我可以使用指针打印出来int数组数组一样，字符串数组？ [英] Can I use pointer to print out arrays of int array just like arrays of string?

问题描述

如果我想打印出像字符串数组：

 字符juices_A [] [12] = {火龙果，waterberry，sharonfruit};
 

我可以简单地使用 juices_A [0] ， juices_A [1] ， juices_A [3] ，分别作为指向字符串火龙果，waterberry ，sharonfruit，并打印出来。

但如果我想打印出来int数组数组这样的：

  int数据[3] [3] = {{1，2，3}，{4，5，6}，{7，8，9}};
 

我不能简单地用数据[0] ，数据[1] ，数据[2] 为指针 {1，2，3} ， {4，5，6} ， {7，8，9} ，我需要采用打印出来一个更复杂的方式。我需要使用内循环回路。所以，我想知道为什么我不能在INT情况下使用指针？我读的书的数组变量就像指针.. 所以我假设int数组变量数据[0]，数据[1]，数据[2] 也像指针...

 的#include＆LT;＆stdio.h中GT;诠释主（）{    INT I_A = 0;
    炭juices_A [] [12] = {火龙果，waterberry，sharonfruit};
    为（; I_A 3;; I_A ++）
        的printf（％S;juices_A [I_A]）;
    卖出期权（）;    INT i_B = 0;
    int数据[3] [3] = {{1，2，3}，{4，5，6}，{7，8，9}};
    为（; i_B 3;; i_B ++）{
        INT i_C = 0;
        为（; i_C 3;; i_C ++）
            的printf（％i中，数据[i_B] [i_C]）;
        的printf（）;
    }    返回0;
}
 

结果是：

 火龙果; waterberry; sharonfruit;
1,2,3; 4,5,6; 7,8,9，;
 

解决方案

您可以像使用指针访问数组，反之亦然。

下面是从下面的程序的输出：

 火龙果; waterberry; sharonfruit;
{{1，2，3，}，{4，5，6，}，{7，8，9，}，}
 

下面是您code的修改：

 的#include＆LT;＆stdio.h中GT;
＃定义ROW_WIDTH 3
＃定义TABLE_HEIGHT 3
诠释主要（无效）{    字符*果汁[] = {火龙果，waterberry，sharonfruit};
    int数据[TABLE_HEIGHT] [ROW_WIDTH] = {{1，2，3}，{4，5，6}，{7，8，9}};    对于（INT N = 0; N＆LT; ROW_WIDTH; N ++）
        的printf（％S;，果汁[N]）;    为int * IP =（INT *）的数据;    的printf（\\ N {）;
    的for（int i = 0; I＆LT; TABLE_HEIGHT;我++）{
       的printf（{）;
       对于（INT J = 0; J＆LT; ROW_WIDTH; J ++）{
           的printf（％d个，* IP +（（我* ROW_WIDTH）+ J））;
       }
       的printf（}）;
    }
    的printf（} \\ n）;
    返回0;
}
 

If I want to print out array of string like:

char juices_A[][12] = { "dragonfruit", "waterberry", "sharonfruit", };

I can simply use juices_A[0],juices_A[1],juices_A[3], respectively as a pointer to string "dragonfruit", "waterberry", "sharonfruit", and print them out.

But what if I want to print out array of int array like:

int data[3][3] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };

I can not simply use data[0],data[1],data[2] as pointer to { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 }, I need to adopt a more complicated way to print them out. I need to use loop inside loop. So I want to know why I can not use pointer in the int situation? I read on book "Array variables are like pointers.." so I assume int array variable data[0],data[1],data[2] are also like pointers...

#include <stdio.h>

int main() {

    int i_A = 0;
    char juices_A[][12] = { "dragonfruit", "waterberry", "sharonfruit", };
    for (; i_A < 3; i_A++)
        printf("%s;", juices_A[i_A]);
    puts("");

    int i_B = 0;
    int data[3][3] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
    for (; i_B < 3; i_B++) {
        int i_C = 0;
        for (; i_C < 3; i_C++)
            printf("%i,", data[i_B][i_C]);
        printf(";");
    }

    return 0;
}

the result is:

dragonfruit;waterberry;sharonfruit;
1,2,3,;4,5,6,;7,8,9,;

解决方案

You can use access arrays like pointers and vice versa.

Here's the output from the program below:

dragonfruit;waterberry;sharonfruit;
{ { 1, 2, 3, }, { 4, 5, 6, }, { 7, 8, 9, }, }

Here's a modification to your code:

#include <stdio.h>
#define ROW_WIDTH 3
#define TABLE_HEIGHT 3
int main(void) {

    char *juices[] = { "dragonfruit", "waterberry", "sharonfruit", };
    int data[TABLE_HEIGHT][ROW_WIDTH] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };

    for (int n = 0; n < ROW_WIDTH; n++) 
        printf("%s;", juices[n]);

    int *ip = (int *)data;

    printf("\n{ ");
    for (int i = 0; i < TABLE_HEIGHT; i++) {
       printf("{ ");
       for(int j = 0; j < ROW_WIDTH; j++) {
           printf("%d, ", *ip + ((i * ROW_WIDTH) + j));
       }
       printf("}, ");
    }
    printf("}\n");
    return 0;
}

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