（字符串）基于迭代器到 int 的转换

Question

There are the atox , strtox and stox families that I know of, but I can't seem to find any iterator based string to int conversions in the Standard Library or Boost.

The reason I need them is because I am having a parser whose match result is a range referencing the input string. I might very well have an input string like

...8973893488349798923475...
     ^begin   ^end

so I need 738934883 as an integer.

Of couse, I could first take begin and end to construct an std::string to use with any of above families, but I would very much like to avoid that overhead.

So my question: Is there anything in the Standard Library or Boost accepting iterators as input, or do I have to write my own.

Answer 1

Boost does actually support this, using the Lexical Cast library. The following code uses a substring range to parse the number without performing any dynamic allocation:

#include <boost/lexical_cast.hpp>
#include <string>
#include <iostream>

int convert_strings_part(const std::string& s, std::size_t pos, std::size_t n)
{
    return boost::lexical_cast<int>(s.data() + pos, n);
}

int main(int argc, char* argv[])
{
    std::string s = "8973893488349798923475";

    // Expect: 738934883

    std::cout << convert_strings_part(s, 2, 9) << std::endl;

    return 0;
}

The output (tested on OS X with Boost 1.60):

738934883

The lexical cast library has some great features for conversion to and from strings, though it isn't as well known as some of the others for some reason.

Answer 2

Until gavinb's answer, I was not aware of any such library function. My try would have been this, using any of atox and strtox as follows (you could avoid a dependency on boost library then, if wanted):

::std::string::iterator b; // begin of section
::std::string::iterator e; // end of section, pointing at first char NOT to be evaluated
char tmp = *e;
*e = 0;
int n = atoi(&*b);
*e = tmp;

If you only had const_iterators available, you would have to apply a const_cast to *e before modifying.

Be aware that this solution is not thread safe, though.

Answer 3

You could do it with strstream but it was depracated. Below two examples, with strstream and boost arrays:

http://coliru.stacked-crooked.com/a/04d4bde6973a1972

#include <iostream>
#include <strstream>

 #include <boost/iostreams/device/array.hpp>
 #include <boost/iostreams/stream.hpp>
 #include <boost/iostreams/copy.hpp>

int main()
{
    std::string in = "8973893488349798923475";
    //                  ^^^^^

    auto beg = in.begin()+2;
    auto end = in.begin()+6; 

    // strstream example - DEPRECATED
    std::istrstream os(&*beg, end-beg);
    int n;
    std::string ss;
    os >> n;
    std::cout << n << "\n";


    // Boost example
    namespace io = boost::iostreams;
    int n2;
    io::array_source src(&*beg, end-beg);
    io::stream<io::array_source> os2(src);

    os2 >> n2;
    std::cout << n2 << "\n";

    return 0;
}

Answer 4

使用现代 STL 实现std::string(begin,end)并没有那么糟糕 - SSO 消除了对小于 ~15 个字符（64 位为 22 个）的字符串的任何分配。