C++二维数组内存结构疑问：相同地址对应的不同指针含义解析

Hey there! No worries at all about your English—your question is totally clear, and it's a great one that trips up a lot of people learning C++ arrays and pointers. Let's break this down step by step.

First off, let's clarify how a 2D array lives in memory. Your int v[2][2] is actually stored as a single block of contiguous memory. The elements are laid out row by row:
v[0][0] → v[0][1] → v[1][0] → v[1][1]

All of these are right next to each other in memory, so the starting address of the entire array (v), the starting address of the first row (v[0]), and the address of the first element (&v[0][0]) all point to the exact same spot in RAM. That's why your print statements show the same hex value for all three.

The key here is pointer type, not just the address value. Even though all three point to the same memory location, their types determine how the compiler interprets them—especially when you do pointer arithmetic (like adding 1). Let's break down each one:

• v: When you use the array name v in most contexts, it decays to a pointer to the first element of the array. Since v is a 2D array (an array of arrays), its first element is the entire first row (v[0], which is an array of 2 ints). So v has the type int (*)[2]—a pointer to an array of 2 ints. If you did v + 1, it would skip an entire row (2 ints, 8 bytes on your system) and point directly to the start of the second row (v[1]).

• v[0]: This is the name of the first row (a 1D array of 2 ints). Like any array name, it decays to a pointer to its first element (v[0][0]), so its type is int*. If you did v[0] + 1, it would skip one int (4 bytes) and point to v[0][1].

• &v[0][0]: This is the direct address of the very first integer element in the array. Its type is also int*, same as v[0] after decay. So &v[0][0] + 1 would also point to v[0][1].

• &v: This is the address of the entire 2D array itself. Its type is int (*)[2][2]—a pointer to a 2x2 array of ints. If you did &v + 1, it would skip the entire 2D array (4 ints, 16 bytes) and point to the memory right after the array ends.

关于你的sizeof结果 #

You noticed all three sizeof calls return 4. That's because on a 32-bit system, all pointer types take up the same amount of memory (4 bytes). Even though the pointers have different types, the size of the variable storing the address is the same. On a 64-bit system, all of these would return 8 instead.

To see the difference in behavior clearly, try modifying your code to print the result of pointer arithmetic:

#include <iostream>
using namespace std;
int main() {
    int v[2][2] = {0,};
    cout << "v: " << v << endl;
    cout << "v + 1: " << v + 1 << endl; // Skips 8 bytes (one row)
    cout << endl;
    cout << "v[0]: " << v[0] << endl;
    cout << "v[0] + 1: " << v[0] + 1 << endl; // Skips 4 bytes (one int)
    cout << endl;
    cout << "&v[0][0]: " << &v[0][0] << endl;
    cout << "&v[0][0] + 1: " << &v[0][0] + 1 << endl; // Skips 4 bytes
    cout << endl;
    cout << "&v: " << &v << endl;
    cout << "&v + 1: " << &v + 1 << endl; // Skips 16 bytes (entire array)
    cout << sizeof(&v) << " " << sizeof(&v[0]) << " " << sizeof(&v[0][0]) << endl;
    return 0;
}

You'll see the addresses jump by different amounts, which shows how the compiler interprets each pointer type.

To wrap it up:

• A 2D array is just a contiguous block of memory, laid out row by row.
• v, v[0], and &v[0][0] all point to the same starting address, but their types are different.
• The pointer type determines how pointer arithmetic works (how much memory is skipped when you add 1).
• All pointers have the same size on a given system, regardless of their type.

内容的提问来源于stack exchange，提问作者이강욱