c – 在不使用&或*的情况下等待两个指针？

int * A; // create a pointer to an int named "A"
int * B; // create a pointer to an int named "B"
A = (int *)malloc( size_t somenumber ); // Allocate A some memory,now B is an 
                                        // uninitialized pointer; A is initialized,// but just to uninitialized memory

概念：

B = A; // Assign B to the value of A (The uninitialized memory)

free(A);

毕竟,我认为你可以看到发生了什么. B被赋予A的值,该值是已分配和未初始化的内存块.所以现在你只有两个指向同一区域的指针.

至于free()问题,正如你可以看到的那样,当你打电话给免费(A);你离开时A和B都指向同一个区域,那里就没有任何东西分配到你的程序了.这就是为什么在调用free()时将指针设置为NULL是很好的.

现在回到你最初的问题.如果你想检查两个指针??==：

int * A; // create a pointer to an int named "A"
int * B; // create a pointer to an int named "B"
A = (int *)malloc( size_t somenumber ); // Allocate A some memory,// but just to uninitialized memory

if(B == A){
   // The pointers are pointing to the same thing!
}
if(*B == *A){
   // The values these pointers are pointing to is the same!
}

UPDATE
因此,要回答您更新的问题,我们需要更改B的定义.

int *A;  // A is a pointer to an int
int **B; // B is a pointer to a pointer to an int
B = &A;  // B is now pointing to A

所以说明一下：

对于B = * A：

int *A;
int B;
A = malloc(sizeof(int));
*A = 5;
B = *A;

这是对A的尊重.所以你只需要拿A指向的东西并将其分配给B,在这种情况下为5