是不是更喜欢预增量而不是后增量？

过去的情况是，预增量是首选，因为类上重载的后增量需要返回表示增量之前对象状态的临时副本。

看来这不再是一个严重的问题（只要内联到位），因为我的旧 C++ 编译器（GCC 4.4.7）似乎将以下两个函数优化为相同的代码：

class Int {
    //...
public:
    Int (int x = 0);
    Int & operator ++ ();
    Int operator ++ (int) {
        Int x(*this);
        ++*this;
        return x;
    }
};

Int & test_pre (Int &a) {
    ++a;
    return a;
}

Int & test_post (Int &a) {
    a++;
    return a;
}

这两个函数的最终汇编结果是：

    .cfi_startproc
    .cfi_personality 0x3,__gxx_personality_v0
    pushq   %rbx
    .cfi_def_cfa_offset 16
    .cfi_offset 3, -16
    movq    %rdi, %rbx
    call    _ZN3IntppEv
    movq    %rbx, %rax
    popq    %rbx
    .cfi_def_cfa_offset 8
    ret
    .cfi_endproc

_{If nothing is inlined, however, there seems to still be a benefit to preferring pre-increment to post-increment, since test_post is forced to call out into operator++(int).}

Let's assume operator++(int) is inlined as an idiomatic copy constructor, call to the pre-increment, and return of the copy, as illustrated above. If the copy constructor is inlined or the default copy constructor implementation, is that sufficient information for the compiler to optimize post-increment so that test_pre and test_post become identical functions? If not, what other information is required?

是的。对于内置类型来说应该不重要。对于这样的类型，编译器可以easily分析语义并优化它们——如果这不会改变行为。

然而，对于类类型来说，may（如果不does）很重要，因为在这种情况下语义可能会更复杂。

class X { /* code */ };

X x;

++x;
x++; 

最后two调用可能完全不同，并且可能执行不同的操作，就像这些调用一样：

x.decrement(); //may be same as ++x (cheating is legal in C++ world!)
x.increment(); //may be same as x++

所以不要让自己陷入语法糖的陷阱。