打印字符数组地址的正确方法

Question

So, I have been digging into character array's lately, and I am trying to print the address of each element of a character array.

char a[4] = {'i','c','e','\0'};
for(int i = 0; i < 3; ++i){
  cout<<(void*)a[i]<<" ";
  cout<<&a[i]<<" ";
  cout<<a[i]<<endl;
 }

The code above gives me the following output:

0x69 ice i
0x63 ce c
0x65 e e
test.cpp: In function ‘int main()’:
test.cpp:29:23: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]
       cout<<(void*)a[i]<<" ";
                       ^

I am not comfortable about the output of (void*)a[i] . Shouldn't the character addresses be 1 byte apart. I am seeing 0x69 followed by 0x63 and then by 0x65 . Is there a reason for that. And is there a relationship between the address representation and the warning sign that it is showing.

Answer 1

I am trying to print the address of each element of a character array

(void*)a[i] is converting the element (the char ) itself to void* , not the address of the element.

You should get the address of each element as:

cout<<(void*)&a[i]<<" ";
//           ^

Or better to use static_cast .

cout<<static_cast<void*>(&a[i])<<" ";

Answer 2

Your print value casted to void* , to print address, you need

cout<< static_cast<void*>(&a[i])<<" ";

Answer 3

Currently, you are not getting the addresses. You are casting the ASCII values of the characters to a void* instead. This is why the values are not correct.

What you want to do is use a static_cast and get the address of the element &a[i] :

cout << static_cast<void*> (&a[i]) << " ";