converting from std::string to uint8

Question

referring to the beautiful solutions provided here, Convert string to int with bool/fail in C++,

I would like to cast a std::string to a 8 bit number (unsigned or signed) The problem is that because 8 bit number is represented as a char so it is parsed wrong
(trying to parse anything above 1 digits - like 10 - fails)

Any ideas?

C++11 has these: en.cppreference.com/w/cpp/string/basic_string/stoul but you'll still need to convert (ie static_cast<uint8_t>) the result and check bounds. — rubenvb
– rubenvb, Commented Mar 7, 2012 at 13:25

Shawnone · Accepted Answer · 2012-03-07 14:37:34Z

4

Use template specialization:

template <typename T>
void Convert(const std::string& source, T& target)
{
    target = boost::lexical_cast<T>(source);
}

template <>
void Convert(const std::string& source, int8_t& target)
{
    int value = boost::lexical_cast<int>(source);

    if(value < std::numeric_limits<int8_t>::min() || value > std::numeric_limits<int8_t>::max())
    {
        //handle error
    }
    else
    {
        target = (int8_t)value;
    }
}

answered Mar 7, 2012 at 14:37

Shawnone

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swegi · Accepted Answer · 2012-03-07 12:57:38Z

1

Parse the number as int and then cast it to uint8_t. You may perform bound checks as well.

answered Mar 7, 2012 at 12:57

swegi

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2 Comments

Boaz Over a year ago

that was my thinking... the problem is it ruins the pretty templating, I'm trying to figure out if there is more elegant solution

Boaz Over a year ago

and adding to the point at hand - I have template function that accepts a <typename T> - What's a good way to exclude int8?

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