除了你自己的程序，我不认为你会破坏任何东西，在最坏的情况下，你会尝试从一个内存地址读取或写入一个页面，内核没有分配给你的进程，产生适当的异常并被杀死(我的意思是，你的进程)。

我只是想为这个问题添加一些实际的例子——想象一下下面的代码:

#include <stdio.h>

int main(void) {
    int n[5];
    n[5] = 1;

    printf("answer %d\n", n[5]);

    return (0);
}

它具有未定义的行为。例如，如果你启用了clang优化(-Ofast)，结果会是这样的:

answer 748418584

(如果你没有编译，可能会输出答案1的正确结果)

这是因为在第一种情况下，赋值给1的函数实际上从未在最终代码中组装(您也可以查看godbolt asm代码)。

(然而，必须注意的是，根据这种逻辑，main甚至不应该调用printf，所以最好的建议是不要依赖于优化器来解决你的UB -而是要知道有时它可能会以这种方式工作)

这里的结论是，现代C优化编译器将假设未定义的行为(UB)永远不会发生(这意味着上面的代码将类似于(但不相同):

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    int n[5];

    if (0)
        n[5] = 1;

    printf("answer %d\n", (exit(-1), n[5]));

    return (0);
} 

相反，这是完全定义的)。

这是因为第一个条件语句从来没有达到它的真状态(0总是假的)。

在printf的第二个参数上，我们有一个序列点，在此之后，我们调用exit，程序在调用第二个逗号操作符中的UB之前终止(因此它的定义很好)。

所以第二个结论是，只要没有求值，UB就不是UB。

此外，我没有看到这里提到有相当现代的未定义行为消毒器(至少在clang)(选项-fsanitize= Undefined)将在第一个例子(但不是第二个)上给出以下输出:

/app/example.c:5:5: runtime error: index 5 out of bounds for type 'int[5]'
SUMMARY: UndefinedBehaviorSanitizer: undefined-behavior /app/example.c:5:5 in 
/app/example.c:7:27: runtime error: index 5 out of bounds for type 'int[5]'
SUMMARY: UndefinedBehaviorSanitizer: undefined-behavior /app/example.c:7:27 in 

以下是godbolt的所有样品:

https://godbolt.org/z/eY9ja4fdh(第一个示例，没有标记)

https://godbolt.org/z/cGcY7Ta9M(第一个示例和-Ofast clang)

https://godbolt.org/z/cGcY7Ta9M(第二个例子和UB消毒剂打开)

https://godbolt.org/z/vE531EKo4(第一个例子和UB消毒剂打开)

一般来说，现在的操作系统(流行的操作系统)使用虚拟内存管理器在受保护的内存区域中运行所有应用程序。事实证明，简单地读取或写入存在于已分配给进程的区域之外的REAL空间中的位置(就其本身而言)并不容易。

直接回答:

Reading will almost never directly damage another process, however it can indirectly damage a process if you happen to read a KEY value used to encrypt, decrypt, or validate a program / process. Reading out of bounds can have somewhat adverse / unexpected affects on your code if you are making decisions based on the data you are reading The only way your could really DAMAGE something by writing to a loaction accessible by a memory address is if that memory address that you are writing to is actually a hardware register (a location that actually is not for data storage but for controlling some piece of hardware) not a RAM location. In all fact, you still wont normally damage something unless you are writing some one time programmable location that is not re-writable (or something of that nature). Generally running from within the debugger runs the code in debug mode. Running in debug mode does TEND to (but not always) stop your code faster when you have done something considered out of practice or downright illegal. Never use macros, use data structures that already have array index bounds checking built in, etc....

ADDITIONAL I should add that the above information is really only for systems using an operating system with memory protection windows. If writing code for an embedded system or even a system utilizing an operating system (real-time or other) that does not have memory protection windows (or virtual addressed windows) that one should practice a lot more caution in reading and writing to memory. Also in these cases SAFE and SECURE coding practices should always be employed to avoid security issues.

二维或多维数组的考虑超出了其他答案中提到的那些。考虑以下函数:

char arr1[2][8];
char arr2[4];
int test1(int n)
{
  arr1[1][0] = 1;
  for (int i=0; i<n; i++) arr1[0][i] = arr2[i];      
  return arr1[1][0];
}
int test2(int ofs, int n)
{
  arr1[1][0] = 1;
  for (int i=0; i<n; i++) *(arr1[0]+i) = arr2[i];      
  return arr1[1][0];
}

The way gcc will processes the first function will not allow for the possibility that an attempt to write arr[0][i] might affect the value of arr[1][0], and the generated code is incapable of returning anything other than a hardcoded value of 1. Although the Standard defines the meaning of array[index] as precisely equivalent to (*((array)+(index))), gcc seems to interpret the notion of array bounds and pointer decay differently in cases which involve using [] operator on values of array type, versus those which use explicit pointer arithmetic.

你写的:

我读过很多“任何事情都可能发生”，“市场细分可能是。 “最不坏的问题”，“你的硬盘可能会变成粉红色，独角兽也可能 在你的窗下唱歌，这是很好的，但真正的 危险吗?

这么说吧，给枪上膛。瞄准窗外，不要瞄准，然后开火。危险在哪里?

The issue is that you do not know. If your code overwrites something that crashes your program you are fine because it will stop it into a defined state. However if it does not crash then the issues start to arise. Which resources are under control of your program and what might it do to them? I know at least one major issue that was caused by such an overflow. The issue was in a seemingly meaningless statistics function that messed up some unrelated conversion table for a production database. The result was some very expensive cleanup afterwards. Actually it would have been much cheaper and easier to handle if this issue would have formatted the hard disks ... with other words: pink unicorns might be your least problem.

认为操作系统会保护你的想法是乐观的。如果可能，尽量避免越界写作。

Objective-C中的nsarray被分配一个特定的内存块。超过数组的边界意味着您将访问没有分配给数组的内存。这意味着:

This memory can have any value. There's no way of knowing if the data is valid based on your data type. This memory may contain sensitive information such as private keys or other user credentials. The memory address may be invalid or protected. The memory can have a changing value because it's being accessed by another program or thread. Other things use memory address space, such as memory-mapped ports. Writing data to unknown memory address can crash your program, overwrite OS memory space, and generally cause the sun to implode.

从程序的角度来看，您总是想知道代码何时超出了数组的边界。这可能导致返回未知值，导致应用程序崩溃或提供无效数据。