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I'm dealing with pointers in C, declaring the integer value as 1025. Then I changed the integer to char. When I increase the char pointer by 1, the char value increases by 3. How is this happening?

#include<stdio.h>
int main(){
    int a = 1025;
    int *p;
    p = &a;

    char *p0;
    p0 = (char*)p;
    printf("Address = %d, value = %d\n",p0,*p0);
    printf("Address = %d, value = %d\n",p0+1,*(p0+1));

    return 0;

When you use a char * to alias an object of another type as you're doing, it allows you to access the byte representation of that object.

The value 1025 can be represented in hex as 0x0401. Your system appears to use little-endian byte ordering to store integers, which means the low order bytes appear first in the representation.

So assuming an int is 32 bits on your system a looks like this in memory:

  -----------------------------
a | 0x01 | 0x04 | 0x00 | 0x00 |
  -----------------------------

The pointer p0 points to the first byte, so *p0 is 1. Then p0+1 points to the next byte so *(p0+1) is 4.

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You are incrementing the pointer to the bytes the integer a is composed from. Incidentally 1025 is composed of two bytes with values 1 and 4 - you can check. 1025 = 4*256 + 1. So once you move from 1 to 4 it looks like it was incremented by 3

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If you want to access the next element it can be accessed by incrementing the pointer value, not the char value. Increasing the pointer value will enable that pointer to point towards the next value's address and hence you will be seeing the value at the address.

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Your output shows what is happening:

Address = 1204440828, value = 1
Address = 1204440829, value = 4

Notice that the address of what you are printing is changing by one. So you are not adding 1 to a you are looking at the value in the location a + 1.

UPDATE: As a clarification: the pointer is pointing to an int but you are printing it as a series of byte values. So the fact that the first byte of the value 1025 happens to decode to 1 If you change a to some other value, you will get somewhat random output. Try changing int a = 1035 and you will get 11 and 4 instead of 1 and 4.

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Comments
  • You can't use %d to format a pointer value. Your compiler should have warned about that. In particular, a pointer may not have the same size as an integer.
  • That's a great explanation. Thank you!
  • This is not random. It becomes 11 since the value is increasing by 10.
  • No, it becomes 11 because 1035 = 256*4 + 11. Not random indeed though.
  • To clarify: I shouldn't have said "random" but "arbitrary" depending on the byte ordering of the system - which is not defined by the C language.