英文:
Unexpected behaviour while referencing and dereferencing array with pointer
问题
这是代码：
int a[5] = { 1, 2, 3, 4, 5 };
int* p = (int*)(&a + 1);
printf("%d, %d", (*a+1), *(p - 1));
// 输出结果 2, 5
因为我相信 p 指向数组 a 的第二个元素，即 2。所以 *(p-1) 的输出应该是 1。但在 VS 中，*(p-1) 输出为 5，原因是什么？谢谢帮助！
我认为输出应该是 2, 1。
英文:
Here is the code:
int a[5] = { 1, 2, 3, 4, 5 };
int* p = (int*)(&a + 1);
printf("%d, %d", (*a+1), *(p - 1));\
// output 2, 5
As I believe p points the second element in array a, which is 2. So the output of *(p-1) is 1.
But in VS *(p-1) output 5, what is the reason?
Thanks for helping!
I believe the output should be 2, 1.
答案1
得分: 4
答案在你插入(int*)强制类型转换以覆盖编译器更好判断之前，你（可能）看到的错误消息中：
int* p = (&a + 1);
temp.cpp:7:10: error: cannot initialize a variable of type 'int *' with an rvalue of type 'int (*)[5]'
int * q = &a;
因为a的类型是int[5]，这意味着&a的类型是指向int[5]的指针，并且这也意味着sizeof(a)等于5*sizeof(int)... 这意味着(&a+1)指向数组结束后的地址（也就是如果你有一个由多个包含5个整数的数组组成的数组，它指向下一个数组的开头），而不是你数组中的第二个int。
所以，当你在你的代码中评估*(p-1)时，你得到一个指向&a[5]（虽然它不存在，但不要紧）的指针，然后减去1个int，得到一个指向&a[4]的指针，然后解引用它，从你的数组中获取值5。
英文:
The answer to your question lies in the error message that you (probably) saw just before you inserted the (int*) cast to override the compiler's better judgement:
int* p = (&a + 1);
temp.cpp:7:10: error: cannot initialize a variable of type 'int *' with an rvalue of type 'int (*)[5]'
int * q = &a;
Since a is of type int[5], that means that &a is of type pointer-to-int[5], and also it means that sizeof(a) is equal to 5*sizeof(int)... which means that (&a+1) is pointing to the address after the end of the array (i.e. where the next array-of-5-ints would be if you had an array of arrays-of-5-ints-each), not to the second int in your array.
So when you evaluate *(p-1) in your code, you are taking a pointer to &a[5] (which doesn't exist, but nevermind that) and then subtracting 1 int from it, which gives you a pointer to &a[4], which you dereference to get the value 5 from your array.
答案2
得分: 0
问题出在数组的标识符（变量a），即a存储了数组起始位置的指针。
你写了(&a + 1)，所以它将整个数组视为指针，+1指向数组大小之后的下一个位置。
例如，数组的地址为100, 104, 108, 112, 116
&a+1 -> 120
请使用a+1而不是&a+1
a+1 -> 104
因此，新的代码将是：
int a[5] = { 1, 2, 3, 4, 5 };
int* p = (int*)(a + 1);
printf("%d, %d", (*a+1), *(p - 1));
这将给出期望的输出2,1。
希望这能够解决疑惑。
另外，要说的是int *a和int a[]在指针方面有些相似，但不完全相同。
英文:
The problem is identifier(variable a) of array i.e. a stores the pointer of starting of array itself
you have written (&a + 1) so it consider whole array as a pointer and +1 points p to next location after size of array
eg. array has addressed 100,104,108,112,116
&a+1 -> 120
use a+1 instead of &a+1
a+1 -> 104
so the new code will be
int a[5] = { 1, 2, 3, 4, 5 };
int* p = (int*)(a + 1);
printf("%d, %d", (*a+1), *(p - 1));
this gives expected output as 2,1
hope this clears doubt
more to say int *a and int a[] are somewhat similar in pointer but not exactly.
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