A correct implementation of hypot( x, y ) avoids the overflow that might otherwise happen if either x or y is large enough that when squared it would overflow, but the answer is small enough not to cause overflow.
![]() | hypot说明float hypot ( float x, float y )hypot() 函数将会跟据直角三角形的两直解边长度 x 和 y 计算其斜边的长度。或者是从标点 (x, y) 到原点的距离。该函数的算法等同于 sqrt(x*x + y*y)。 ![]()
R. Victor Klassen
24-Jun-2005 11:03
A correct implementation of hypot( x, y ) avoids the overflow that might otherwise happen if either x or y is large enough that when squared it would overflow, but the answer is small enough not to cause overflow.
robinv at ecosse dot net
08-Jan-2004 01:18
A simpler approach would be to allow an arbitrary number of parameters. That would allow for whatever number of dimensions you want *and* it would be backwards compatible with the current implementation. | ![]() |