michelbi@oregon.uoregon.edu (Michel Biedermann) (05/23/89)
Here is the most elegant answer to the problem of rounding-off an arbitrary number to an arbitrary decimal level. #define round(x,y) ((x < 0.0F? ceil((x)/(y)-.5F)*(y) : floor((x)/(y)+.5F)*(y))) round(1.5555,.1) will return 1.6 round(-1.555,.1) will return -1.6 round(.502,.01) will return .50 Credit is due to Steve Emmerson (steve@umigw.miami.edu). Thanks Steve... Michel Biedermann U. of Oregon