增量运算阵列上的错误指针？ [英] Increment operator on pointer of array errors?

问题描述

我想很简单的东西，好应该是简单的，但不知何故与我搞乱...

I'm trying something very simple, well supposed to be simple but it somehow is messing with me...

我想，当为数组处理时作为指针和指针处理，以了解++阵列上的效果。

I am trying to understand the effect of ++ on arrays when treated as pointers and pointers when treated as arrays.

因此​​，

int main()
{
    int a[4] = { 1, 4, 7, 9 };
    *a = 3;
    *(a+1) = 4;
    *++a = 4; //compiler error
}

1 因此，在 *（A + 1）= 4 我们设置一个[1] = 4; //快乐
但是，当 * + A = 4; ，我期望指针以递增之一，因为++是precedent为*然后*在踢我们使其等于4。但这code刚刚为什么不工作...？

1: So at *(a+1)=4 we set a[1]=4; //Happy But when *++a = 4;, I'd expect pointer a to be incremented one since ++ is precedent to * and then * kicks in and we make it equal to 4. But this code just does not work... Why is that?

另一个问题：

int main()
{

    int* p = (int *)malloc(8);
    *p = 5;
    printf("%d", p[0]);

    *++p = 9; //now this works!
    printf("%d", p[1]); //garbage
    printf("%d", p[0]); //prints 9

}

2 立即* ++ P = 9;工作正常，但它不是真正的表现就像一个数组。如何是两个不同的？这只是递增p和使它等于9。如果我打印P [0]，它现在打印9和我看到，虽然不能经由p访问[0]了，*（P-1）示出了5仍然存在。所以，用索引[0]，正是它指向一个指针？发生了什么变化？

2: Now *++p = 9; works fine but it's not really behaving like an array. How are two different? This is just incrementing p, and making it equal to 9. If I print p[0], it now prints 9 and I see that though can't access it via p[0] anymore, *(p-1) shows 5 is still there. So indexing a pointer with [0], where exactly does it point to? What has changed?

非常感谢各位专家！

推荐答案

数组名是不可修改的左值那么操作++不会因此适用 ++中的试图修改 A 是编译时错误（其中 A 是数组名）。

The array names is not modifiable lvalue so operation ++ is not applied hence ++a that try to modify a is compilation time error (where a is array name).

请注意 *（A + 1）和 *一个++ 不一样的， A + 1 是因为它只需添加一个有效的指令 1 ，但不修改 A 本身，而 ++中的（即equvilent到 A = A + 1 ）尝试修改，因此错误。

Note *(a + 1) and *a++ are not same, a + 1 is a valid instruction as it just add 1 but doesn't modify a itself, Whereas ++a (that is equvilent to a = a + 1) try to modify a hence error.

请注意数组名'是的不的指针。指针是可变的，但数组名都没有。的，当然，当你在最前pressions数组名指定数组名的指针，然后衰变为第一个元素的地址。例如。

Note 'array names' are not pointer. Pointers are variable but array names are not. Of-course when you assign array name to a pointer then in most expressions array names decays into address of first element. e.g.

int *p = a;

请注意 P 点（ A [0] ）。

看了一些例外情况数组名不衰变成一个指向第一个元素？

这是前pression A [I] 等同于 *（A + I），其中 A 可以是指针或数组名。因此，在你的第二个例子 P [I] 有效期的前pression。

An expression a[i] is equivalent to *(a + i), where a can be either a pointer or an array name. Hence in your second example p[i] is valid expression.

此外， * ++ P 是有效的，因为，因为 P 是第二个$指针（变量） C $ç例子。

Additionally, *++p is valid because because p is a pointer (a variable) in second code example.

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