Is there some way to define a variable as a function such that calling the variable at some given time will return the function's output at that time?

Essentially, I'm trying to do something like

#define foobar foo.bar()

But without the use of #define, so I can write something along the lines of

double foobar = foo.bar();

Obviously, compiling the code above will just define foobar as whatever foo.bar() returns at the time of definition. What I want to do is the above in such a way that using foobar at some time in the code will just use whatever foo.bar() returns at that time, and not whatever it was at definition of foobar.

Solution

Obviously, compiling the code above will just define foobar as whatever foo.bar() returns at the time of definition. What I want to do is the above in such a way that using foobar at some time in the code will just use whatever foo.bar() returns at that time, and not whatever it was at definition of foobar.

You want a function not a variable:

auto foobar() { return foo.bar(); }

If foo is not a global (i hope so) and you want to declare the callable on the fly just as you can declare a double, you can use a lambda expression:

Foo foo;
auto foobar = [&foo](){ return foo.bar(); };

// call it:
foobar();

To call the function without the function call syntax () you could use a custom type that calls the function when converted to the returned type. However, as this is non-idiomatic obfuscation, I am not going into more details.