Loss of precision on adding doubles

Question

I have add the two double number.

double a=4.0;
double b=4.0;
cout<<a+b;

Answer is 8 only but required 8.0.

Please be aware that 8.0 and 8 are the exact same numbers - they would have the same bitwise representation. They are indistinguishable. If you want to specify an *output format* that's a different matter. — Jon Skeet, Jul 14 '17 at 10:35
but i have solve the hackerrank problem . where test case accept the 8.0 value not 8. he is not submit the answer — Bilal Siddiq, Jul 14 '17 at 10:39
@BilalSiddiq I will repeat what already was said: you are having issue with specifying **output format**, and **not** with the actual precision of the number, as is suggested by the answer. In addition: if you can't understand what comments are saying, there's no need to tell us, that we don't know the answer. — Algirdas Preidžius, Jul 14 '17 at 10:54
Here is my code int main() { double d = 4.0; double doubleValue; cin>>doubleValue; cout< — Bilal Siddiq, Jul 14 '17 at 10:57
The problem is **not** the addition; all of the values involved can be exactly represented as floating-point values. The problem is in how you **display** the result. — Pete Becker, Jul 14 '17 at 11:27

score 2 · Accepted Answer · edited Jul 14 '17 at 10:41

2

I believe what you want is something like this:

cout << std::setprecision(1) << a+b;

edited Jul 14 '17 at 10:41

msc

answered Jul 14 '17 at 10:39

Hashibuto

1 Answers1