I need help with a code in C ++, the exercise asks me to do a calculation, if the result is, for example 1.57, in the output it should show 1.5, (the variable can be float or double) without making the rounding.
#include <iostream>
#include <stdio.h>
using namespace std;
int main()
{
int c=0;
cin >> c;
while(c--){
int Vkh=0;
double Vms=0, Vm=0;
cin >> Vkh;
Vms=Vkh/3.6;
Vm=Vms/340;
printf(" %.1f\n", Vm);
}
return 0;
}
Thanks in advance.
The cmath library ( math.h for those coming from C), has a trunc function that allows you to truncate a float or a double . The truncation function eliminates the decimal part of the number. To eliminate a certain number of decimals you can choose to temporarily include those decimals in the whole part of the number, apply truncated and reposition them as decimals:
#include <cmath>
#include <iomanip>
#include <iostream>
int main()
{
float numero = 1.57;
std::cout << std::fixed << std::setprecision(1) << std::trunc(numero*10)/10;
}
Example in wandbox
This can be left more nice if we divide the code into functions:
constexpr double Exp(int numero)
{
return ( numero <= 0 )? 1 : 10 * Exp(numero-1);
}
double Truncar(double numero, int decimales)
{
return std::trunc(numero*Exp(decimales))/Exp(decimales);
}
void ImprimirDouble(double numero, int decimales)
{
std::cout << std::fixed << std::setprecision(decimales) << Truncar(numero,decimales) << '\n';
}
int main()
{
double numero = 3.777;
ImprimirDouble(numero,0);
ImprimirDouble(numero,1);
ImprimirDouble(numero,2);
}
Example in wandbox
Although I personally like more the option of giving a strong typing to the variable ... the code is cleaner. The example can be abstracted with templates and so on, but I think that it would lose the focus of what I'm trying to teach:
namespace MathUtils
{
constexpr double Exp(int numero)
{
return ( numero <= 0 )? 1 : 10 * Exp(numero-1);
}
double Truncar(double numero, int decimales)
{
return std::trunc(numero*Exp(decimales))/Exp(decimales);
}
}
class Double
{
double value;
public:
explicit Double(double value)
: value(value)
{ }
operator double() const
{ return value; }
friend std::ostream& operator<<(std::ostream& os, Double const& value);
};
std::ostream& operator<<(std::ostream& os, Double const& value)
{
auto precision = os.precision();
return os << std::fixed << MathUtils::Truncar(value,precision);
}
int main()
{
Double numero(3.777);
std::cout << std::setprecision(0) << numero << '\n';
std::cout << std::setprecision(1) << numero << '\n';
std::cout << std::setprecision(2) << numero << '\n';
}
Example in wandbox
This solution has several advantages:
This is a simple and fast way, I'm not proud of it, there must be a better one ...: - (
#include <cstdio>
using namespace std;
int main() {
printf("%f\n", 3.0/7);
printf("%.4f\n", 3.0/7);
printf("%.4f\n", (float)((int)(3.0/7*10000))/10000);
return 0;
}
$ g++ stack.cpp && ./a.out
0.428571
0.4286
0.4285