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 unsig
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?
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