C ++函数调用无对象初始化 [英] C++ function called without object initialization

问题描述

为什么会运行以下代码？

  #include< iostream> 
 class A {
 int num; 
 public：
 void foo（）{num = 5; std :: cout<< num =; std :: cout<< num;} 
}; 
 
 int main（）{
 A * a; 
 a-> foo（）; 
 return 0; 
} 
  

输出为

  num = 5 
  

在第10行只获取以下编译器警告：

（ 警告：a在此函数中未初始化 >）

但是根据我的理解，这个代码不应该运行吗？当num不存在，因为没有创建类型A的对象时，如何将值5分配给num？

解决方案

您尚未初始化 * a 。

试试这个：

  #include< iostream> 
 
 class A 
 {
 int num; 
 public：
 void foo（）{std :: cout< num =; num = 5; std :: cout<< num;} 
}; 
 
 int main（）
 {
 A * a = new A（） 
 a-> foo（）; 
 return 0; 
} 
  

---

未初始化指针）可能导致未定义的行为。如果你幸运的话，你的指针指向堆中的一个位置，用于初始化*。 （假设没有异常被抛出，当你这样做）。如果你不幸运，你会覆盖内存的一部分用于其他目的。

这是不安全的代码;一个黑客可能会利用它。

*当然，即使你访问该位置，也不能保证以后不会被初始化 p>

---

Lucky（实际上，lucky使您更难调试程序）：

  //未初始化的内存0x00000042到0x0000004B 
 A * a; 
 // a = 0x00000042; 
 * a =lalalalala; 
 //无发生
  

---

Unlucky（让你更容易调试你的程序，所以我不认为它不幸，真的）：

  void * a; 
 // a =& main; 
 * a =lalalalala; 
 //不好。 *可能*导致崩溃。 
 //也许有人可以告诉我到底会发生什么？ 
  

Why does the following code run?

#include <iostream>
class A {
    int num;
    public:
        void foo(){ num=5; std::cout<< "num="; std::cout<<num;}
};

int main() {
    A* a;
    a->foo();
    return 0;
}

The output is

num=5

I compile this using gcc and I get only the following compiler warning at line 10:

(warning: 'a' is used uninitialized in this function)

But as per my understanding, shouldn't this code not run at all? And how come it's assigning the value 5 to num when num doesn't exist because no object of type A has been created yet?

解决方案

You haven't initialized *a.

Try this:

#include <iostream>

class A
{
    int num;
    public:
        void foo(){ std::cout<< "num="; num=5; std::cout<<num;}
};

int main()
{
    A* a = new A();
    a->foo();
    return 0;
}

---

Not initializing pointers (properly) can lead to undefined behavior. If you're lucky, your pointer points to a location in the heap which is up for initialization*. (Assuming no exception is thrown when you do this.) If you're unlucky, you'll overwrite a portion of the memory being used for other purposes. If you're really unlucky, this will go unnoticed.

This is not safe code; a "hacker" could probably exploit it.

*Of course, even when you access that location, there's no guarantee it won't be "initialized" later.

---

"Lucky" (actually, being "lucky" makes it more difficult to debug your program):

// uninitialized memory 0x00000042 to 0x0000004B
A* a;
// a = 0x00000042;
*a = "lalalalala";
// "Nothing" happens

---

"Unlucky" (makes it easier to debug your program, so I don't consider it "unlucky", really):

void* a;
// a = &main;
*a = "lalalalala";
// Not good. *Might* cause a crash.
// Perhaps someone can tell me exactly what'll happen?

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