c++ - C++ (gcc) 中是否有等效的 TryParse？

Question

在 C++(gcc) 中是否有等效的 TryParse？

我想解析一个可能包含 (+31321) 的字符串并将其存储很长时间。我知道电话号码存储为字符串和匹配的字符串，但出于我的需要，我想将它们存储很长时间，有时它们可​​能包含加号 (+)。什么会在 C++ 中解析它？

Answer 1

strtoul() 及其系列的问题是没有真正的方法来测试失败。
如果解析失败，则返回 0 而不设置 errno(仅在溢出时设置)。

提高词汇量

#include <boost/lexical_cast.hpp>


int main()
{
    try
    {
        long x = boost::lexical_cast<long>("+1234");
        std::cout << "X is " << x << "\n";
    }
    catch(...)
    {
        std::cout << "Failed\n";
    }
}

使用流来实现

int main()
{
    try
    {
        std::stringstream stream("+1234");
        long x;
        char test;

        if ((!(stream >> x)) || (stream >> test))
        {
            // You should test that the stream into x worked.
            // You should also test that there is nothing left in the stream
            //    Above: if (stream >> test) is good then there was content left after the long
            //           This is an indication that the value you were parsing is not a number.
            throw std::runtime_error("Failed");
        }
        std::cout << "X is " << x << "\n";
    }
    catch(...)
    {
        std::cout << "Failed\n";
    }
}

使用扫描:

int main()
{
    try
    {
        char integer[] = "+1234";
        long x;
        int  len;

        if (sscanf(integer, "%ld%n", &x, &len) != 1 || (len != strlen(integer)))
        {
            // Check the scanf worked.
            // Also check the scanf() read everything from the string.
            // If there was anything left it indicates a failure.
            throw std::runtime_error("Failed");
        }
        std::cout << "X is " << x << "\n";
    }
    catch(...)
    {
        std::cout << "Failed\n";
    }
}