Re: [問題] 一個二維陣列的問題

[s3748679 (冷羽憶塵)]

※ 引述《dendrobium (石斛蘭)》之銘言：
: ※ 引述《Arim (Arim5566)》之銘言：
: : 小弟愚昧,請問一下
: : int dis[20][20]; //這是一個二維陣列,dis指向一個二維陣列
: : int **p;
: : 一般做動態陣列的時候,可以使p指向一個二維的陣列
: : 現在我不做動態配置,直接利用assignment: p=dis;
: : 將dis的位址assign給 p,可是compile不會給我過
: : 如果同樣的問題用一維的陣列就可以過
: : 請問這是為什麼呢??
: 簡單的來說, 這是因為 type 不同所導致的
: 20有點長 , 用 int dis[2][2]; 來說好了
: 他的記憶體配置是這樣的連續分佈
:            ┌───┐
:         dis│      │(type: int (*)[2] )
:            └─┼─┘
:                │
:                ↓
:              [0][0]  [0][1]  [1][0]  [1][1]
:          ─┬───┬───┬───┬───┬───┬───┬─
:         .. │      │      │      │      │      │      │ ...
:          ─┴───┴───┴───┴───┴───┴───┴─
:             (type: int)
: 而如果是 int **p; 的話, 則是像下圖那樣
:                     [0]   [1]
:      ┌─┐       ┌──┬──┐ (type: int*)
:    p │  ┼──→ │    │    │
:      └─┘       └─┼┴─┼┘           [0]     [1]
:   (type: int**)       │    │          ┌───┬───┐
:                       ╰──┼────→│      │      │ (type: int)
:                             │          └───┴───┘
:                             │             [0]     [1]
:                             │          ┌───┬───┐
:                             ╰────→│      │      │ (type: int)
:                                         └───┴───┘
: 這三塊記憶體都不一定是連續的
: 所以雖然兩個都是二維陣列
: 但是其實記憶體配置方式差滿多的
orz...原來就是出在加法上，害我傻了那麼久QQ。 (角落劃圈圈中...)

提供一下測試結果:
        x[0][0] = 11
        x[0][1] = 21
        x[1][0] = 12
        x[1][1] = 22

        示意圖:
        11      21
        12      22

        x = 0x22ff60
        &x = 0x22ff60
        *x = 0x22ff60

        x[0] = 0x22ff60
        &x[0] = 0x22ff60
        *x[0] = 11

        x[0][0] = 11
        &x[0][0] = 0x22ff60
        x[0][1] = 21
        &x[0][1] = 0x22ff64

        x[1] = 0x22ff68
        &x[1] = 0x22ff68
        *x[1] = 12

        x[1][0] = 12
        &x[1][0] = 0x22ff68
        x[1][1] = 22
        &x[1][1] = 0x22ff6c

        x + 0 = 0x22ff60
                *(x+0) = 0x22ff60
                        **(x+0) = 11
        x + 1 = 0x22ff68
                *(x+1) = 0x22ff68
                        **(x+1) = 12
        x + 2 = 0x22ff70
                *(x+2) = 0x22ff70
                        **(x+2) = 59
        x + 3 = 0x22ff78
                *(x+3) = 0x22ff78
                        **(x+3) = 2293680

        *x + 0 = 0x22ff60
                *x+0 = 0x22ff60
                        *(*x+0) = 11
        *x + 1 = 0x22ff64
                *x+1 = 0x22ff64
                        *(*x+1) = 21
        *x + 2 = 0x22ff68
                *x+2 = 0x22ff68
                        *(*x+2) = 12
        *x + 3 = 0x22ff6c
                *x+3 = 0x22ff6c
                        *(*x+3) = 22

        請按任意鍵繼續 . . .

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忘了附上程式碼....
#include <iostream>
using namespace std;

int main()
{
        int x[2][2];

        for (int i=0; i<2; i++)
                for (int j=0; j<2; j++)
                        x[j][i] = (i+1)*10 + (j+1);

        for (int i=0; i<2; i++)
                for (int j=0; j<2; j++)
                        cout << "x[" << i << "]"
                                 << "[" << j << "] = " << x[i][j] << endl;
        cout << endl;

        cout << "示意圖:" << endl;
        for (int i=0; i<2; i++){
                for (int j=0; j<2; j++)
                        cout << x[i][j] << "\t";
                cout << endl;
        }
        cout << endl;

        cout << "x = " << x << endl;
        cout << "&x = " << &x << endl;
        cout << "*x = " << *x << endl;
        cout << endl;

        for (int i=0; i<2; i++){
                //cout << "i = " << i << ":" << endl;

                cout << "x[" << i << "] = " << x[i] << endl;
                cout << "&x[" << i << "] = " << &x[i] << endl;
                cout << "*x[" << i << "] = " << *x[i] << endl;
                cout << endl;
                for (int j=0; j<2; j++){
                //      cout << "j = " << j << ":" << endl;

                        cout << "x["<< i << "][" << j << "] = " << x[i][j]
                                 << endl;
                        cout << "&x[" << i << "][" << j << "] = "
                                 << &x[i][j] << endl;
                }
                cout << endl;
        }

        for (int i = 0; i<4; i++)
                cout << "x + " << i << " = " << x + i << endl
                         << "\t" << "*(x+" << i << ") = " << *(x+i) << endl
                         << "\t\t" << "**(x+" << i <<") = " << **(x+i) << endl;
        cout << endl;

        for (int i = 0; i<4; i++)
                cout << "*x + " << i << " = " << *x + i << endl
                         << "\t" << "*x+" << i << " = " << *x+i << endl
                         << "\t\t" << "*(*x+" << i <<") = " << *(*x+i) << endl;
        cout << endl;

        system("pause");
        return 0;
}
※ 編輯: s3748679        來自: 218.164.83.22        (02/15 19:22)