Why my code greater output d1>d2

Question

double d1=100 - 99.99 ;
double d2=10 - 9.99 ;
the output d1>d2

Answer 1

Write this and you will understand the problem.

cout.precision(17);
cout<<d1<<endl<<d2;

Basically you can't do much about it . To store floating values as binary value there is some rounding off of value after some digits due to which there is some precision loss .

Maybe there are libraries out there that can fix the value at after some decimal digits but i am not aware of any .

Additionally take a look at this thread for other ways to print the full double value .

https://stackoverflow.com/questions/554063/how-do-i-print-a-double-value-with-full-precision-using-cout

Answer 2

https://code.sololearn.com/c4193UFx86gq/?ref=app

Answer 3

Thanks