我想在将 input_string
传递给 if
语句中的 regex_match
之前向左/向右修剪,但我没有成功尝试boost trim_right
and trim_left
,但不确定如何解决。
如何简单地修剪输入,使前两个输入 "0 "
和 "0.1 "
成为有效数字?
#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;
}
当前输出
[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.
期望的输出
[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.
最佳答案
最大的问题是std::regex_match
需要一个持久的字符串对象供结果引用。如果你没有,那么在检查结果后将访问一个不存在的字符串,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
喜欢
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;
}
这将花费您一份拷贝,但这是在不修改源数据的情况下执行此操作的唯一方法。如果您可以修改源数据,那么您可以使用
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。
关于c++ - 力扣 65 : Valid Number (C++),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/58417404/