声明一个负长度的数组

What happens in C when you create an array of negative length?

当创建一个长度为负的数组时，C中会发生什么?

For instance:

例如:

int n = -35;

int testArray[n];

for(int i = 0; i < 10; i++)
    testArray[i]=i+1;

This code will compile (and brings up no warnings with -Wall enabled), and it seems you can assign to testArray[0] without issue. Assigning past that gives either a segfault or illegal instruction error, and reading anything from the array says "Abort trap" (I'm not familiar with that one). I realize this is somewhat academic, and would (hopefully) never come up in real life, but is there any particular way that the C standard says to treat such arrays, or is does it vary from compiler to compiler?

这段代码将会编译(并且没有启用-Wall的警告)，并且看起来您可以毫无问题地分配给testArray[0]。赋值给过去，要么给出分段错误，要么给出非法的指令错误，并且从数组中读取任何内容都表示“中止陷阱”(我不熟悉这个)。我意识到这有点学术性，并且(希望)在现实生活中永远不会出现，但是C标准是否有任何特定的方法来处理这样的数组，或者它是否因编译器而异?

3 个解决方案

#1

It's undefined behaviour, because it breaks a "shall" constraint:

它是未定义的行为，因为它打破了“应该”的约束:

C99 §6.7.5.2:

C99§6.7.5.2:

If the size is an expression that is not an integer constant expression... ...each time it is evaluated it shall have a value greater than zero.

如果大小不是整数常量表达式……每次计算时，它的值都应该大于零。

#2

Undefined behavior, I believe, though don't quote me on that.

不明确的行为，我相信，尽管不要引用我的话。

This gives the error error: size of array 'testArray' is negative in gcc:

这就产生了误差:在gcc中，'testArray'数组的大小为负:

int testArray[-35];

though, as you've seen:

不过,正如您所看到的:

int n = -35;
int testArray[n];

does not give an error even with both -Wall and -W.

即使对-Wall和-W都不产生错误。

However, if you use -pedantic flag, gcc will warn that ISO C90 forbids variable length array.

但是，如果您使用-pedantic标志，gcc将警告说ISO C90禁止可变长度数组。

#3

Visual studio erro message for compilation, you can use -1 to say an empty array. It expects int and you are passing int, so no compiler error.

Visual studio erro消息用于编译，您可以使用-1表示一个空数组。它期望int并且您正在传递int，所以没有编译错误。

#1