CODEWITHSUNDEEP
c++macrosc-preprocessor
Somnium
Somnium

Reputation: 1145

C++ define changeable macro

Is it possible in C++ to create variable, which will be expanded after each time it used to different value?

For example, I want that following

#define mytype [smth here]
void foo(mytype a,mytype b,mytype c)

will be expanded into

void foo(mytype1 a,mytype2 b,mytype3 c)

or

void foo(mytype1 a,mytype11 b,mytype111 c)

Upvotes: 1

Views: 300

Answers (4)

user2992792
user2992792

Reputation: 1

#define mytype [smth here]
#define TYPE(a, b) _TYPE(a, b)
#define _TYPE(a, b) a##b
void foo(TYPE(mytype, 1) a, TYPE(mytype, 2) b, TYPE(mytype, 3) c)

Upvotes: -1

Angew is no longer proud of SO
Angew is no longer proud of SO

Reputation: 171097

Here is an example of using Boost.Preprocessor to achieve the function parameter thing:

#include <boost/preprocessor/repetition/enum_binary_params.hpp>

void foo(BOOST_PP_ENUM_BINARY_PARAMS(3, mytype, p));

That will expand to

void foo(mytype0 p0, mytype1 p1, mytype2 p2);

Upvotes: 3

king_nak
king_nak

Reputation: 11513

If it should only increase when outputting, you can define a struct overriding the operator<<, e.g.:

struct myvarstruct {
  int i;
  myvarstruct() : i(1){}
} myvar;

ostream &operator << (ostream &o, myvarstruct &s) {
  return o << "myvar" << s.i++;
}

Upvotes: 0

Luchian Grigore
Luchian Grigore

Reputation: 258548

You can't have a macro that changes what it expands to, but you can make what it expands to mean different things:

std::string foo()
{
    static int x;
    return std::string("myvar") + std::to_string(x++) + "\n";
}

#define myvar foo()

and

cout<<myvar;
cout<<myvar;

will print out 

myvar1
myvar2

Upvotes: 1

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