No, C has no support for lambda expressions.

If you're willing to use C++, Boost has a library that emulates lambdas. Also, C++0x will have built-in support for lambda expressions.

There wasn't a huge demand for lambda expression support in C at the time, so the language didn't support it.

This behavior seems to be specfic to newer versions of Clang, and is a language extension called "blocks".

The Wikipedia article on C "blocks" also provides information which supports this claim:

Blocks are a non-standard extension added by Apple Inc. to Clang's implementations of the C, C++, and Objective-C programming languages that uses a lambda expression-like syntax to create closures within these languages. Blocks are supported for programs developed for Mac OS X 10.6+ and iOS 4.0+, although third-party runtimes allow use on Mac OS X 10.5 and iOS 2.2+ and non-Apple systems.

Emphasis above is mine. On Clang's language extension page, under the "Block type" section, it gives a brief overview of what the Block type is:

Like function types, the Block type is a pair consisting of a result value type and a list of parameter types very similar to a function type. Blocks are intended to be used much like functions with the key distinction being that in addition to executable code they also contain various variable bindings to automatic (stack) or managed (heap) memory.

GCC also has something similar to blocks called lexically scoped nested functions. However, there are some key differences also note in the Wikipedia articles on C blocks:

Blocks bear a superficial resemblance to GCC's extension of C to support lexically scoped nested functions. However, GCC's nested functions, unlike blocks, must not be called after the containing scope has exited, as that would result in undefined behavior.

GCC-style nested functions also require dynamic creation of executable thunks when taking the address of the nested function. [...].

Emphasis above is mine.

The problem

C++ includes useful generic functions like std::for_each and std::transform, which can be very handy. Unfortunately they can also be quite cumbersome to use, particularly if the functor you would like to apply is unique to the particular function.

#include <algorithm>
#include <vector>

namespace {
  struct f {
    void operator()(int) {
      // do something
    }
  };
}

void func(std::vector<int>& v) {
  f f;
  std::for_each(v.begin(), v.end(), f);
}

If you only use f once and in that specific place it seems overkill to be writing a whole class just to do something trivial and one off.

In C++03 you might be tempted to write something like the following, to keep the functor local:

void func2(std::vector<int>& v) {
  struct {
    void operator()(int) {
       // do something
    }
  } f;
  std::for_each(v.begin(), v.end(), f);
}

however this is not allowed, f cannot be passed to a template function in C++03.

The new solution

C++11 introduces lambdas allow you to write an inline, anonymous functor to replace the struct f. For small simple examples this can be cleaner to read (it keeps everything in one place) and potentially simpler to maintain, for example in the simplest form:

void func3(std::vector<int>& v) {
  std::for_each(v.begin(), v.end(), [](int) { /* do something here*/ });
}

Lambda functions are just syntactic sugar for anonymous functors.

Return types

In simple cases the return type of the lambda is deduced for you, e.g.:

void func4(std::vector<double>& v) {
  std::transform(v.begin(), v.end(), v.begin(),
                 [](double d) { return d < 0.00001 ? 0 : d; }
                 );
}

however when you start to write more complex lambdas you will quickly encounter cases where the return type cannot be deduced by the compiler, e.g.:

void func4(std::vector<double>& v) {
    std::transform(v.begin(), v.end(), v.begin(),
        [](double d) {
            if (d < 0.0001) {
                return 0;
            } else {
                return d;
            }
        });
}

To resolve this you are allowed to explicitly specify a return type for a lambda function, using -> T:

void func4(std::vector<double>& v) {
    std::transform(v.begin(), v.end(), v.begin(),
        [](double d) -> double {
            if (d < 0.0001) {
                return 0;
            } else {
                return d;
            }
        });
}

"Capturing" variables

So far we've not used anything other than what was passed to the lambda within it, but we can also use other variables, within the lambda. If you want to access other variables you can use the capture clause (the [] of the expression), which has so far been unused in these examples, e.g.:

void func5(std::vector<double>& v, const double& epsilon) {
    std::transform(v.begin(), v.end(), v.begin(),
        epsilon -> double {
            if (d < epsilon) {
                return 0;
            } else {
                return d;
            }
        });
}

You can capture by both reference and value, which you can specify using & and = respectively:

• [&epsilon, zeta] captures epsilon by reference and zeta by value
• [&] captures all variables used in the lambda by reference
• [=] captures all variables used in the lambda by value
• [&, epsilon] captures all variables used in the lambda by reference but captures epsilon by value
• [=, &epsilon] captures all variables used in the lambda by value but captures epsilon by reference

The generated operator() is const by default, with the implication that captures will be const when you access them by default. This has the effect that each call with the same input would produce the same result, however you can mark the lambda as mutable to request that the operator() that is produced is not const.