负数组索引 [英] Negative array index

问题描述

我有一个定义如下的指针:

I have a pointer which is defined as follows:

A ***b;

按以下方式访问它有什么作用:

What does accessing it as follows do:

A** c = b[-1]

是否因为我们对数组使用了负索引而导致访问冲突?还是类似于*--b的合法操作?

Is it an access violation because we are using a negative index to an array? Or is it a legal operation similar to *--b?

EDIT 请注意，负数组索引在 C 和 C++ 中具有不同的支持.因此，这不是骗人的.

EDIT Note that negative array indexing has different support in C and C++. Hence, this is not a dupe.

推荐答案

X[Y] is identical to *(X + Y)> 只要 X 和 Y 之一是指针类型，另一个是整数类型.所以 b[-1] 与 *(b - 1) 相同，后者是一个表达式，在格式良好的程序中可能会也可能不会被计算——这一切都取决于b的初始值！例如，以下完全没问题:

X[Y] is identical to *(X + Y) as long as one of X and Y is of pointer type and the other has integral type. So b[-1] is the same as *(b - 1), which is an expression that may or may not be evaluated in a well-formed program – it all depends on the initial value of b! For example, the following is perfectly fine:

int q[24];
int * b = q + 13;

b[-1] = 9;
assert(q[12] == 9);

一般来说，作为程序员，保证指针在您执行操作时具有允许的值是您的责任.如果你弄错了，你的程序有未定义的行为.例如:

In general, it is your responsibility as a programmer to guarantee that pointers have permissible values when you perform operations with them. If you get it wrong, your program has undefined behaviour. For example:

int * c = q;   // q as above
c[-1] = 0;     // undefined behaviour!

最后，只是为了加强原来的陈述，以下也很好:

Finally, just to reinforce the original statement, the following is fine, too:

std::cout << 2["Good morning"] << 4["Stack"] << 8["Overflow
"];

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