将字符范围转换为int

Question

Suppose we have a character range:

// For example
const char *valueBegin="123blahblah"; // Beginning of value
const char *valueEnd=valueBegin+3;    // One past end of value

..., and I want to convert it to an int:

int value=...// Given valueBegin and valueEnd, calculate 
             // the number stored starting at valueBegin

What are some good C++11 ways to do that?

Obviously you can create an std::string and use stoi , or copy it to a temporary NUL-terminated character array and then it's easy (eg, via atoi or strtol ).

Think about a way that doesn't involve copying the characters to some temporary array/object - in other words a function that works on the character data in-place.

Update :

Lots of answers, but please think before you answer. The range is not NUL terminated, hence the need for valueEnd . You don't know what is beyond the value (ie, perhaps valueEnd is beyond the buffer containing the value), so if your answer does not use valueEnd , it is wrong. Also, if your answer creates a temporary std::string object, it is not within the guidelines of this question.

Answer 1

Use boost::lexical_cast :

std::cout << boost::lexical_cast<int>(sBegin, 3) << std::endl;

This does not create any temporaries and supports any kind of character range. It's also quite fast .

If you want to avoid the length specifier then you can use boost::iterator_range :

std::cout << boost::lexical_cast<int>(boost::make_iterator_range(begin, end)) << std::endl;

Answer 2

Without any error checks, this seems simple enough.

int number = 0;
for ( char* cp = valueBegin; cp != valueEnd; ++cp )
{
   number += number*10 + (cp-'0');
}

If you need to want to stop at the first non-digit,

int number = 0;
for ( char* cp = valueBegin; cp != valueEnd && isdigit(*cp); ++cp )
{
   number += number*10 + (cp-'0');
}

If you need to be able extract negative numbers and be able to gracefully deal with leading +/- signs, the code will become a little bit more complex.

int number = 0;
char* cp = valueBegin;
int sign = 1;
if ( *cp == '-' )
{
   sign = -1;
   ++cp;
}
if ( *cp == '+' )
{
   ++cp;
}

for ( ; cp != valueEnd && isdigit(*cp); ++cp )
{
   number += number*10 + (cp-'0');
}
number *= sign;

Answer 3

If you don't want to use Boost and don't mind needing to copy the digits to a temporary string, you can do this with std::stoi :

std::cout << std::stoi(std::string(valueBegin, valueEnd)) << std::endl;

There's other converters for different numeric types as well - stol , stoul , stoll , stoull , stof , stod , stold .

These are defined in C++11.

(Personally, I would take the Boost option, but some people might not want the additional dependency.)