什么是字面值的范围，以及编译器如何为其分配内存？(What is the scope of a literal value, and how does the compiler allocate me

什么是字面值的范围，以及编译器如何为其分配内存？(What is the scope of a literal value, and how does the compiler allocate memory to it?) int x = 12;

12被认为是整数字面量，因此不能在LValue中使用。

编译器如何为文本分配内存？ 文字的范围是什么？ 为什么我们不能在它的范围内得到一个＆12的地址？ int x = 12;

12 is said to be integer literal, and therefore can't be used in the LValue.

How does the compiler allocate memory to a literals? What is the scope of a literals? Why can't we get its address with an &12 in its scope?

最满意答案

好的问题中的坏榜样。 但问题仍然有效： 咱们试试吧：

Foo getFoo() {return Foo();} int func() { getFoo().bar(); // Creates temporary. // before this comment it is also destroyed. // But it lives for the whole expression above // So you can call bar() on it. } int func2() { Foo const& tmp = getFoo(); // Creates temporary. // Does not die here as it is bound to a const reference. DO STUFF } // tmp goes out of scope and temporary object destroyed. // It lives to here because it is bound to a const reference.

编译器如何将内存分配给临时对象？

未定义的编译器。 _{但是，将更多的内存分配到堆栈帧并将其保留在那里真的很容易。 然后摧毁它并减小堆栈框架的大小（尽管这个答案对底层硬件做了大量的你永远不应该做的假设（最好只是把它看作是编译器在做魔术））。}

什么是临时对象的范围？

临时对象一直存在到表达式的末尾（通常是; ），除非它被绑定到一个const引用。 如果它绑定到一个常量引用，那么它就存在于引用所属范围的末尾（有一些例外（如构造函数））。

为什么我们不能在它的范围内得到一个＆12的地址？

在问题12中不是一个临时对象。 它是一个整数文字。

OK Bad example in the question. But the question is still valid: Lets try:

Foo getFoo() {return Foo();} int func() { getFoo().bar(); // Creates temporary. // before this comment it is also destroyed. // But it lives for the whole expression above // So you can call bar() on it. } int func2() { Foo const& tmp = getFoo(); // Creates temporary. // Does not die here as it is bound to a const reference. DO STUFF } // tmp goes out of scope and temporary object destroyed. // It lives to here because it is bound to a const reference.

How does the compiler allocate memory to a temporary object?

Undefined up-to the compiler. _{But it would be real easy to allocate a tiny bit more memory onto the stack frame and hold it there. Then destroy it and reduce the size of the stack frame (though this answer makes a whole lot of assumptions about the underlying hardware that you should never do (best just to think of it as the compiler doing magic)).}

What is the scope of a temporary object?

The temporary object lives until the end of the expression (usually the ;) unless it is bound to a const reference. If it is bound to a const reference then it lives to then end of the scope that the reference belongs too (with a few exceptions (like constructors)).

Why can't we get its address with an &12 in its scope?

In the question 12 is not a temporay object. It is an integer literal.