Why is this output when compare a float and a double
This question already has an answer here:
When you do the first assignment, the constant is truncated to fit the float . When you do the second assignment, the float -precision literal 1.1111...11F is converted to double . Since c contains the value of the 1.1111...11F literal, initialization of d is equivalent to
double d = ((double)c);
Both assignments change the precision of the constant from the literal, but they change it differently. That is why you see different printouts in the first two WriteLine s.
When you compare c and d , the value with the lower precision, ie c , is converted to the type with the higher precision, ie to double . That is the same conversion as the conversion that has been performed when you assigned the 1.1111...11F literal to variable d , so the values compare the same in the == operation. In other words, when you do this
Console.WriteLine(c == d);
the compiler does this:
Console.WriteLine(((double)c) == d);
That is why the comparison returns true .
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