3 ms·
Julia goes the ADA way, supporting both. There's div(x,y) # truncated division; quotient rounded towards zero fld(x,y) # floored division; quotient
by FabHK 8y ago
Julia goes the ADA way, supporting both. There's
div(x,y) # truncated division; quotient rounded towards zero
fld(x,y) # floored division; quotient rounded towards -Inf
and correspondingly
rem(x,y) # remainder; satisfies x == div(x,y)*y + rem(x,y); sign matches x
mod(x,y) # modulus; satisfies x == fld(x,y)*y + mod(x,y); sign matches y
and the convenience functions
divrem(x,y) # returns (div(x,y),rem(x,y)) ## Lotus 123 convention
fldmod(x,y) # returns (fld(x,y),mod(x,y)) ## Excel convention
(There's also mod1 and fld1, and there's mod2pi(x) which is more precise than mod(x, 2pi).)