How to covert int to double in c++

Question

I want to convert 3 to 3.00. I have done explicit type conversion. but guess it doesn't seem to work!whats wrong?

my code:

#include<iostream>


using namespace std;

int main(){
    int PI = 3;
    double a = static_cast<double>(PI);
    cout<<a;


}

output: 3

@jabaa yes. but i would like to covert to a decimal form with two zeros after it — Abhinav K, May 26 '21 at 14:37
you are confusing the value with its representation on your screen. For showing decimal places there are iomanipulators — 463035818_is_not_an_ai, May 26 '21 at 14:38
That's `cout << PI << ".00";` For this simple task (integer with two zeros after it) you don't even need `iomanip`. `iomanip` is very useful for floating point numbers but overkill for integers. — , May 26 '21 at 14:39
@jabaa thanks, but i wanted to use the concept of type coversion. — Abhinav K, May 26 '21 at 14:47
The problem is that type conversion doesn't help here. You don't want to change the type/value but the string representation and that's mostly unrelated to the type. Therefore `cout << PI;` (integer) and `cout << a;` (double) print the same string. — , May 26 '21 at 15:11

score 4 · Accepted Answer · answered May 26 '21 at 14:43

4

This may help.

#include<iostream>

using namespace std;

int main(){
    int PI = 3;
    double a = static_cast<double>(PI);
    cout.precision(2);
    cout <<  std::fixed << a;
}

answered May 26 '21 at 14:43

0xdw

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How to covert int to double in c++

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