Is it possible to inline a lambda expression?

It is possible that a lambda expression might be inlined. Under the hood, a lambda expression is no different than any other function object.

Whether a particular lambda expression is inlined is entirely dependent upon the compiler and whether it decides it is worth inlining.

The compiler will inline it if it can. For example, in g++ 4.5 with -O2,

#include <vector>
#include <algorithm>

int main () {
    std::vector<int> a(10);
    for (int i = 0; i < 10; ++ i) a[i] = i;

    asm ("Ltransform_begin: nop; nop; nop; nop; nop; nop; ");
    std::transform(a.begin(), a.end(), a.begin(), [] (int x) { return 2*x; });
    asm ("Lforeach_begin: nop; nop; nop; nop; nop; nop; ");
    std::for_each(a.begin(), a.end(), [] (int x) { printf("%d\n", x); });
    asm ("Lforeach_done: nop; nop; nop; nop; nop; nop; ");

    return 0;
}

generates the assembly that the 2*x and printf lambdas are completely inlined.

# 9 "x.cpp" 1
    Ltransform_begin: nop; nop; nop; nop; nop; nop; 
# 0 "" 2
    .align 4,0x90
L13:
    sall    (%rax)
    addq    $4, %rax
    cmpq    %rax, %r12
    jne L13
# 13 "x.cpp" 1
    Lforeach_begin: nop; nop; nop; nop; nop; nop; 
# 0 "" 2
    .align 4,0x90
L14:
    movl    (%rbx), %esi
    leaq    LC0(%rip), %rdi
    xorl    %eax, %eax
LEHB1:
    call    _printf
LEHE1:
    addq    $4, %rbx
    cmpq    %r12, %rbx
    jne L14
# 17 "x.cpp" 1
    Lforeach_done: nop; nop; nop; nop; nop; nop; 
# 0 "" 2

If you have a regular struct functor, the compiler will almost certainly inline it. If you have a C++0x style lambda, the compiler will almost certainly inline it. If you're using a boost::lambda, then it might do, depending on how the lambda works underneath the scenes. Short version: You can't guarantee it's inlining or non-inlining, but you should trust your compiler and if in doubt, make it easy and simple to inline.

The inline keyword does not actually cause functions to be inlined. Any recent compiler is going to make better decisions with regards to inlining than you will.

In the case of a short lambda, the function will probably be inlined.

If you're trying to use the inline keyword with a lambda, the answer is no, you can't use that.