指针的存储内存地址的增加不应该取决于系统的体系结构吗?

考虑以下内容 C 密码

#include <stdio.h>

int main() {
    int arr[] = {10, 20, 30, 40};
    int *ptr = arr; // ptr points to the internal pointer variable of `arr`, that is, the address of its first array element, 10.

    printf("The address of the first int array element is                : %p\n"
           "The stored value (.i.e., the memory address) of the pointer is: %p\n"
           &arr[0], ptr);

    printf("The memory size the a int variable is: %zu bytes, which is equal to %lu bits (each byte has 8 bits).\n"
           "Since `ptr` is a int pointer, the command `ptr = ptr + 1` shifts stored memory address of `ptr` in %zu bytes.\n\n",
            sizeof(int), 8*sizeof(int), sizeof(int));

    ptr = ptr + 1; // Move ptr to the memory address of the next integer (20) (instead, you could use `ptr++`)
    printf("The address of the first int array element is                : %p\n"
           "The stored value (.i.e., the memory address) of the pointer is: %p\n\n",
           &arr[0], ptr);

    return 0;
}

它打印:

The address of the first int array element is                : 0x7ffe100ee500
The store value (.i.e., the memory address) of the pointer is: 0x7ffe100ee500

The memory size the a int variable is: 4 bytes, which is equal to 32 bits (each byte has 8 bits).
Since `ptr` is a int pointer, the command `ptr = ptr + 1` shifts stored memory address of `ptr` in 4 bytes.

The address of the first int array element is                : 0x7ffe100ee500
The store value (.i.e., the memory address) of the pointer is: 0x7ffe100ee504

然而,这与我对内存地址和系统内存架构的推理方式相矛盾:

• 我使用的是64位系统,这意味着每个内存地址存储64位,或者相当于8个字节。
• 我的第一个整数元素数组的内存地址是 0x7ffe100ee500
• 由于每个内存地址最多可存储8个字节 C int 变量大小为4字节,一个内存地址就足以存储它(我想剩下的4个字节 0x7ffe100ee500 被简单地忽略)。
• 代码 ptr = ptr + 1 使指针的存储值(即,内存地址 0x7ffe100ee500 )向前移动4个字节。
• 由于每个存储器地址最多可存储8个字节,因此将存储器地址移位到 0x7ffe100ee501 就足够了。

然而,我得到了 0x7ffe100ee504 ,就好像每个存储器地址只包含1个字节。有人能帮我理解一下吗?