c中的& a，& a [0]，a有什么区别 [英] what is the difference between &a,&a[0],a in c

问题描述

我得到以下代码的输出-- 6

I got the output for the following code as -> 6

我对输出感到困惑，所以我更改了一小部分代码并进行了检查.

I was confused by the output, so I changed a small portion of code and checked it.

我将此 int * ptr =(int *)(a + 1)替换为Not clear语句，我的输出为-> 1 .怎么样?

I substituted this int * ptr=(int*)(a+1) for the Not clear statement, I got the output as --> 1. How?

我听说& a ， a ， a [0] 都是相同的地址?是什么让它与众不同?

I heard &a, a, a[0] all are same address? what makes it different?

#include<stdio.h>

int main()
{
    int a[] = {1, 2, 3, 4, 5, 6};
    int *ptr = (int*)(&a+1); //NOT CLEAR
    printf("%d ", *(ptr-1) );
    getchar();
    return 0;
} 

推荐答案

和许多其他开发人员一样，您对衰落感到困惑.以下代码是有效的，因为 a 将隐式地 decay 指向指针:

Like many other developers, you are confused by decay. The following code is valid, because a will implicitly decay to a pointer:

int a[] = {1,2,3};
int* b = a;

衰减是值得注意的，它允许数组隐式转换为指向其第一个元素的指针.

Decay is what notably allows arrays to implicitly transform into pointers to their first elements.

但是，以下代码无效，因为衰减与获取数组的地址不同:

However, the following code is not valid, because decay is not the same as taking the address of the array:

int a[] = {1,2,3};
int* b = &a; // INVALID

第一个示例为您提供了一个指向整数(数组中的第一个整数)的指针.在第二个示例中，& a 为您提供指向3个整数的数组的指针.如果必须将其分配给变量，则其类型为:

The first example gets you a pointer to an integer (the first integer in the array). In the second example, &a gets you a pointer to an array of 3 integers. If you had to assign it to a variable, its type would be:

int a[] = {1,2,3};
int (*b)[3] = &a;

外星人的形式应该告诉你，这根本不是你所期望的.

The alien form should show you that this is not what you expect at all.

a (一旦衰减)，& a 和& a [0] 确实都应该指向同一地址.但是，它们并不指向同一类型的事物(& a 指向数组，而其他指向整数).

It is true that a (once decayed), &a and &a[0] should all point to the same address. However, they do not point to the same kind of thing (&a points to an array, while the others point to integers).

指针操作的结果至少取决于指针的类型和指针的值.每次将& a 指针增加1时，您将在内存中移动三个整数，而不仅仅是移动一个，因为类型的大小为(3 *整数).这是您执行(& a + 1)时发生的事情，这就是为什么您得到意外结果的原因.在此内存地址中未定义任何值，因此您会得到未定义的行为.

The result of pointer operations depend at least as much on the type of the pointer as to the value of the pointer. Each time you increase that &a pointer by 1, you move by three integers in memory, not just one, because the size of the type is (3 * size of an integer). This is what happens when you do (&a + 1), and this is why you get an unexpected result. There is no value defined at this memory address, so you get undefined behavior.

它之所以编译是因为将指针强制转换为int指针，并且如果需要我的建议，则应该从不使有关强制转换指针的编译器警告保持沉默.

It compiles because you cast the pointer to an int pointer, and if you want my advice, you should never silence compiler warnings about pointers with casts.

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