜0 <＆lt; 5）给出二进制011111
• 解析〜（~0 <＆lt; 5）＆lt;＆lt; 2给出二进制111100
• 将二进制111100转换为十进制为60

我做错了什么？

Answer 1

if int is 4 bytes then:
~0        = 11111111111111111111111111111111 = -1        
-1 << 5   = 11111111111111111111111111100000 = -32        
~-32      = 00000000000000000000000000011111 =  31          
31 << 2   = 11111111111111111111111000000000 = 124        

if int is 2 bytes then:
~0        = 1111111111111111                 = 65535
65535 << 5= 1111111111100000                 = 65504
~65504    = 0000000000011111                 = 31
31 << 2   = 0000000001111100                 = 124


int is guaranteed to be able to hold -32767 to 32767,
which requires 16 bits.
In that case, int , is 2 bytes.
However, implementations are free to go beyond that minimum,
as you will see that many modern compilers make int 32-bit 
(which also means 4 bytes pretty ubiquitously).

Answer 2

~0是二进制的全1。

左移5，将有5 0，然后全1。

〜这是5 1;这是31。

左移2位等于乘以4，留下124作为最终答案。