在 C 中打印出一个字节的值及其地址

Question

I'm working on pointers right now and I'm trying to get an understanding for how things look in memory so I'm attempting to build this program. The goal of it is to take an integer, print its address in memory and then its integer value at each byte. So for an integer 4 pairs of (address, value). However, the result of my work is rather confusing as I'm getting strange results such as this. If someone could explain what I'm doing wrong that would be much appreciated. For context, I understand a little bit about computer architecture and memory from a class

Enter an integer: 287

0x7ffd455a7474 31

0x7ffd455a7475 1

0x7ffd455a7476 0

0x7ffd455a7477 0

#include <stdio.h>

void byte_value(int *);
int main(){

    int n =1;
    printf("Enter an integer: ");

    if (scanf("%d", &n) == 1)

        byte_value(&n);

    

    return 0;  

}

void byte_value(int *p) {
    char *p2=((char *)p);
    char *max=p2+sizeof(int);
    for(;p2<max;p2++){
            printf("%p %u\n",p2,*p2);
    }

    return;

}

Answer 1

I think this version may explain a bit more. (Hex values are very convenient)

void byte_value(void *, size_t);
int main(void)
{

    int n =1;
    printf("Enter an integer: ");

    if (scanf("%d", &n) == 1)
        printf("\n------------\nentered: %d (0x%x)\n", n, (unsigned)n);
        
        byte_value(&n, sizeof(n));
}

void byte_value(void *ptr, size_t size) 
{
    unsigned char *p = ptr;
    while(size--)
    {
        printf("%p - %03u (0x%02x)\n", (void *)p, *p, *p);
        p++;
    }
}