Anonymous functions using GCC statement expressions

Question

This question isn't terribly specific; it's really for my own C enrichment and I hope others can find it useful as well.

Disclaimer: I know many will have the impulse to respond with "if you're trying to do FP then just use a functional language". I work in an embedded environment that needs to link to many other C libraries, and doesn't have much space for many more large shared libs and does not support many language runtimes. Moreover, dynamic memory allocation is out of the question. I'm also just really curious.

Many of us have seen this nifty C macro for lambda expressions:

#define lambda(return_type, function_body) \
({ \
      return_type __fn__ function_body \
          __fn__; \
})

And an example usage is:

int (*max)(int, int) = lambda (int, (int x, int y) { return x > y ? x : y; });
max(4, 5); // Example

Using gcc -std=c89 -E test.c, the lambda expands to:

int (*max)(int, int) = ({ int __fn__ (int x, int y) { return x > y ? x : y; } __fn__; });

So, these are my questions:

1. What precisely does the line int (*X); declare? Of course, int * X; is a pointer to an integer, but how do these two differ?

2. Taking a look at the exapnded macro, what on earth does the final __fn__ do? If I write a test function void test() { printf("hello"); } test; - that immediately throws an error. I do not understand that syntax.

3. What does this mean for debugging? (I'm planning to experiment myself with this and gdb, but others' experiences or opinions would be great). Would this screw up static analyzers?

Answer 1

This declaration (at block scope):

int (*max)(int, int) =     ({     int __fn__ (int x, int y) { return x > y ? x : y; }     __fn__;     }); 

is not C but is valid GNU C.

It makes use of two gcc extensions:

1. nested functions
2. statement expressions

Both nested functions (defining a function inside a compound statement) and statement expressions (({}), basically a block that yields a value) are not permitted in C and come from GNU C.

In a statement expression, the last expression statement is the value of the construct. This is why the nested function __fn__ appears as an expression statement at the end of the statement expression. A function designator (__fn__ in the last expression statement) in a expression is converted to a pointer to a function by the usual conversions. This is the value used to initialize the function pointer max.

Answer 2

Your lambda macro exploits two funky features. First it uses nested functions to actually define the body of your function (so your lambda is not really anonymous, it just uses an implicit __fn__ variable (which should be renamed to something else, as double-leading-underscore names are reserved for the compiler, so maybe something like yourapp__fn__ would be better).

All of this is itself performed within a GCC compound statement (see http://gcc.gnu.org/onlinedocs/gcc/Statement-Exprs.html#Statement-Exprs), the basic format of which goes something like:

({ ...; retval; }) 

the last statement of the compound statement being the address of the just-declared function. Now, int (*max)(int,int) simply gets assigned the value of the compound statement, which is now the pointer to the 'anonymous' function just declared.

Debugging macros are a royal pain of course.

As for the reason why test; .. at least here, i get the 'test redeclared as different type of symbol', which I assume means GCC is treating it as a declaration and not a (useless) expression. Because untyped variables default to int and because you have already declared test as a function (essentially, void (*)(void)) you get that.. but I could be wrong about that.

This is not portable by any stretch of the imagination though.

Answer 3

Partial answer: It isn't int(*X) you are interested in. It is int (*X)(y,z). That is a function pointer to the function called X which takes (y,z) and returns int.

For debugging, this will be really hard. Most debuggers can't trace through a macro. You would most likely have to debug the assembly.