我试图寻找一种使用宏使变量的位表示起作用的方法。该代码向后打印二进制字符串,因为我对 c++
比较陌生,我认为索引是问题所在,因为它需要从高位开始,然后倒数。
#include <iostream>
#include <memory>
#include <climits>
#define EXTRACTBIT(ith,ch)(std::cout<<((*ch >> ith) & 1 ? 1 : 0))
template < typename T > void printlnbits (T v)
{
const int s = sizeof (T) * CHAR_BIT;
auto const bytes_begining{reinterpret_cast<unsigned char const *>(std::addressof(v))};
auto byte{bytes_begining};
auto bit_index_in_byte{0};
for (int n = s - 1; n >=0; --n)
{
EXTRACTBIT(bit_index_in_byte, &(*byte));
++bit_index_in_byte;
if (CHAR_BIT == bit_index_in_byte){
std::cout << " ";
bit_index_in_byte = 0;
++byte;
}
}
std::cout << " " << s << " bits" << std::endl;
}
int main ()
{
const char a = 'a';
const char b = 2;
printlnbits (a);
printlnbits (b);
return 0;
}
结果:
bit_index_in_byte = 0;
++bit_index_in_byte;
10000110 8 bits //correct 01100001
01000000 8 bits //correct 00000010
bit_index_in_byte = 8;
--bit_index_in_byte;
00110000 8 bits //correct 01100001
00000001 8 bits //correct 00000010
我尝试过失败,第一个是正确的但落后,另一个是不正确的。我还想补充一点,任何帮助或其他简化解决方案的建议将不胜感激
最佳答案
正如评论中提到的 -500 ,如果你像这样以 7
开头,你应该得到正确的结果
auto bit_index_in_byte{7};
for (int n = s - 1; n >=0; --n)
{
EXTRACTBIT(bit_index_in_byte, &(*byte));
--bit_index_in_byte;
if (-1 == bit_index_in_byte)
{
std::cout << " ";
bit_index_in_byte = 7;
++byte;
}
}`
结果:
01100001 8 bits
00000010 8 bits
关于c++ - 二进制字符串向后打印,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/69521679/