Exponential Operator in C++

Question

I am taking a class in C++ and I noticed there are only a few math operators to use. I also noticed that C++ does not come with an exponential operator within its math libr

Answer 1

You don't write a function for this (unless you're insane, of course). There's a perfectly good pow function defined in the <cmath> header.

Aside: if you try to use ^ as a power operator, as some people are wont to do, you'll be in for a nasty surprise. It's the exclusive-or (XOR) operator (see here).

Answer 2

Most C operations readily intended to mapped to a single processor instruction when C was invented. At the time, exponentiation was not a machine instruction, thus the library routine.

Answer 3

What platform and which compiler are you using? My guess is TurboC. Mostly cmath file has most of the mathematical functions covered in other compilers.

Answer 4

Python has the ** operator for exponentiation. In C++ you can actually define an operator like that with some trickery. By combining the unary * operator with the binary * operator, like this:

#include <cmath>
#include <iostream>

struct Num {
    double value;

    Num(double value) : value(value) { }

    typedef Num* HalfStar;

    HalfStar operator*() const { return HalfStar(this); }
    Num operator*(const HalfStar& rhs) const
    {
        return Num(std::pow(value, rhs->value));
    }
    Num operator*(const Num& rhs) const
    {
        return Num(value * rhs.value);
    }

    friend std::ostream& operator<<(std::ostream& os, const Num& n)
    {
        return os << n.value;
    }
};

int main(int argc, char**argv)
{
    Num a = 10;
    Num b = 9;

    std::cout << "a*b = " << (a*b) << "\n";
    std::cout << "a**b = " << (a**b) << "\n";

    return 0;
}

This will not work with non-class types though, so you can not do: x**2.

See here for a live example.

Answer 5

According to Bjarne Stroustrup in his book The design and evolution of C++. They decided to avoid exponential operator because :

• An operator provides notational convenience, but does not provide any new functionality. Members of the working group, representing heavy users of scientific/engineering computation, indicated that the operator syntax provides minor syntactic convenience.
• Every user of C++ must learn this new feature
• Users have stressed the importance of susbtituting their own specialized exponentiation functions for the system default, which would not be possible with an intrinsic operator
• The proposal is not sufficiently well motivated. In particular, by looking at one 30000 line Fortran program one cannot conclude that the operator would be widely used in C++
• The proposal requires adding a new operator and adding another precedence level thus increasing the complexity of the language