Question

我想在input_string语句中将regex_match传递到if之前先左右修剪" 0 "，但我不确定boost trim_right and trim_left尝试是否成功不过如何解决。

如何简单地修剪输入，以使前两个输入" 0.1 "和#include <string> #include <boost/algorithm/string.hpp> #include <iostream> #include <regex> #include <vector> using namespace std; int main() { vector<string> string_vector = {" 0 "," 0.1 ","abc","1 a","2e10","-90e3","1e","e3","6e-1","99e2.5","53.5e93","--6","-+3","95a54e53"}; regex expression_two("^[+-]?(?:[0-9]*\\.[0-9]+|[0-9]+\\.[0-9]*|[0-9]+)[Ee][+-]?[0-9]+$|^[+-]?(?:[0-9]*\\.[0-9]+|[0-9]+\\.[0-9]*|[0-9]+)$|^[+-]?[0-9]+$"); for (const auto &input_string: string_vector) { if (std::regex_match(input_string, expression_two)) cout << "[0-9] Char Class: '" << input_string << "' is a valid number." << endl; } return 0; }成为有效数字？

[0-9] Char Class: '2e10' is a valid number.
[0-9] Char Class: '-90e3' is a valid number.
[0-9] Char Class: '6e-1' is a valid number.
[0-9] Char Class: '53.5e93' is a valid number.

电流输出

[0-9] Char Class: '  0  ' is a valid number.
[0-9] Char Class: '  0.1  ' is a valid number.
[0-9] Char Class: '2e10' is a valid number.
[0-9] Char Class: '-90e3' is a valid number.
[0-9] Char Class: '6e-1' is a valid number.
[0-9] Char Class: '53.5e93' is a valid number.

所需的输出

Route::resource('reports/{id}', 'ReportController');

Answer 1

最大的问题是std::regex_match需要一个持久的字符串对象来引用结果。如果没有，那么在检查结果之后，将访问不存在的字符串，这是未定义的行为。这意味着您需要修改循环以使用What's the best way to trim std::string?中的trim_copy，例如

for (const auto &input_string: string_vector) {
    auto trimmed = trim_copy(input_string);
    if (std::regex_match(trimmed, expression_two))
        cout << "[0-9] Char Class: '" << input_string << "' is a valid number." << endl;
}

这将使您花费一个副本，但这是不修改源数据的唯一方法。如果可以修改源数据，则可以使用

for (auto &input_string: string_vector) {
    if (std::regex_match(trim(input_string), expression_two))
        cout << "[0-9] Char Class: '" << input_string << "' is a valid number." << endl;
}

相反。

LeetCode 65：有效数字（C ++）

电流输出

所需的输出

1 个答案: