Question

I have a std::map<string, double> whose members look something like:

X = [{"N", 200}, {"sigma", 1.0}, {"T", .2}]

Now, given the struct foo Y:

struct foo {
    int N;
    double T;
};

Can I programmatically map the key/value pairs X -> Y without writing a custom class for each X -> Y type mapping? Note that:

1. X["sigma"] is not in Y, i.e. the mapping is not necessarily one-to-one
2. The type of Y.N is an int while X["N"] is a double.

I suspect the answer is no, unless some trickery is done at compile time.

Edit: It may not be clear what I’m looking for. A pseudo-code version for this example would look something like:

if("N" in X) -> Y.N = X["N"];
if("T" in X) -> Y.T = X["T"];

Or programmatically:

for key in Y:
    if (key in X) -> Y.key = X[key]

Answer 1

No. C++ has no concept of reflection. At compile time, there’s no "foo::N" string anymore. The compiler has converted all occurances of foo::N in the source code to a 0 byte offset within Foo objects. Also, you cannot enumerate class members at compile time.