Are lambdas inlined like functions in C++?

Question

Can/does the compiler inline lambda functions to increase efficiency, as it might with simple standard functions?

e.g.

std::vector vd;
         


        

Answer 1

It depends on the optimisation level given to the compiler. Take for example, these two functions, which are semantically identical. One is C++11 style, the other C style.

void foo1 (void)
{
    int arr[100];
    std::generate(std::begin(arr), std::end(arr), [](){return std::rand()%100;});
}

void foo2 (void)
{
    int arr[100];
    for (int *i = arr; i < arr+100; i++) *i = std::rand()%100;
}

Compiling this with gcc -O4 emits code which is extremely similar (not identical, but equivalent complexity) for the two functions.

However the lambda is not inlined when compiling unoptimised (and neither are the std::begin and std::end calls).

So although the compiler can (and does) do an excellent job at optimizing the modern style code when asked to do so, there is possibly or probably going to be a performance penalty for this kind of code in an unoptimized debug build.

Answer 2

First off: the whole point of the design of lambdas in C++ is that they don’t have an overhead compared to function calls. That notably includes the fact that calls to them can be inlined.

But there’s a confusion of concepts here: in the C++ standard, “inline” is the linkage of a function, i.e. it is a statement about how a function is defined, not how it gets called. Functions that are defined inline can benefit from a compiler optimisation by which calls to such functions are inlined. It’s a different but highly related concepts.

In the case of lambdas, the actual function being called is a member operator() that is implicitly defined as inline in an anonymous class created by the compiler for the lambda. Calls of the lambda are translated to direct calls to its operator() and can therefore be inlined. I’ve explained how the compiler creates lambda types in more detail in another answer.

Answer 3

In simple cases, like your example, you should expect better performance with lambdas than with function pointers, see

Why can lambdas be better optimized by the compiler than plain functions?

As others have already pointed out, there is no guarantee that your call will be inlined but you have better chances with lambdas. One way of checking whether the call has been inlined is to check the generated code. If you are using gcc, pass the -S flag to the compiler. Of course, it assumes that you can understand the assembly code.

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Update on Sep 11, 2018: Vipul Kumar pointed out two compiler flags in his edit.

GCC -Winline

Warn if a function that is declared as inline cannot be inlined. Even with this option, the compiler does not warn about failures to inline functions declared in system headers.

The compiler uses a variety of heuristics to determine whether or not to inline a function. For example, the compiler takes into account the size of the function being inlined and the amount of inlining that has already been done in the current function. Therefore, seemingly insignificant changes in the source program can cause the warnings produced by -Winline to appear or disappear.

As I understand this, if your function is not declared inline, this compiler flag is most likely not helpful. Nevertheless it is good to know it exists and it partly answers your second question.

The other flag that he pointed out is:

Clang -Rpass=inline

Options to Emit Optimization Reports

Optimization reports trace, at a high-level, all the major decisions done by compiler transformations. For instance, when the inliner decides to inline function foo() into bar() [...]

I haven't used this one myself but based on the documentation it might be useful for your use case.

I personally check the generated assembly whenever it is that important.