c++ - 检查尺寸值及其内容的功能

Question

我正在尝试编写一个程序来检查矩形的尺寸是否大于零。在 void 函数 Check 中，我尝试使用数组来检查值并使用字符串来显示用户错误的维度。我收到一个错误，它“无法将参数 1 从‘double[6]’转换为‘double’。

#include <iostream>
#include <string>

using namespace std;

void Check(double, string);

int main()
{
    const int size = 3;
    double DimArray[size];
    string MyArray[size] = { "Height", "Length", "Width"};

    cout << "Enter the height, length and width of rectangle: ";
    cin >> DimArray[0] >> DimArray[1] >> DimArray[2];

    Check(DimArray, MyArray);

    return 0;
}

void Check(double arr1[], string arr2[])
{
    int i;
    for (i = 0; i < 4; i++)
    {
        if (arr1[i] <= 0)
          cout << "Your entered " << arr2[i] << "is less than zero!";
        cout << "Please enter a valid number --> ";
        cin >> arr1[i];
    }
}

Answer 1

您应该正确声明函数。而不是

void Check(double, string);

应该有

void Check( double[], const std::string[], size_t );

也代替了函数体中的循环

for (i = 0; i < 4; i++)

必须有

for (i = 0; i < 3; i++)

函数可以定义为

void Check( double arr1[], const std::string arr2[], size_t n )
{
   for ( size_t i = 0; i < n; i++ )
   {
       while ( arr1[i] <= 0 )
       {
            std::cout << "Your entered " << arr2[i] << "is not positive!\n";
            std::cout << "Please enter a valid number --> ";
            std::cin >> arr1[i];
       }
   }
}

或者如果你将定义文件范围的常量

const size_t SIZE = 3;

然后可以简化函数定义(以及相应的声明)

void Check( double arr1[], const std::string arr2[] )
{
   for ( size_t i = 0; i < SIZE; i++ )
   {
       while ( arr1[i] <= 0 )
       {
            std::cout << "Your entered " << arr2[i] << "is not positive!\n";
            std::cout << "Please enter a valid number --> ";
            std::cin >> arr1[i];
       }
   }
}

此外，最好定义一个 const char *

数组而不是 std::string(s) 数组

const char * MyArray[size] = { "Height", "Length", "Width"};

因为据我了解您不会更改它。