On the std::abs function

Question

Is the std::abs() function well defined for ALL arithmetic types in C++11 and will return |x| with no problem of approximation?

A weird thing is that with g++4.7, std::abs(char), std::abs(short int), std::abs(int), std::abs(long int) and std::abs(long long int) seem to return a double (on the contrary of : http://en.cppreference.com/w/cpp/numeric/math/abs). And if the number is casted to a double, we could have some approximation error for very large number (like -9223372036854775806LL = 2^63-3).

So do I have the guarantee that std::abs(x) will always return |x| for all arithmetic types ?

EDIT : here is an example program to make some tests

#include <iostream>
#include <iomanip>
#include <cmath>
#include <typeinfo>

template<typename T>
void abstest(T x)
{
    static const unsigned int width = 16;
    const T val = x;
    if (sizeof(val) == 1) {
        std::cout<<std::setw(width)<<static_cast<int>(val)<<" ";
        std::cout<<std::setw(width)<<static_cast<int>(std::abs(val))<<" ";
    } else {
        std::cout<<std::setw(width)<<val<<" ";
        std::cout<<std::setw(width)<<static_cast<T>(std::abs(val))<<" ";
    }
    std::cout<<std::setw(width)<<sizeof(val)<<" ";
    std::cout<<std::setw(width)<<sizeof(std::abs(val))<<" ";
    std::cout<<std::setw(width)<<typeid(val).name()<<" ";
    std::cout<<std::setw(width)<<typeid(std::abs(val)).name()<<std::endl;
}

int main()
{
    double ref = -100000000000;
    abstest<char>(ref);
    abstest<short int>(ref);
    abstest<int>(ref);
    abstest<long int>(ref);
    abstest<long long int>(ref);
    abstest<signed char>(ref);
    abstest<signed short int>(ref);
    abstest<signed int>(ref);
    abstest<signed long int>(ref);
    abstest<signed long long int>(ref);
    abstest<unsigned char>(ref);
    abstest<unsigned short int>(ref);
    abstest<unsigned int>(ref);
    abstest<unsigned long int>(ref);
    abstest<unsigned long long int>(ref);
    abstest<float>(ref);
    abstest<double>(ref);
    abstest<long double>(ref);
    return 0;
}

Answer 1

You cannot guarantee that std::abs(x) will always return |x| for all arithmetic types. For example, most signed integer implementations have room for one more negative number than positive number, so the results of abs(numeric_limits<int>::min()) will not equal |x|.

Answer 2

It's not weird that g++ (with C++11 standard) returns a double when you use std::abs from <cmath> with an integral type: From http://www.cplusplus.com/reference/cmath/abs/:

Since C++11, additional overloads are provided in this header (<cmath>) for the integral types: These overloads effectively cast x to a double before calculations (defined for T being any integral type).

This is actually implemented like that in /usr/include/c++/cmath:

template<typename _Tp>
inline _GLIBCXX_CONSTEXPR
typename __gnu_cxx::__enable_if<__is_integer<_Tp>::__value,
                                double>::__type
abs(_Tp __x)
{ return __builtin_fabs(__x); }

Answer 3

The correct overloads are guaranteed to be present in <cmath>/<cstdlib>:

C++11, [c.math]:

In addition to the int versions of certain math functions in <cstdlib>, C++ adds long and long long overloaded versions of these functions, with the same semantics.

The added signatures are:

long abs(long);            // labs()
long long abs(long long);  // llabs()

[...]

In addition to the double versions of the math functions in <cmath>, overloaded versions of these functions, with the same semantics. C++ adds float and long double overloaded versions of these functions, with the same semantics.

float abs(float);
long double abs(long double);

So you should just make sure to include correctly <cstdlib> (int, long, long long overloads)/<cmath> (double, float, long double overloads).

Answer 4

Check that you're in fact using std::abs from <cstdlib> and not std::abs from <cmath>.

PS. Oh, just saw the example program, well, there you go, you are using one of the floating point overloads of std::abs .