problems in floating point comparison
c, floating-point, floating-point-conversion
Solution
`f` is using `float` precision, but 0.98 is in `double` precision by default, so the statement `f <= 0.98` is compared using `double` precision.
The `f` is therefore converted to a `double` in the comparison, but may make the result slightly larger than 0.98.
Use
if(f <= 0.98f)
or use a `double` for `f` instead.
In detail... assuming `float` is IEEE single-precision and `double` is IEEE double-precision.
These kinds of floating point numbers are stored with base-2 representation. In base-2 this number needs an infinite precision to represent as it is a repeated decimal:
0.98 = 0.1111101011100001010001111010111000010100011110101110000101000...
A `float` can only store 24 bits of significant figures, i.e.
0.111110101110000101000111_101...
^ round off here
= 0.111110101110000101001000
= 16441672 / 2^24
= 0.98000001907...
A `double` can store 53 bits of signficant figures, so
0.11111010111000010100011110101110000101000111101011100_00101000...
^ round off here
= 0.11111010111000010100011110101110000101000111101011100
= 8827055269646172 / 2^53
= 0.97999999999999998224...
So the 0.98 will become slightly larger in `float` and smaller in `double`.
Problem
``` void main() { float f = 0.98; if(f <= 0.98) printf("hi"); else printf("hello"); getch(); } ``` I am getting this problem here.On using different floating point values of f i am getting different results. Why this is happening?