c ++尝试初始化二维数组中的值

Question

我已经创建了一个名为 digits 的二维数组，并且想一个一个地初始化每个子数组以使我的代码更清晰。我了解以下代码有效：

string digits[2][5] = { { " - ","| |","   ","| |"," - " },{ " - ","| |","   ","| |"," - " } };

但我想知道为什么以下内容不起作用：

string digits[2][5];
digits[0] = { " - ","| |","   ","| |"," - " };
digits[1] = { " - ", "| |", "   ", "| |", " - " };

Answer 1

第二个不是初始化，它是赋值（digits的元素）。

string digits[2][5];                               // initialization
digits[0] = { " - ","| |","   ","| |"," - " };     // assignment of the 1st element of digits
digits[1] = { " - ", "| |", "   ", "| |", " - " }; // assignment of the 2nd element of digits

digits的元素是数组，原始的array不能整体赋值

Objects of array type cannot be modified as a whole: even though they are lvalues (e.g. an address of array can be taken), they cannot appear on the left hand side of an assignment operator

您可以使用 std::array 或 std::vector 来执行此操作，它们可以使用花括号初始化程序进行分配。

std::array<std::array<std::string, 5>, 2> digits;
digits[0] = { " - ","| |","   ","| |"," - " };
digits[1] = { " - ", "| |", "   ", "| |", " - " };

Answer 2

初始化与赋值有很大不同。初始化是在声明（调用构造函数）时为变量赋值，而赋值是声明然后赋值（调用赋值运算符）。要正确分配，请删除括号，然后使用循环或手动，例如：

#include <iostream>
#include <string>
using namespace std;


int main()
{
    string s[2][3];
    string Hi = "Hi there!";

    s[0][0] = "Hello there!";
    //....

    for(int i(0); i < 2; i++)
        for(int j(0); j < 3; j++)
            s[i][j] = Hi;

    for(int i(0); i < 2; i++)
        for(int j(0); j < 3; j++)
            cout << s[i][j] << endl;

    return 0;
}

Answer 3

如果您为了清楚起见而更改此行，还请考虑：

string digits[2][5] = {
        {" - ", "| |", "   ", "| |", " - "},
        {" - ", "| |", "   ", "| |", " - "}
    };

注意：考虑。缩进是一个强大的工具，但它可能会被滥用。

c ++尝试初始化二维数组中的值

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