将指针类型转换为＆（int *）和（int *）＆并将指针分配给某个指针变量时有什么区别？

Question

If we have a void pointer and we try to assign it the address of some pointer variable (for instance, a pointer to int ), then what is the difference between the statements 1 and 2.

For statement 2 , I am getting lvalue error , but I am not getting the reason behind this - ptr is lvalue only so why does it show an error?

int a;
int *ptr =&a ;
1. void *p =(int*)&ptr;
2. void *p=&(int*)ptr;

Answer 1

According to section 6.5.4 of the C99 specification, line 4, footnote 89, available online here :

A cast does not yield an lvalue

When we look at your code:

int a;
int *ptr =&a ;
1. void *p =(int*)&ptr;
2. void *p=&(int*)ptr;

In case of 1: A void pointer is assigned what has been typecast to int* . This is valid. In fact the following statements also are valid:

3. void *p = (int*)123;
4. void *p = (int*)a;
5. void *p = (int*)ptr;

But in case of 2: (int*)ptr has become an rvalue and then you are using & on that rvalue. It fails stating: lvalue required as unary '&' operand

For 2., it is not the assignment to void* that is causing the problem but the operator & on an rvalue.

Answer 2

In your example the second one void *p=&(int*)ptr; has incorrect syntax as & requires an lvalue.

int *n;
void *p = &n;
void *q = &(void *)n;  //incorrect syntax will not compile at all
void *w = &(int *)n;   //incorrect syntax will not compile at all

So the difference is: &(some_type *)n will not compile at all, and does not have any sense.
Sample code .

&(int*)ptr is just not valid in C.

Why? Because p and (type *)p are completely different. p is object which reference can be obtained (type *)p is only the value without the object behind it and consequently it does not have the address.

Answer 3

It is the same behaviour while casting with int :

int a;
int *ptr = &(int)a;