c++ Converting a std::string to int, double, etc WITHOUT using a char array

Question

In C++ I have only seen this done by converting the string object into an array of characters. The tutorials with an array are a bit hard for me to understand. But I want to do the conversion without the array.

I do have an idea how to do it: the string is "1234". After that I convert this text to an integer this way:

if (symol4 == "4") int_var += 4 * 1;
if (symol3 == "3") int_var += 3 * 10;
if (symol3 == "2") int_var += 2 * 100;
if (symol3 == "1") int_var += 1 * 1000; //Don't worry, I'm familiar with cycles, this code is only for explaining my algorithm

I hope you can understand the idea.

But I don't know if this is the best way. I don't know if there is a library that has a function that allows me to do that (I won't be surprised if there is one).

I don't know if not using a char array is a good idea. But that's a different question that I'm going to ask later.

What's the best way to convert a string to an integer, double, etc WITHOUT using an array of characters.

Answer 1

boost::lexical_cast to the rescue: int result = boost::lexical_cast<int>(input)

If you don't want to rely on boost, you can use a stringstream, something like:

std::stringstream ss;
int result;
ss << input;
ss >> result;

but that's rather roundabout imo

And no don't use atoi - that function was flawed even back in C and it hasn't gotten better with time. It returns 0 when an error happened while parsing - which has the obvious problem how you distinguish an error from parsing the string "0" .

Answer 2

I really can't get what your pasted code is about, but in C++ the best way to convert string to integer or float is to use stringstream .

const char* str = "10 20.5";
std::stringstream ss(str);
int x;
float y;

ss >> x >> y;

Answer 3

There is a function atoi which you can use. This converts it to a character array, but you don't have to do the math involved with indexing the array in a for loop.

#include <stdlib.h>
...
String number = "1234";    
int value = atoi(number.c_str());
std::cout << number;
...

For the atoi nay sayers, hopefully he'll understand this >.>

#include <boost/lexical_cast.hpp>

try {
    int x = boost::lexical_cast<int>( "123" );
} catch( boost::bad_lexical_cast const& ) {
    std::cout << "Error: input string was not valid" << std::endl;
}

The best way is the most efficient way, I don't think you'll find a better alternative to this, or using a character array.

Answer 4

The standard string class already has a member function that gives you access to the internal character array, c_str(), so you can just pass this to one of the standard C library functions that parse integers, such as strtol():

string s = "1234";
long n = strtol(s.c_str(), 0, 10);

That's the simplest code if you already know the string is a valid integer and don't care about error checking. If you want full error checking you would do something like this:

char* end = 0;
errno = 0;
long n = strtol(s.c_str(), &end, 10);
if (end == 0 || *end == 0)
    throw invalid_argument("Not a number");
else if (errno == ERANGE)
    throw overflow_error("Number is out of range");
else if (errno != 0)
    throw invalid_argument("Not a number");

Alternatively you could use C++ streams if you want to avoid C style character arrays completely (or rather, hide them completely inside the classes):

istringstream in(s);
int n;
in >> n;

You could also use boost::lexical_cast, which does basically the same thing.

Answer 5

• I recommend Boost.Lexical_Cast

• Or see the upcoming Boost.Conversion

• Can also be achieved using Boost.Spirit, but is somewhat more complex

See "The String Formatters of Manor Farm" article by Herb Sutter.

Answer 6

您可能需要查看atoi函数。