没有越界错误 [英] No out of bounds error

问题描述

我在 C 中有这段代码，它接受一堆 chars

I have this code in C which takes in bunch of chars

#include<stdio.h> 
# define NEWLINE '
'
int main()
{

char c;
char str[6];
int i = 0;
while( ((c = getchar()) != NEWLINE))
{
        str[i] = c;
        ++i;
        printf("%d
", i);
}

return 0;
}

输入为:testtesttest

Input is: testtesttest

输出:12345678117118119120

Output: 1 2 3 4 5 6 7 8 117 118 119 120

我的问题是:

1. 为什么我没有得到越界(分段错误)异常，尽管我明显超出了数组的容量?

1. Why don't I get an out of bounds (segmentation fault) exception although I clearly exceed the capacity of the array?

为什么输出中的数字突然跳到非常大的数字?

Why do the numbers in the output suddenly jump to very big numbers?

我在 C++ 中尝试了这个并得到了相同的行为.谁能解释一下这是什么原因?

I tried this in C++ and got the same behavior. Could anyone please explain what is the reason for this?

推荐答案

1. C 不检查数组边界.仅当您尝试取消引用指向您的程序无权访问的内存的指针时，才会发生分段错误.简单地越过数组的末尾不太可能导致这种行为.未定义的行为就是 - 未定义.它可能看起来工作得很好，但你不应该依赖它的安全性.
2. 您的程序通过访问数组末尾之后的内存来导致未定义的行为.在这种情况下，看起来您的 str[i] = c 写入之一覆盖了 i 中的值.
3. 在这种情况下，C++ 具有与 C 相同的规则.

1. C doesn't check array boundaries. A segmentation fault will only occur if you try to dereference a pointer to memory that your program doesn't have permission to access. Simply going past the end of an array is unlikely to cause that behaviour. Undefined behaviour is just that - undefined. It may appear to work just fine, but you shouldn't be relying on its safety.
2. Your program causes undefined behaviour by accessing memory past the end of the array. In this case, it looks like one of your str[i] = c writes overwrites the value in i.
3. C++ has the same rules as C does in this case.

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