Why isn't this unused variable optimised away?

问题

I played around with Godbolt's CompilerExplorer. I wanted to see how good certain optimizations are. My minimum working example is:

#include <vector>

int foo() {
    std::vector<int> v {1, 2, 3, 4, 5};
    return v[4];
}

The generated assembler (by clang 5.0.0, -O2 -std=c++14):

foo(): # @foo()
  push rax
  mov edi, 20
  call operator new(unsigned long)
  mov rdi, rax
  call operator delete(void*)
  mov eax, 5
  pop rcx
  ret

As one can see, clang knows the answer, but does quite a lot of stuff before returning. It seems to my that even the vector is created, because of "operator new/delete".

Can anyone explain to me what happens here and why it does not just return?

The code generated by GCC (not copied here) seems to construct the vector explicitly. Does anyone know GCC is not capable to deduce the result?

回答1:

std::vector<T> is a fairly complicated class that involves dynamic allocation. While clang++ is sometimes able to elide heap allocations, it is a fairly tricky optimization and you should not rely on it. Example:

int foo() {
    int* p = new int{5};
    return *p;
}

foo():                                # @foo()
        mov     eax, 5
        ret

---

As an example, using std::array<T> (which does not dynamically allocate) produces fully-inlined code:

#include <array>

int foo() {
    std::array v{1, 2, 3, 4, 5};
    return v[4];
}

foo():                                # @foo()
        mov     eax, 5
        ret

---

As Marc Glisse noted in the other answer's comments, this is what the Standard says in [expr.new] #10:

An implementation is allowed to omit a call to a replaceable global allocation function ([new.delete.single], [new.delete.array]). When it does so, the storage is instead provided by the implementation or provided by extending the allocation of another new-expression. The implementation may extend the allocation of a new-expression e1 to provide storage for a new-expression e2 if the following would be true were the allocation not extended: [...]

回答2:

As the comments note, operator new can be replaced. This can happen in any Translation Unit. Optimizing a program for the case it's not replaced therefore requires Whole-Program Analysis. And if it is replaced, you have to call it of course.

Whether the default operator new is a library I/O call is unspecified. That matters, because library I/O calls are observable and therefore they can't be optimized out either.

回答3:

N3664's change to [expr.new], cited in one answer and one comment, permits new-expressions to not call a replaceable global allocation function. But vector allocates memory using std::allocator<T>::allocate, which calls ::operator new directly, not via a new-expression. So that special permission doesn't apply, and generally compilers cannot elide such direct calls to ::operator new.

All hope is not lost, however, for std::allocator<T>::allocate's specification has this to say:

Remarks: the storage is obtained by calling ​::​operator new, but it is unspecified when or how often this function is called.

Leveraging this permission, libc++'s std::allocator uses special clang built-ins to indicate to the compiler that elision is permitted. With -stdlib=libc++, clang compiles your code down to

foo():                                # @foo()
        mov     eax, 5
        ret

来源：https://stackoverflow.com/questions/47072261/why-isnt-this-unused-variable-optimised-away