问题
I have this code (very basic):
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
float a = 0.0,
b = 0.0,
c = 0.0;
cout<<"Input a: ";
cin>>a;
cout<<"input b: ";
cin>>b;
cout<<endl;
c = a / b;
cout<<"Result: "<<fixed<<setprecision(2)<<c<<endl;
return 0;
}
When I enter two numbers (say, a = 513 and b = 791) I get 0.65. Calculator shows that the correct answer is 0.648. I understand that my code rounds up the last decimal number but this is not what I want.
How can I get it to where it just stays as 0.64 and not 0.65?
回答1:
If you would like to truncate the value to two decimal places, you can multiply it by 100, truncate to integer, and then divide by 100, like this:
c = a / b;
c = floor(100 * c) / 100;
cout<<"Result: "<<fixed<<setprecision(2)<<c<<endl;
Demo on ideone.
回答2:
You can use trunc
to truncate to a certain number of digits:
c = a / b;
// truncate past two decimals:
c = trunc(c * 100) / 100;
cout<<"Result: "<<fixed<<setprecision(2)<<c<<endl;
of for a generic function:
int trunc(double val, int digits)
{
double pow10 = pow(10,digits);
return trunc(val * pow10) / pow10;
}
then use
cout << "Result: " << fixed << setprecision(2) << trunc(c,2) << endl;
来源:https://stackoverflow.com/questions/19759131/how-to-display-a-fixed-number-of-digits-in-c-without-rounding