11_以一維指標表示二維、三維陣列－藍影

關於這個問題之前有不少人提過，只是換個型式又出現而已。這部份沒有想像中那麼難，只是一樣要拿個紙筆畫一下圖就是了。這次我不多說，大家仔細看範例還有執行結果就知道是要幹嘛的了…

1. 原始碼

// ====================================
// FileName: Ptr1To2.cpp
// Author : Edison.Shih.
// Complier: VC 2008

#include <stdio.h>
#include <stdlib.h>
// ====================================

#define X        2
#define Y        3
#define Z        4

#define ROW 2
#define COL 4

int main(int argc, char **argv)
{
int i=0, j=0, k=0;
int *ptr1 = NULL;

// === 一維陣列 ptr1 表示二維陣列 Array2[ROW][COL]=== //

        int Array2[ROW][COL] = {{0,1,2,3},        {4,5,6,7}};
        // set the arry , method1
        ptr1 = (int*)malloc(sizeof(int)*ROW*COL);
        for(i=0; i<ROW; i++){
                for(j=0; j<COL; j++){
                        ptr1[i*COL+j] = Array2[i][j];
                }
        }
        // read the array , method2
        for(i=0; i<ROW*COL; i++) printf("ptr1[%d] = %d\n", i, ptr1[i]);
        for(i=0; i<ROW; i++){
                for(j=0; j<COL; j++){
                        printf("*(ptr1+%d*COL+%d)=%d\n", i, j, *(ptr1+i*COL+j));
                }
        }
        free(ptr1);

        // === 一維陣列 ptr1 表示三維陣列 Array[X][Y][Z]=== //
        int Array[X][Y][Z] = {
                { {0,1,2,3},{4,5,6,7}, {8,9,10,11} },
                { {12,13,14,15},{16,17,18,19}, {20,21,22,23} }
        };

        // set the array, method1
        printf("\n");
        ptr1 = (int*)malloc(sizeof(int)*X*Y*Z);
        for(i=0; i<X; i++){
                for(j=0; j<Y; j++){
                        for(k=0; k<Z; k++){
                                ptr1[i*Y*Z+j*Z+k] = Array[i][j][k];
                        }
                }
        }

        // read the array, method2
        for(i=0; i<X*Y*Z; i++) printf("*(ptr1+%d)=%d\n", i, *(ptr1+i));
        for(i=0; i<X; i++){
                for(j=0; j<Y; j++){
                        for(k=0; k<Z; k++){
                                printf("ptr1[%d*Y*Z+%d*Z+%d] = %d\n", i, j, k, *(ptr1+i*Y*Z+j*Z+k));
                        }
                }
        }

return 0;
}

2. 執行結果

ptr1[0] = 0
ptr1[1] = 1
ptr1[2] = 2
ptr1[3] = 3
ptr1[4] = 4
ptr1[5] = 5
ptr1[6] = 6
ptr1[7] = 7
*(ptr1+0*COL+0)=0
*(ptr1+0*COL+1)=1
*(ptr1+0*COL+2)=2
*(ptr1+0*COL+3)=3
*(ptr1+1*COL+0)=4
*(ptr1+1*COL+1)=5
*(ptr1+1*COL+2)=6
*(ptr1+1*COL+3)=7

*(ptr1+0)=0
*(ptr1+1)=1
*(ptr1+2)=2
*(ptr1+3)=3
*(ptr1+4)=4
*(ptr1+5)=5
*(ptr1+6)=6
*(ptr1+7)=7
*(ptr1+8)=8
*(ptr1+9)=9
*(ptr1+10)=10
*(ptr1+11)=11
*(ptr1+12)=12
*(ptr1+13)=13
*(ptr1+14)=14
*(ptr1+15)=15
*(ptr1+16)=16
*(ptr1+17)=17
*(ptr1+18)=18
*(ptr1+19)=19
*(ptr1+20)=20
*(ptr1+21)=21
*(ptr1+22)=22
*(ptr1+23)=23
ptr1[0*Y*Z+0*Z+0] = 0
ptr1[0*Y*Z+0*Z+1] = 1
ptr1[0*Y*Z+0*Z+2] = 2
ptr1[0*Y*Z+0*Z+3] = 3
ptr1[0*Y*Z+1*Z+0] = 4
ptr1[0*Y*Z+1*Z+1] = 5
ptr1[0*Y*Z+1*Z+2] = 6
ptr1[0*Y*Z+1*Z+3] = 7
ptr1[0*Y*Z+2*Z+0] = 8
ptr1[0*Y*Z+2*Z+1] = 9
ptr1[0*Y*Z+2*Z+2] = 10
ptr1[0*Y*Z+2*Z+3] = 11
ptr1[1*Y*Z+0*Z+0] = 12
ptr1[1*Y*Z+0*Z+1] = 13
ptr1[1*Y*Z+0*Z+2] = 14
ptr1[1*Y*Z+0*Z+3] = 15
ptr1[1*Y*Z+1*Z+0] = 16
ptr1[1*Y*Z+1*Z+1] = 17
ptr1[1*Y*Z+1*Z+2] = 18
ptr1[1*Y*Z+1*Z+3] = 19
ptr1[1*Y*Z+2*Z+0] = 20
ptr1[1*Y*Z+2*Z+1] = 21
ptr1[1*Y*Z+2*Z+2] = 22
ptr1[1*Y*Z+2*Z+3] = 23