how this convertion done in c++ from unsigned char* to unsigned short? x= (unsigned char)&x - (unsigned char) 0x8888;

Question

how this convertion done from unsigned char* to unsigned short?

#include<iostream>
using namespace std;

int main()
{
    unsigned short x;
    x= (unsigned char*)&x - (unsigned char*) 0x8888;
    cout<<us<<endl;
    return 0;
}

when I am trying to add like this

x= (unsigned char*)&x + (unsigned char*) 0x8888;

Then error comes up. it says "invalid operands of types 'unsigned char*' and 'unsigned char*' to binary 'operator+' " please help me to understand this syntax.

There is no conversion from `unsigned char *` to `unsigned short` in your code. The subtraction has higher precedence than the assignment. — M.M, Feb 13 '17 at 09:05

score 0 · Answer 1 · edited Jun 20 '20 at 09:12

0

Subtraction operation can be performed on two address data type but not addition.
For addition, either both operand of + shall be of arithmetic type or one operand shall be a pointer and other shall be an integer type.

C++11-§ 5.7/1:

For addition, either both operands shall have arithmetic or unscoped enumeration type, or one operand shall be a pointer to a completely-defined object type and the other shall have integral or unscoped enumeration type.

edited Jun 20 '20 at 09:12

Community

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answered Feb 12 '17 at 19:48

haccks

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Brother this not working x= (unsigned char*)&x + 88 i have changed second operand to int. – Sirazul Islam Feb 12 '17 at 20:19
I am trying this code "x= (unsigned char*)&x + 88". Then "error: invalid conversion from 'short unsigned int*' to 'short unsigned int' [-fpermissive]|" – Sirazul Islam Feb 12 '17 at 20:25
`(unsigned char*)&x + 88` will give an `unsigned char*` value. `x` is `unsigned short` type. Error/warning is legit. – haccks Feb 12 '17 at 20:27
x= (unsigned short*)&x + 88; i changed (unsigned char*) to (unsigned short*) Then "error: invalid conversion from 'short unsigned int*' to 'short unsigned int' [-fpermissive]|" – Sirazul Islam Feb 12 '17 at 20:32
Do you know the difference between pointer and `int` data types? – haccks Feb 12 '17 at 20:34

score 0 · Answer 2 · edited May 23 '17 at 11:53

You could convert pointer to int/short/long using cstdint C++11 as below, But I'm not certain whether it is advisable to add pointers as mentioned by @haccks

#include<iostream>
#include <cstdint>

using namespace std;

int main()
{
    unsigned short x;
    //x= (unsigned char*)&x + (unsigned char*) 0x8888;
    x = static_cast<unsigned short>(reinterpret_cast<uintptr_t>(&x)) + static_cast<unsigned short>(0x8888);
    cout<<x<<endl;
    return 0;
}

Refer to this link for more information.

how this convertion done in c++ from unsigned char* to unsigned short? x= (unsigned char*)&x - (unsigned char*) 0x8888;

2 Answers2

C++11-§ 5.7/1:

how this convertion done in c++ from unsigned char* to unsigned short? x= (unsigned char)&x - (unsigned char) 0x8888;