using namespace std;
int main()
{
return 0;
}
double C2F()
{
f = value * 9 / 5 + 32;
return f;
}
double K2F()
{
f = (value - 273.15) * 1.8 + 32.0;
return f;
}
double N2F()
{
f = value * 60 / 11 + 32;
return f;
}
我无法调用这些函数来计算温度转换,而不是从案例中计算温度转换。添加这些功能后,程序甚至不会编译。 “错误:期待”;“”
答案 0 :(得分:3)
您无法在另一个函数中声明或定义函数。将您的定义移到int main(){ ... }
之外。
答案 1 :(得分:1)
首先,您无法在 main() 函数中声明另一个函数。
其次你的所有函数都有一个返回类型,但令人惊讶的是你调用它们是因为它们是无效的。使函数为void而不是返回类型。例如....
void C2F()
{
f = value * 9 / 5 + 32;
}
然后
case 'C':
C2F();
cout << value << "C is " << f << " in Farenheit" << endl;
break;
OR。 您可以在double类型变量中接收返回值,然后打印该值。
case 'C':
cout << value << "C is " << C2F() << " in Farenheit" << endl;
break;
答案 2 :(得分:1)
这就是你想要的。
using namespace std;
double C2F(double f)
{
return f * 9 / 5 + 32;
}
double K2F(double f)
{
return ((f - 273.15) * 1.8 + 32.0);
}
double N2F(double f)
{
return (f * 60 / 11 + 32);
}
int main()
{
char function;
double value;
cout << "This temperature Conversion program converts other temperatures to farenheit" << endl;
cout << "The temperature types are" << endl;
cout << "" << endl;
cout << "C - Celcius" << endl;
cout << "K - Kelvin" << endl;
cout << "N - Newton" << endl;
cout << "X - eXit" << endl;
cout << "" << endl;
cout << "To use the converter you must input a value and one of the temperature types." << endl;
cout << "For example 32 C converts 32 degrees from Celsius to Fahrenheit" << endl;
cin >> value >> function;
function = toupper(function);
while (function != 'X')
{
switch (function)
{
case 'C':
cout << value << "C is " << C2F(value) << " in Farenheit" << endl;
break;
case 'K':
cout << value << "K is " << K2F(value) << " in Farenheit" << endl;
break;
case 'N':
cout << value << "N is " << N2F(value) << " in Farenheit" << endl;
break;
default:
cout << "Correct choices are C, K, N, X" << endl;
}
cout << "Please enter a value and it's type to be converted" << endl;
cin >> value >> function;
function = toupper(function);
}
return 0;
}