Question

#include <iostream>
#include <string>



using namespace std;

/*
Function Name: weightConv
Purpose: To take the weight and convert the following number to the coressponding weight unit
Return : 0
*/
  double weightConv(double w, string weightUnit)
{

      if (weightUnit == "g" || weightUnit == "G" )
        cout << " Mass = " <<  w * 0.035274 << "oz";
    else if (weightUnit == "oz"||weightUnit == "OZ"||weightUnit == "oZ" ||weightUnit == "Oz")
        cout << " Mass = " <<  w / 28.3495 << "g";
    else if (weightUnit == "kg"||weightUnit == "KG"||weightUnit == "Kg" ||weightUnit == "kG")
        cout << " Mass = " <<  w * 2.20462 << "lb";
    else if (weightUnit == "lb" ||weightUnit == "LB" ||weightUnit== "Lb" ||weightUnit == "lB")
        cout << " Mass = " <<  w * 0.453592 << "kg";
    else if (weightUnit == "Long tn" ||weightUnit == "LONG TN"|| weightUnit == "long tn" || weightUnit == "long ton")
        cout << " Mass = " <<  w * 1.12 << "sh tn";
    else if (weightUnit == "sh tn" || weightUnit == "SH TN")
        cout << " Mass = " << w / 0.892857 << " Long tons";
    else if (weightUnit == "s" || weightUnit == "S")
        cout << " Mass = " << w * 6.35029 << "stones";
    else
        cout << "Is an unknown unit and cannot be converted";

    return 0;   
}// end of weightCov function


int main()
{
    for (;;)
    {

        double mass;
        string unitType;
        cout << "Enter a mass and its unit type indicator(g,kg,lb,oz,long tn,or sh tn)" << endl;
        cin >> mass >> unitType;

            // Output Results
            cout << weightConv(mass, unitType) << endl;

    }// end of for loop
}// end of main

没有空间的重量单位效果很好。问题是Long tn（Long ton）和sh tn（短吨）单元不起作用，我假设这是因为弦之间的空间。谁能帮忙。提前谢谢。

Answer 1

您在此处使用的

std::istream＆＃39; operator>>(std::string &)：

cin >> mass >> unitType;

读取以空格分隔的标记。这意味着，当您在输入流中输入"12 long tn"时，mass将为12.0，unitType将为"long"。

问题的解决方案可能涉及std::getline，如

std::cin >> mass;
std::getline(std::cin, unitType);

这将读到下一个换行符。但是，这并不像operator>>那样剥离前导空格，因此您需要使用" long tn"而不是"long tn"。你需要明确地忽略那些空格：

std::cin >> std::ws;

这最终会让你

std::cin >> mass >> std::ws;      // read mass, ignore whitespaces
std::getline(std::cin, unitType); // the rest of the line is the unit

请注意，这不会删除尾随空格，因此如果您的用户键入"12 long tn "，它将无法识别该单位。如果这是一个问题，您必须从unitType的末尾手动删除它们。

Answer 2

istream::operator>>对空格字符进行标记。当您输入sh tn时，只有sh会保存在unitType中，tn将保留在信息流中。

当您在流中获得sh或long时，请检查另一个令牌，看看它是否为tn。一个强大的解决方案可能是一个Unit类，它有一个自由函数std::istream& operator>>(std::istream, Unit&)来进行流提取。

为什么我的质量计算器无法正常使用空格？

2 个答案: