Redefining for loops with the preprocessor

Question

I am looking to do something somewhat evil. I want to redefine for loops to modify the conditional. Is there any way to do this?

From the GCC documentation, redefining keywords is supported: http://gcc.gnu.org/onlinedocs/cpp/Macros.html

What I'm trying to do is make a "probabilistic" wrapper for C++, to see if I can make anything interesting with it.

#include <iostream>
#include <cstdlib>

#define P 0.85

#define RANDOM_FLOAT  ((float)rand()/(float)RAND_MAX)

#define RANDOM_CHANCE (RANDOM_FLOAT < P)

#define if(a) if((a) && RANDOM_CHANCE)

#define while(a) while((a) && RANDOM_CHANCE)


// No more for loops or do-while loops
#define do
#define goto

// Doesn't work :(
//#define for(a) for(a) if(!RANDOM_CHANCE) { break; } 


int main() {
    srand(time(NULL));

    //Should output a random list of Y's and N's
    for(int i=0; i < 100; ++i) {
        if(i < 100) {
            std::cout << "Y";
        } else {
            std::cout << "N";
        }
    }

    std::cout << std::endl;

    // Will loop for a while then terminate
    int j = 0;
    while(j < 100) {
        ++j;
        std::cout << j << "\n";
    }

   std::cout << std::endl;

   return 0;
}

A perhaps more legitimate use would be to count the number of loop iterations in a running program, for example by mapping

for(int i=0; i < 10; ++i)

to

for(int i=0; i < 10; ++i, ++global_counter)

Is it possible to accomplish what I am trying to do?

EDIT: Thanks for the responses - I really appreciate it!

One of the things I can do with this is simulate a fair coin (effectively force P to be 0.5), by noting that if you have a biased coin, Pr(HT) == Pr(TH). I didn't discover this trick, but it's useful. So that means that you can approximate any probability distribution for P.

bool coin() {
  bool c1 = false;
  bool c2 = false;

  if(true) {
    c1 = true;
  }
  if(true) {
    c2 = true;
  }

  //If they have different faces.
  bool valid = (c1 && !c2) || (!c1 && c2);
  bool result = c1;

  if(valid) { return result; }

  return coin();
}

If anyone else might need to maintain your code, make sure they don't know where you live. — Mike Seymour, Mar 06 '12 at 16:51

score 8 · Accepted Answer · answered Mar 06 '12 at 17:04

Have you tried adding an else to your for loop? That should make it work.

#define for(a) for(a) if(!RANDOM_CHANCE) { break; } else

for (int i = 0; i < 10; ++i)
{
    // do something
}

/* should roughly compile to:
for (int i = 0; i < 10; ++i) if(!RANDOM_CHANCE) { break; } else
{
    // do something
}

or (differing only in whitespace):

for (int i = 0; i < 10; ++i)
    if(!RANDOM_CHANCE)
    {
        break;
    }
    else
    {
        // do something
    }
*/

+1 I was about to click the *Post* button with the exact same answer. — netcoder, Mar 06 '12 at 17:07

Sebastian Mach · Answer 2 · 2012-03-06T17:24:31.300

4

You showed it already. Here's a compilable example:

#include <iostream>

unsigned globalCounter = 0;

#define for(x)  for(x, ++globalCounter)

int main () {
    for (int i=0; i<10; ++i);
    for (int i=0; i<10; ++i);

    std::cout << globalCounter << '\n';
}

This outputs 20. However, existing code like

for (int i=0; i<x; ++i, foobar+=20)

will break as it now passes two arguments to the for-macro.

What you would need is a macro overloads or variadic macros. The latter is supported in C99 and C++11 derives it, so if you want the evil to happen:

unsigned globalCounter = 0;
#define for(...)  for(__VA_ARGS__, ++globalCounter)

#include <iostream>

void main () {
    int another = 0;
    for (int i=0; i<10; ++i, ++another);
    for (int i=0; i<10; ++i, ++another);

    std::cout << globalCounter << ' ' << another << '\n';
}

then

g++ --std=c++0x source.cc
./a.out
20 20

I do not support this. Macros may harm your children.

edited Mar 06 '12 at 17:24

answered Mar 06 '12 at 16:57

Sebastian Mach

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Make it variadic and the last snippet won't break it. `#define for(...) for(__VA_ARGS__, ++globalCounter)`. (C++11). – R. Martinho Fernandes Mar 06 '12 at 17:11
@R. Martinho Fernandes: But C++ does to my knowledge not have variadic macros, that's why I added a C99 answer. – Sebastian Mach Mar 06 '12 at 17:15
Like I mentioned, variadic macros were added in C++11. – R. Martinho Fernandes Mar 06 '12 at 17:16
@R.MartinhoFernandes: Cannot find it in n3242, through. – Sebastian Mach Mar 06 '12 at 17:17
@R.MartinhoFernandes: Oh, didn't know this. [Reference](http://stackoverflow.com/a/9588701/76722), thanks for the pointer. – Sebastian Mach Mar 06 '12 at 17:23

score 1 · Answer 3 · answered Mar 06 '12 at 16:59

It doesn't work because the block does not belong to the for statement anymore. The transformation becomes

for (int i=0; i < 100; ++i)
{
   if (!RANDOM_CHANCE)
   {
      break;
   }
}

// unrelated to the 'for' above.
{
    if(i < 100) {
        std::cout << "Y";
    } else {
        std::cout << "N";
    }
}

Instead, you could just write

#define for(...) for(__VA_ARGS__) if(RANDOM_CHANCE)

So the code becomes

for (int i=0; i < 100; ++i)
    if (RANDOM_CHANCE)
    {
        if(i < 100) {
            std::cout << "Y";
        } else {
            std::cout << "N";
        }
    }

(Note that I have used variadic macros.)

And for the global_counter thing, if you can guarantee that the increament part will never be empty, then it's easy:

#define for(...) for(__VA_ARGS__, ++ global_counter)

Otherwise (e.g. there is for(;;) in the code), I don't think that's possible.

I think you can fix the first issue by adding an `else` to the macro (see my answer). — Platinum Azure, Mar 06 '12 at 17:07

Redefining for loops with the preprocessor

3 Answers3