Question

我想拥有一个存储2个项目的数组，所以这是最好的方法吗？

char array[5][2]= {{"data1","1"},
                {"data2","2"},
                {"data3","3"},
                {"data4","4"},
                {"data5","5"}};

如何在“data2”，“2”中存储内容？

Answer 1

2D char数组在这里不起作用，但是一个struct数组可以满足你的需要：

    struct string_pairs
    {
        char str1[10];
        char str2[10];
    } array[] = {{"data1","x"},
                {"data2","x"},
                {"data3","x"},
                {"data4","x"},
                {"data5","x"}};

Answer 2

char

的数组

char arr1[6] = "data1";               // 1D array

char

的数组数组

char arr2[2][6] = { {"data1"},        // 2D array
                    {"1"}      };

char

数组的数组

char arr3[5][2][6] = {{"data1","1"},  // 3D array
                      {"data2","2"},
                      {"data3","3"},
                      {"data4","4"},
                      {"data5","5"}};

将"dataX","X"分配给数组的第二个元素

strcpy(arr[1][0], "dataX");
strcpy(arr[1][1], "X");

Answer 3

您可以将其设为2d char*数组

char *array[5][2]= {{"data1","1"},
                   {"data2","2"},
                   {"data3","3"},
                   {"data4","4"},
                   {"data5","5"}};

然后要更改字符串数据，您只需访问各个组件（编辑{"data2","2"}）：

array[1][0] = "words", array[1][1] = "a";

基本上，当您使用第一组方括号array[1]，访问行{"data2","2"}，然后使用第二组方括号时，您正在访问行内部的组件。因此，array[1][0]最初为"data2"，array[1][1]最初为"2"。就像array[2][0] "data3"和array[2][1] "3"一样。

Answer 4

我认为这是结构的典型任务。在我看来，这应该有效：

struct strings  // definition of a struct calld strings
{
char[] a;
char[] b;
}

struct strings arrayofstrings[n];  // instancing an array of structs

arrayofstrings[i].a = {"data"};  // reference to the element "a" of the i-part of the array
arrayofstrings[i].b = {"1"};

你可以创建n个结构实例＆＃34;字符串＆＃34;

如何存储2个char数组的数组？

4 个答案: