Question

我找到了以下帖子：
Calculate broadcast address from ip and subnet mask以及指向http://lpccomp.bc.ca/netmask/netmask.c

的链接

有人可以解释下面一行，我不明白：

for ( maskbits=32 ; (mask & (1L<<(32-maskbits))) == 0 ; maskbits-- )

尤其是mask & (1L<<(32-maskbits))

Answer 1

<<是bitwise left shift operator;它将剩余给定数量的值的位移位。因此1L<<(32-maskbits)将值1移至左32-maskbits次。

&是bitwise AND operator。

因此循环表达式mask & (1L<<(32-maskbits)) == 0测试mask值内的所有位，从低到高。循环将在mask的第一个（最低）非零位停止，此时maskbits将包含该位以上（和包括）位的位数。

E.g。

如果 ~~mask == 0xFFFF~~ mask == 0xFFFFFFFF (== binary 11111111111111111111111111111111)，循环将在第一次迭代时停止，maskbits将为32
如果 ~~mask == 0x0001~~ mask == 0x00000001 (== binary 00000000000000000000000000000001)，循环将在第一次迭代时再次停止，maskbits将为32
如果 ~~mask == 0x1000~~ mask == 0x01000000 (== binary 00000001000000000000000000000000)，循环将在第24次迭代时停止，maskbits将为8

Answer 2

看一下按位运算符，特别是左移。

http://en.wikipedia.org/wiki/Bitwise_operation#Shifts_in_C.2C_C.2B.2B_and_Java

Answer 3

要了解发生了什么：运行它。

#include <stdio.h> 
#include <iostream>
using namespace std;

char *binary (unsigned int v) {
static char binstr[33] ;
int i ;

binstr[32] = '\0' ;
for (i=0; i<32; i++) {
binstr[31-i] = v & 1 ? '1' : '0' ;
v = v / 2 ;
}

return binstr ;
}

int main(void){  

  unsigned long maskbits,mask;  

mask = 0x01000000;
cout << "MASK IS: " << binary(mask) << "\n";
cout << "32 is: " << binary(32) << "\n\n";
for ( maskbits=32 ; (mask & (1L<<(32-maskbits))) == 0 ; maskbits-- ) {
cout << "maskbits: " << binary(maskbits) << "\n";
cout << "\t(32-maskbits): " << binary((32-maskbits)) << "\n";
cout << "\t1L<<(32-maskbits): " << binary((1L<<(32-maskbits))) << "\n";
cout << "\t(mask & (1L<<(32-maskbits))): " << binary((mask & (1L<<(32-maskbits)))) << "\n\n";

}

cout << "\nFinal maskbits: " << maskbits;

return 0;
}

http://ideone.com/eB8Kp

我不明白以下C代码行

3 个答案: