60

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?

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12
  • Also, I don't see int (*X); anywhere. Commented May 1, 2012 at 22:49
  • It's not in there as such. While trying to figure out the syntax, I was a little stumped that int (*X); compiles, but not really sure what it defined.. Commented May 1, 2012 at 22:51
  • 4
    int (*X)(<arguments>); is a prototype for a function pointer by the name of X. In other words: X is a pointer to a function accepting <arguments> and returning a int. Commented May 1, 2012 at 22:52
  • So, let's say we just have int (*x) - this is a pointer to a function with... a variable list of arguments, no arguments? Commented May 1, 2012 at 22:54
  • 2
    To make gcc a strict C89 (or C99) compiler, use -std=c89 -pedantic. Commented May 1, 2012 at 22:56

4 Answers 4

Reset to default
48

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.

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6 Comments

Such a useful functionality, When will this be implemented into ANSI C?
It is syntactically valid GNU C, but is it really semantically valid? When the block containing the function definition exits (which is before the any actual use of the "lambda expression" can occur), wouldn't the function trampoline be invalidated along with it?
@Dolda2000 Data created in a statement expression probably has static duration. After all, it's no longer ANSI C. It's GNU C. GNU defines that standard.
@BradenBest: It certainly does not have static duration, because a pointer to a nested function is invalid after the containing function returns. Since a nested function has access to all variables that are lexically in its scope, it wouldn't be strange if its trampoline is allocated with the same duration as those variables. The manual is a bit unclear on the issue, though.
@Pacerier statement expression is supported by gcc and clang so it's quite portable and you can safely use it (even linux kernel code uses it)
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7

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.

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Comments

1

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.

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Comments

0

  1. int (*max)(int, int) is the type of variable you are declaring. It is defined as a function pointer named max which returns int, and takes two ints as parameters.

  2. __fn__ refers to the function name, which in this case is max.

  3. I don't have an answer there. I would imagine you can step through it if you have run it through the preprocessor.

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3 Comments

wrt 2, I just don't understand why does int (*max)(int, int) = ({ int __fn__ (int x, int y) { return x > y ? x : y; } __fn__; }); compile, but void test() { puts("hello"); } test; not compile?
@B.VB. I would make the code look exactly alike. void (*test)() = ({ void test() {puts("hello");} test; }); or if that doesn't work, void (*test)() = ({ void __fn__() { puts("hello");} __fn__; });
__fn__ is just an arbitrary name for the function defined inside the block. The naming is unfortunately confusing.

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