C ++ SegFault，当解除引用cout的指针时 [英] C++ SegFault when dereferencing a pointer for cout

问题描述

我是C ++的新手，只是试图搞定它。它通常似乎不太糟糕，但我偶然发现了这种奇怪的/病理性的错误行为：

I'm new to C++ and just trying to get a hang of it. It generally seems not too bad, but I stumbled upon this weird/pathological segfaulting behavior:

int main () {
  int* b;
  *b = 27;
  int c = *b;
  cout << "c points to " << c << endl; //OK                                                                                                                                      
  printf( "b points to %d\n", *b); //OK                                                                                                                                          
  // cout << "b points to " << (*b) << endl; - Not OK: segfaults!                                                                                                               
  return 0;
}

这个程序产生了你期望的结果：

This program, as given, produces what you'd expect:

c points to 27
b points to 27

另一方面，如果你取消第二行到最后一行的注释，你会得到一个程序在运行时崩溃（seg-fault）。为什么？这是一个有效的指针。

On the other hand, if you uncomment the second-to-last line, you get a program that crashes (seg-fault) in runtime. Why? This is a valid pointer.

推荐答案

int * b 指向未知的内存地址，初始化。如果你把它初始化为你的编译器存在的任何空值（C ++中为0或NULL，C ++ 0x中为nullptr），你最好能早一些获得segfault。问题在于你为指针分配空间，而不是它指向的数据。如果您改为这样做：

int* b points to an unknown memory address because it wasn't initialized. If you initialized it to whatever null value exists for your compiler (0 or NULL in C++, nullptr in C++0x), you'd most certainly get a segfault earlier. The problem lies in the fact that you allocated space for the pointer but not the data it points to. If you instead did this:

int c = 27;
int* b = &c;

cout << "c points to " << c << endl;
printf ("b points to %d\n", *b);
cout << "b points to " << (*b) << endl;

事情会工作，因为 int * b 到您的程序可访问的内存位置（因为内存实际上是您的程序的一部分）。

Things would work because int* b refers to a memory location that is accessible by your program (since the memory is actually a part of your program).

如果您保留一个指针未初始化或分配空值它，你不能使用它，直到它指向一个内存地址，你知道你可以访问。例如，使用动态分配与 new 运算符将为您保留数据的内存：

If you leave a pointer uninitialized or assign a null value to it, you can't use it until it points to a memory address that you KNOW you can access. For example, using dynamic allocation with the new operator will reserve memory for the data for you:

int* b = new int();
*b = 27;
int c = *b;

//output

delete b;

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