为什么我的质量计算器不能正确处理空格?
Why does my mass calculator not work properly with spaces?
#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 space 的重量单位效果很好。问题是 Long tn(Long ton) 和 sh tn (short ton) 单位不起作用,我假设这是因为字符串之间的 space。谁能帮忙。提前致谢。
istream::operator>>
标记空白字符。当您输入 sh tn
时,只有 sh
会保存在 unitType
中,tn
会留在流中。
当您在流中获得 sh
或 long
时,检查另一个标记并查看它是否是 tn
。一个可靠的解决方案可能是 Unit
class 带有一个自由函数 std::istream& operator>>(std::istream, Unit&)
来进行流提取。
std::istream
的 operator>>(std::string &)
你在这里使用:
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
将是 12.0
,而 unitType
将是 "long"
。
您的问题的解决方案可能涉及 std::getline
,如
std::cin >> mass;
std::getline(std::cin, unitType);
这将一直读到下一个换行符。但是,这不会像 operator>>
那样去除前导空格,因此您会得到 " long tn"
而不是 "long tn"
。您需要像这样显式地忽略这些空格:
std::cin >> std::ws;
这最终会给您留下
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 trailing 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
#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 space 的重量单位效果很好。问题是 Long tn(Long ton) 和 sh tn (short ton) 单位不起作用,我假设这是因为字符串之间的 space。谁能帮忙。提前致谢。
istream::operator>>
标记空白字符。当您输入 sh tn
时,只有 sh
会保存在 unitType
中,tn
会留在流中。
当您在流中获得 sh
或 long
时,检查另一个标记并查看它是否是 tn
。一个可靠的解决方案可能是 Unit
class 带有一个自由函数 std::istream& operator>>(std::istream, Unit&)
来进行流提取。
std::istream
的 operator>>(std::string &)
你在这里使用:
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
将是 12.0
,而 unitType
将是 "long"
。
您的问题的解决方案可能涉及 std::getline
,如
std::cin >> mass;
std::getline(std::cin, unitType);
这将一直读到下一个换行符。但是,这不会像 operator>>
那样去除前导空格,因此您会得到 " long tn"
而不是 "long tn"
。您需要像这样显式地忽略这些空格:
std::cin >> std::ws;
这最终会给您留下
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 trailing 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
.