Static array of lambda functions (C++)

Question

I'd like to do something like this (inside a class):

static constexpr MyStruct ops[6] = {
    {'+', [&] (double a, double b) { return a+b; } },
    {'-', [&] (double a, double b) { return a-b; } },
    ...
};

Where MyStruct is defined as:

typedef double (*binOp)(double, double);
struct MyStruct {
    char c;
    binOp fn;
};

I also tried:

std::function <double(double,double)> fn;

for the definition of fn, but no luck.

The error I get for the first case is "error: field initializer is not constant" which I don't really get. If I try with std::function it gets worse, since it says: "cannot be initialized by a non-constant expression when being declared".

Why is the lambda function non-constant? Am I missing something?

lambda-expressions currently may not occur inside constant expressions, but that restriction might get removed eventually: isocpp.org/files/papers/N4487.pdf — dyp
– dyp, Commented May 29, 2015 at 11:11
From C++17, please note that lambdas can now be constexpr: stackoverflow.com/questions/6420085/… — Guillaume D
– Guillaume D, Commented Jan 17, 2019 at 15:27
Also note your lambda's capture everything in their scope by reference. Currently they don't use anything so it's not a problem, but capturing lambda's are not convertible to a raw function pointer. — rubenvb
– rubenvb, Commented Mar 17, 2020 at 12:56

Community · Accepted Answer · 2020-03-17 12:49:52Z

When you construct constexpr object, everything you pass into it needs to be a core constant expression, [decl.constexpr]/9:

A constexpr specifier used in an object declaration declares the object as const. Such an object shall have literal type and shall be initialized. If it is initialized by a constructor call, that call shall be a constant expression (5.19).

and, from [expr.const] lambdas are not constant expressions¹:

A conditional-expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine (1.9), would evaluate one of the following expressions:

[...]

a lambda-expression (5.1.2);

[...]

However, that applies only to constexpr and not to const, so you could simply do that instead:

static const MyStruct ops[6] = {
    {'+', [] (double a, double b) { return a+b; } },
    {'-', [] (double a, double b) { return a-b; } },
};

Note: your lambdas don't need to capture anything, so you should just empty the capture list [].

_{¹As dyp points out, there is a proposal to change this: N4487}

Jarod42 · Accepted Answer · 2015-05-29 11:10:51Z

1

capturing lambda cannot decay to function pointer.

and operator to return the function pointer from a (non-capturing) lambda is not constexpr.

answered May 29, 2015 at 11:10

Jarod42

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