Is there a TryParse equivalent in C++ (gcc)?

2019-02-20 05:50发布

Is there a equivalent of TryParse in C++(gcc) ?

I would like to parse a string which may contain (+31321) and store it as long. I know phone numbers are stored as strings and string matched but for my needs , I would like to store them as long and sometimes they might contain pluses (+). What would parse it in C++?

标签: c++ parsing gcc
4条回答
虎瘦雄心在
2楼-- · 2019-02-20 06:19

In fact, before convert the string into a number, the numbers should be "normalized" to a common format. This requires removing all symbols, and replace them with proper representation.

But you must be very care about representing telephone numbers (that are NOT numbers: they are not subjected to regular arithmetic) as string: Number beginning with one or more zeroes are not the same as the ones with the zeroes removed:

00 is the typical replacement for +, but a number without the 00 (or +) in front should be prefixed by 00c where c is the country code.

You need to do some pre-processing to come to a uniform string representation, before converting, otherwise you risk to "alias" different things.

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萌系小妹纸
3楼-- · 2019-02-20 06:25

The trouble with strtoul() and family there is no real way to test for failure.
If it fails to parse then it returns 0 without setting errno (which is only set on overflow).

boost lexical cast

#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";
    }
}

Doing it using streams

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";
    }
}

Using scanf:

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";
    }
}
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We Are One
4楼-- · 2019-02-20 06:27

the input extraction operator >> (i hope it's an acceptable name) applies and returns a stream&, that has a bool operator, meaning the extraction has been succesfully attempted. For instance, from the Cubbi answer:

...
    std::istringstream s("+31321");
    long n;
    if (s >> n)
      std::cout << n << '\n';
....

This will succed, of course, given the appropriate content of s.

Somewhat different (easier but not type safe) also the scanf family has practical tools available in c++ as well c. You could of course write the example this way:

...
    long n;
    if (sscanf("+31321", "%d", &n) == 1)
      std::cout << n << '\n';
...

A subset of regular expressions make this rather powerful: for instance to match a comma separed multi fields with left space trimming:

   if (sscanf("a,b,c", " [^,], [^,], [^,]", a,b,c) == 3) ...
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Melony?
5楼-- · 2019-02-20 06:29

+31321 can be parsed as a long with the usual stream extraction operators.

#include <iostream>
#include <sstream>
int main()
{
    std::istringstream s("+31321");
    long n;
    s >> n;
    std::cout << n << '\n';
}

demo: http://ideone.com/4rmlp

Although parsing an actual phone number (with parentheses, dashes, extensions, etc) may not be as simple.

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