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std :: string的C位运算符

作者:互联网

我的问题是关于如何在C std :: string上使用按位运算符.通过重载或功能无关紧要.

std :: string的工作XOR / ^函数示例:

std::string XOR(std::string value, std::string key)
{
  std::string retval(value);
  long unsigned int klen = key.length();
  long unsigned int vlen = value.length();
  unsigned long int k = 0;
  unsigned long int v = 0;
  for (; v < vlen; v++) {
    retval[v] = value[v] ^ key[k];
    k = (++k < klen ? k : 0);
  }
  return retval;
}

我现在缺少的是NOT /〜,AND /&的替代品.加上OR / |.
示例C代码(最后两行想要它…… :-P):

//note: matrix_content[][] holds unsigned, *checked* integers...
//      char(foo) works... 
std::string vertical_master   = "";
for (unsigned short int k = 0; k < axis_max; k++) {
  for (unsigned short int l = 0; l < axis_max; l++) {
    horizontal_master += char(matrix_content[l][k]);
    vertical_master   += char(matrix_content[k][l]);
  }
}

std::string vertical_shift1_0 = vertical_master;
usigned short int bit = "@"; //<- just an example... chatched via
                             //   std::string::substr() and std::string::c_str()
for (unsigned long int x = 0; x < axis_max; x++) {
  vertical_shift1_0 += char(bit);
}
std::string vertical_shift2_0;
for (unsigned long int x = 0; x < axis_max; x++) {
  vertical_shift2_0 += char(0);
}
vertical_shift2_0 += vertical_master;

std::string vertical_or  = ~(vertical_shift1 | vertical_shift2);
std::string vertical_and = ~(vertical_shift1_0 & vertical_shift2_0);

在PHP / Perl中,我可以做坏事:-像这样的事情

$vertical_shift1_0 = $vertical_master.str_repeat(chr(0), $axis_max);
$vertical_shift2_0 = str_repeat(chr(0), $axis_max).$vertical_master;
$vertical_or       = chunk_split(~($vertical_shift1 | $vertical_shift2), $axis_max, chr(170));
$vertical_and      = chunk_split(~($vertical_shift1_0 & $vertical_shift2_0), $axis_max, chr(170));

……我很想念它;-).然而XOR工作.我怎样才能得到AND / OR / NOT?

解决方法:

通过[]运算符从字符串中取出char后,可以使用所需的所有按位运算符. C/C++将有符号/无符号字符视为数字类型(不幸的是).

如果您正在执行许多按位操作,则可以考虑使用其中一个专用位数据结构

> ST000> bitset,用于固定长度的位集
> bit_vector来自动态长度位集的提升

标签:c,string,bitwise-operators,stdstring
来源: https://codeday.me/bug/20190730/1583365.html