指针未递增（移动）到预期位置或指针算法未提供预期答案

Question

In the following snippet of code I expect pointer to move to next location ie current location + sizeof(datatype) but not happening unless I type cast to int .

#include <iostream>
#include <string.h>

int sizeOf()
{
    int a = 0;
    int* b = &a;
    int* c = &a;
    c++;// expect pointer to move "current location + sizeof(int)"
    return c - b;
}


int main()
{
    std::cout << "Size of int is " << sizeOf() << ", actual size of int is  "<< sizeof(int) << "\n";
    return 0;
}

Output:

Size of int is 1, actual size of int is  4

Expectation: if I increment pointer of type int then pointer should move to current location + sizeof(int) but not happening

But it actually works when I use following line

return (int)c - (int)b;

Compiler warns

my-pc $ g++ test.cpp -fpermissive
test.cpp: In function ‘int sizeOf()’:
test.cpp:10:17: warning: cast from ‘int*’ to ‘int’ loses precision [-fpermissive]
   10 |     return (int)c - (int)b;
      |                 ^
test.cpp:10:26: warning: cast from ‘int*’ to ‘int’ loses precision [-fpermissive]
   10 |     return (int)c - (int)b;
      |                          ^

my-pc $ ./a.out 
Size of int is 4, actual size of int is  4

Now I got the output what I expect. I know int* is typecasted to int . I want to know the reason why it did not worked in first case and worked later.

Answer 1

The difference is in terms of the size of the pointed object.

So a difference of 1 means the pointers point to 1 * sizeof(int) bytes apart.

For example,

&( a[2] ) - &( a[0] )

always gives 2 for a C array. It doesn't matter if it's an array of char , int , or some struct .

And it's easy to show why this is necessary, given that a[i] is equivalent to *(a+i) for a C array.

   &( a[2] ) - &( a[0] )
=  &( *( a + 2 ) ) - &( *( a + 0 ) ) 
=  ( a + 2 ) - ( a + 0 )
=  2