메모리 하나당 차지하는 바이트 수에 대해 궁금한 점이 있습니다.

#include <iostream>
 
 
int main(int argc, char** argv)
{
 
{
	int a = 10;
	int b = (int)&a;	printf("int b = (int)&a;         b:%x\n", b);
	b = (int)b+1;		printf("b = (int)b+1;            b:%x\n", b);
	int * c = (int*)b;	printf("int * c = (int*)b;       c:%x\n", c);
	*c = 11;			printf("*c = 11;                 c:%x\n", c);
	*c = 12;			printf("*c = 12;                 c:%x\n", c);
						printf("\n");
}
 
{
	int a = 10;
	int b = (int)&a;		printf("int b = (int)&a;       b  :%x\n", b);		//4
	b = (int)b+1;			printf("b = (int)b+1;          b  :%x\n", b);		//5
	int * c = (int*)b;		printf("int * c = (int*)b;   c+1:%x\n", c+1);	//6
	*c = 11;				printf("*c = 11;               c+2:%x\n", c+2);	//7
	*c = 12;				printf("*c = 12;               c  :%x\n", c);		//5
	c = (int*)b+1;			printf("c = (int*)b+1;         c  :%x\n", c);		//5
							printf("\n");
}
 
{
	int a = 10;
	int b = (int)&a;		printf("int b = (int)&a;       b  :%x\n", b);		//4
	b = (int)b+1;			printf("b = (int)b+1;          b  :%x\n", b);		//5
	char * c = (char*)b;	printf("char * c = (char*)b;   c+1:%x\n", c+1);	//6
	*c = 11;				printf("*c = 11;               c+2:%x\n", c+2);	//7
	*c = 12;				printf("*c = 12;               c  :%x\n", c);		//5
	c = (char*)b+1;			printf("c = (char*)b+1;        c  :%x\n", c);		//5
	b = (int)b+1;			printf("b = (int)b+1;          b  :%x\n", b);		//5
	c = (char*)b;			printf("c = (char*)b;          c+1:%x\n", c+1);	//6
	*c = 11;				printf("*c = 11;               c+2:%x\n", c+2);	//7
	*c = 12;				printf("*c = 12;               c  :%x\n", c);		//5
	c = (char*)b+1;			printf("c = (char*)b+1;        c  :%x\n", c);		//5
}
 
 
	return 0;
}
 
 
출력
int b = (int)&a;         b:22fed4
b = (int)b+1;            b:22fed5
int * c = (int*)b;       c:22fed5
*c = 11;                 c:22fed5
*c = 12;                 c:22fed5
 
int b = (int)&a;       b  :22fed0
b = (int)b+1;          b  :22fed1
int * c = (int*)b;   c+1:22fed5
*c = 11;               c+2:22fed9
*c = 12;               c  :22fed1
c = (int*)b+1;         c  :22fed5
 
int b = (int)&a;       b  :22fecc
b = (int)b+1;          b  :22fecd
char * c = (char*)b;   c+1:22fece
*c = 11;               c+2:22fecf
*c = 12;               c  :22fecd
c = (char*)b+1;        c  :22fece
b = (int)b+1;          b  :22fece
c = (char*)b;          c+1:22fecf
*c = 11;               c+2:22fed0
*c = 12;               c  :22fece
c = (char*)b+1;        c  :22fecf