指向C ++中对象的指针-堆栈/堆中有什么? [英] Pointers to objects in C++ - what's on the stack/heap?

问题描述

我从Java开始，所以我对以下行中的堆栈/堆发生了什么感到困惑:

I started out with Java, so I am a bit confused on what's going on with the stack/heap on the following line:

string *x = new string("Hello");

其中，x是局部变量.在C ++中，关于该语句，堆栈上是否发生任何事情?我从阅读中知道对象在堆上，但是x呢?在Java中，x将仅在堆栈中保存指向该对象的内存地址，但是我还没有找到一个清晰的源代码来说明C ++中发生的事情.

where x is a local variable. In C++, does ANYTHING happen on the stack at all in regards to that statement? I know from reading it says that the object is on the heap, but what about x? In Java, x would be on the stack just holding the memory address that points to the object, but I haven't found a clear source that says what's happening in C++.

推荐答案

您刚刚创建的任何对象，例如您的示例中的x在堆栈中.但是，对象x只是一个指针，它指向分配的string堆，您可以使用new string("Hello")将其放置在堆上.通常，您不会在C ++中创建像这样的字符串.相反，您将使用

Any object you just created, e.g. x in your example is on the stack. The object x is just a pointer, though, which points to a heap allocated string which you put on the heap using new string("Hello"). Typically, you wouldn't create a string like this in C++, however. Instead you would use

string x("Hello");

这仍将在堆栈上分配x.表示x值的字符是否还驻留在堆栈上还是驻留在堆上，取决于string实现.作为一个合理的模型，您应该假定它们在堆上(某些std::string实现将短字符串放入堆栈对象中，避免了任何堆分配并有助于局部化).

This would still allocate x on the stack. Whether the characters representing x's value also live on the stack or rather on the heap, depends on the string implementation. As a reasonable model you should assume that they are on the heap (some std::string implementation put short string into the stack object, avoiding any heap allocations and helping with locality).

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