有趣的位運算

Question

我遇到了一些相當有趣的代碼來到今天在檢討SHA1的實施C.

temp = SHA1CircularShift(5,A) + (B^C^D) + E + W[t] + K[1]; 
temp &= 0xFFFFFFFF; 

我覺得有趣的部分是temp &= 0xFFFFFFFF;。請注意0​​已被宣佈爲無符號整數。這個操作不會簡單地沒有效果嗎？我能想到的唯一的事情是設計者試圖強制使用32位整數，但這不是在編譯時完成的嗎？
我很想知道人們的想法。

Answer 1

在某些機器上，int（因此也是unsigned int）可能是64位類型。掩碼在機器上沒有任何操作，其中int是32位類型，但對於64位類型的機器很重要。編譯器會知道並且在沒有任何用處時會優化操作。

此外，過去有機器36-bit int types和其他與60-bit int types;這對於這樣的機器也是很重要的。

Answer 2

SHA1的reference implementation，有評論如下注意事項：

/* 
* sha1.c 
* 
* Description: 
*  This file implements the Secure Hashing Algorithm 1 as 
*  defined in FIPS PUB 180-1 published April 17, 1995. 
* 
*  The SHA-1, produces a 160-bit message digest for a given 
*  data stream. It should take about 2**n steps to find a 
*  message with the same digest as a given message and 
*  2**(n/2) to find any two messages with the same digest, 
*  when n is the digest size in bits. Therefore, this 
*  algorithm can serve as a means of providing a 
*  "fingerprint" for a message. 
* 
* Portability Issues: 
*  SHA-1 is defined in terms of 32-bit "words". This code 
*  uses <stdint.h> (included via "sha1.h" to define 32 and 8 
*  bit unsigned integer types. If your C compiler does not 
*  support 32 bit unsigned integers, this code is not 
*  appropriate. 
* 
* Caveats: 
*  SHA-1 is designed to work with messages less than 2^64 bits 
*  long. Although SHA-1 allows a message digest to be generated 
*  for messages of any number of bits less than 2^64, this 
*  implementation only works with messages with a length that is 
*  a multiple of the size of an 8-bit character. 
* 
*/ 

的Portability Issues是在此實現SHA1的這種操作的情況下，允許其與大int小號機器正常運行。