c++ 中范围的开始大括号处的存储分配？
在阅读 BRUCE ECKEL'S THINKING IN C++ 时，我遇到了以下文本
在 C++ 中，变量可以在作用域中的任何点定义，因此变量的存储可能会在其定义点之前才被定义。实际上，编译器更有可能遵循 c 中的分配作用域左大括号处的所有存储空间。
疑问：我猜它只用于在堆栈上分配存储，但我的疑问是编译器如何知道在达到其定义之前在 main （或其他 fn ）内部定义了多少个对象（或什至不是一个）以便分配存储在范围的左大括号中。
while reading from BRUCE ECKEL'S THINKING IN C++..i came across the following text
In c++,a variable can be define at any point in a scope,so it might be seem that storage for variable may not be define until its point of definition .It's actually more likely that the compiler will follow in c of allocating all the storage for a scope at the opening brace of scope.
Doubt:I guess its only for storage allocating at stack but my doubt is how compiler get to know that how many objects(or not even a one) is defined inside the main(or other fn) before reaching to its definition in order to allocate storage at the opening brace of scope.
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编译器可以在实际发出任何代码之前分析整个函数。通常，编译器将计算出每个花括号段需要多少存储空间，并在函数序言中添加一条汇编指令，以将堆栈指针调整那么多字节。
然而，实际初始化变量发生在代码中的适当位置。
The compiler can analyze the entire function before actually emitting any code. Typically, the compiler will work out, for each braced segment, how much storage is required, and add a single assembly instruction in the function prologue to adjust the stack pointer by that many bytes.
Actually initializing the variables, however, occurs at the appropriate point in the code.
编译器不一定会逐行解析代码。在编译期间，它将遍历整个范围/函数并确定要在堆栈上分配的内存量。
这就是编译存在的原因之一:)
The compiler does not necessarily parse the code in line by line fashion. During compilation it would have gone through the entire scope/function and determined the amount of memory to allocate on the stack.
That is one of the reason why compilation exists :)