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I have been looking at the new constexpr feature of C++ and I do not fully understand the need for it.

For example, the following code:

constexpr int MaxSize()
{
    ...

    return ...;
}

void foo()
{
    int vec[MaxSize()];
}

can be replaced by:

int MaxSize()
{
    ...

    return ...;
}

static const int s_maxSize = MaxSize();

foo()
{
    int vec[s_maxSize];
}

Update

The second example is actually not standard ISO C++ (thanks to several users for pointing this out) but certain compilers (e.g. gcc) support it. So it is not const that makes the program valid, but the fact that gcc supports this non-standard feature. (To my knowledge, this is possible only when the array is defined as local to a function or method, since the size of a global array must still be known at compile time.) If I compile without the options -std=c++98 -pedantic-errors, even the code

int MaxSize()
{
    return 10;
}

void foo()
{
    int vec[MaxSize()];
}

will compile with gcc.

So I will try to rephrase my question taking into account the feedback that came so far (and also some further reading I have done in the mean time).

I use the const keyword heavily. With const I can define a constant that has a certain value during its whole lifetime. A constant can be initialized with any expression, which is evaluated once, namely when the constant is created. For these cases, I think that constexpr is pretty useless: it would introduce a very small optimization in that the expression defining the constant value would be computed at compile time instead of run time. Every time I need a run-time constant with a complex initialization I use the keyword const.

So constexpr may come in handy in situations where we need to initialize a constant at compile time. One example is a vector definition: the standard does not support defining the size at runtime. Another example is a template with one or more non-type parameters.

In such cases I normally use macros:

#define MAX_SIZE (10)

void foo()
{
    int vec[MAX_SIZE];
}

but, if I understand correctly, constexpr functions are more powerful than macros, since they allow recursive calls of constexpr functions in their definition. However, I cannot think of any practical application in which I ever wanted to use such a complex computation to define a compile-time constant.

So, even if it may be an interesting feature, I still wonder if it is needed (i.e. how often it can solve situations where macros are not enough). Maybe looking at a few real-life examples that cannot be solved with macros would help me to change this opinion.

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int vec[s_maxSize]; is actually illegal in the second example, so that is not possible to do in C++. But your first example is perfectly legal C++0x.

So there's your answer. You can't actually do what you propose in C++. It can only be done in C++0x with constexpr.

I would also like to point out, that this code also works in C++0x. Doing this in C++ would require some really fancy class templates.

constexpr unsigned int gcd(unsigned int const a, unsigned int const b)
{
   return (a < b) ? gcd(b, a) : ((a % b == 0) ? b : gcd(b, a % b));
}

char vec[gcd(30, 162)];

Of course, in C++0x you still have to use the ternary operator instead of if statements. But, it works and is still a lot easier to understand than the template version you'd be force to use in C++:

template <unsigned int a, unsigned int b>
class gcdT {
 public:
   static unsigned int const value = gcdT<b, a % b>::value;
};

template <unsigned int a>
class gcdT<a, 0> {
 public:
   static unsigned int const value = a;

};

char vec[gcdT<30, 162>::value];

And then, of course, in C++ you'd still have to write the gcd function if you needed to compute things at runtime because the template can't be used with arguments that vary at runtime. And C++0x would have the additional optimization boost of knowing that the result of the function is completely determined by the passed in arguments, which is a fact that can only be expressed with compiler extensions in C++.


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