Question

a[2][3] = {{-3,14,5},{1,-10,8}}

*(a[j]+k) 
*(a[j+k-2]) 
(*(a+j))[k]) 

(*(a+k-1))[j] 
*((*(a+j))+k)) 

(**(a+j)+k) 
*(&a[0][0]+j+k)

当我打印这些我得到的输出： 8 1 8 -10 8 3 1 分别如果有人能详细解释这些价值的来源，我是新的首发，请原谅我这里的格式错误，并且还要为这么多工作烦恼：）

Answer 1

我假设j == 1且k == 2：

*(a[j]+k) == *(a[1]+2) :

    a[1] = {1, -10, 8};
    So a[1]+2 is a pointer to 8, and *(a[1]+2) == 8

*(a[j+k-2]) == *(a[1+2-2]) == *(a[1]):

    a[1] = {1, -10, 8}
    Since *a[1] is the value of the first element in a[1], the expression evaluates to 1

(*(a+j))[k] == (*(a+1))[2]:

    a+1 is a pointer to the second element in a, so *(a+1) == a[1] = {1, -10, 8}
    a[1][2] == 8

(*(a+k-1))[j]  == (*(a+2-1))[1] == (*(a+1))[1]:

    *(a+1) == a[1] (see the last answer)
    a[1][1] == -10

*((*(a+j))+k) == *((*(a+1))+2):

     *(a+1) == a[1], so the expressions is equivalent to *(a[1]+2)
     *(a[1]+2) is equivalent to a[1][2] for the same reasoning as above, which is 8

(**(a+j)+k)  == (**(a+1)+2):

    *(a+1) = a[1]
   **(a+1) = a[1][0] == 1
     Adding 2 to that gives 3

*(&a[0][0]+j+k) == *(&a[0][0]+1+2) == *(&a[0][0]+3):

    &a[0][0] == a[0]
    *(a[0]+3) == a[0][3];

最后一个返回1，因为它在内存中延伸超过[0]的末尾到[1]，并且你得到[1] [0]。我认为这种行为实际上是未定义的。这取决于C标准是否保证初始化的2D数组将在内存中顺序分配，并且可能导致分段错误或更糟。

2维数组解引用，如何通过指针进行评估

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