指针所存储的内存地址的增加不是应该取决于系统的架构吗？

Question

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;
}

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：

• 我使用的是 64 位系统，这意味着每个内存地址存储 64 位，即 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存储 8 个字节，而我的

  C

  int

  变量是 4 字节大小，因此一个内存地址足以存储它（我想

  0x7ffe100ee500

  中剩余的 4 个字节会被忽略）。
• 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。
• 由于每个内存地址最多可存储 8 个字节，因此将内存地址移至

  0x7ffe100ee501

  即可。

但是，我得到

0x7ffe100ee504

，就好像每个内存地址只包含 1 个字节。有人可以帮我理解吗？

Answer 1

我使用的是 64 位系统，这意味着每个内存地址存储 64 位，即 8 个字节。

错误。每个可寻址内存地址，根据定义，存储 1 个字节。拥有 64 位系统意味着地址是 64 位值。

因此，以 32 位

int

为例，其起始地址为 0x7ffe100ee500，该值占用地址 0x7ffe100ee500、0x7ffe100ee501、0x7ffe100ee502 和 0x7ffe100ee503。