Why is there no std::stou?

Question

C++11 added some new string conversion functions:

http://en.cppreference.com/w/cpp/string/basic_string/stoul

It includes stoi (string to int), stol (string t

Answer 1

The most pat answer would be that the C library has no corresponding “strtou”, and the C++11 string functions are all just thinly veiled wrappers around the C library functions: The std::sto* functions mirror strto*, and the std::to_string functions use sprintf.

---

Edit: As KennyTM points out, both stoi and stol use strtol as the underlying conversion function, but it is still mysterious why while there exists stoul that uses strtoul, there is no corresponding stou.

Answer 2

I've no idea why stoi exists but not stou, but the only difference between stoul and a hypothetical stou would be a check that the result is in the range of unsigned:

unsigned stou(std::string const & str, size_t * idx = 0, int base = 10) {
    unsigned long result = std::stoul(str, idx, base);
    if (result > std::numeric_limits<unsigned>::max()) {
        throw std::out_of_range("stou");
    }
    return result;
}

(Likewise, stoi is also similar to stol, just with a different range check; but since it already exists, there's no need to worry about exactly how to implement it.)

Answer 3

unsigned long ulval = std::stoul(buf);
unsigned long mask = ~0xffffffffl;
unsigned int uival;
if( (ulval & mask) == 0 )
    uival = (unsigned int)ulval;
else {
    ...range error...
}

Using masks to do this with the expected value size in bits expressed in the mask, will make this work for 64-bit longs vs 32-bit ints, but also for 32-bit longs vs 32-bit ints.

In the case of 64-bit longs, ~0xffffffffl will become 0xffffffff00000000 and will thus see if any of the top 32 bits are set. With 32-bit longs, it ~0xffffffffl becomes 0x00000000 and the mask check will always be zero.