考虑以下C代码

#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，它与我对内存地址和系统内存架构的推理方式相矛盾：

• 我在一个64位系统上，这意味着每个内存地址存储64位，或相当于8个字节。
• 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存储8个字节，而我的Cint变量是4个字节，一个内存地址就足以存储它（我假设0x7ffe100ee500中剩下的4个字节被忽略了）。
• 代码ptr = ptr + 1使指针的存储值（即存储器地址0x7ffe100ee500）向前移动4个字节。
• 由于每个内存地址最多存储8个字节，因此将内存地址移位到0x7ffe100ee501就足够了。

然而，我得到了0x7ffe100ee504，好像每个内存地址只包含1个字节。有没有人能给予给我一个帮助来理解它？