2
$\begingroup$
Compile[{}, Module[{n}, n = Length[{1, 2, 3, 4}]; n = n/2]]

The above code will emit errors

Compile::cset: Variable n of type _Integer encountered in assignment of type _Real. >>

Compile::extscalar: n=n/2 cannot be compiled and will be evaluated externally. The result is assumed to be of type Real. >>

Compile::cset: Variable n of type _Integer encountered in assignment of type _Real. >>

Compile::extscalar: n=n/2 cannot be compiled and will be evaluated externally. The result is assumed to be of type Real. >>

Why? n = Length[{1, 2, 3, 4}] already make sure that n is integer, Why n=n/2 cannot be compiled?

$\endgroup$
2
  • 3
    $\begingroup$ When it divides by 2, there's no guarantee that the result will be an integer (even though it is in this case). You can use n = Round[n/2] to keep the type as an integer $\endgroup$
    – Jason B.
    Feb 5, 2016 at 14:51
  • $\begingroup$ @JasonB Thanks, Round works $\endgroup$
    – matheorem
    Feb 5, 2016 at 23:10

1 Answer 1

4
$\begingroup$

If you want integer division, use Quotient:

Compile[{}, Module[{n}, n = Length[{1, 2, 3, 4}]; n = Quotient[n, 2]]]

If you want n to be real instead of an integer, you can coerce it's type in a number of ways. For example, these will both result in n being real:

Compile[{}, Module[{n = 0.0}, n = Length[{1, 2, 3, 4}]; n = n/2]]

Compile[{}, Module[{n}, n = N[Length[{1, 2, 3, 4}]]; n = n/2]]

If you want to see how Compile has interpreted types, you can view the compiled routine like so:

<< CompiledFunctionTools`;

cf = Compile[{}, Module[{n = 0.0}, n = Length[{1, 2, 3, 4}]; n = n/2]];

CompiledFunctionToString[cf]

(*
"
        No argument
        2 Integer registers
        4 Real registers
        1 Tensor register
        Underflow checking off
        Overflow checking off
        Integer overflow checking on
        RuntimeAttributes -> {}

        T(I1)0 = {1, 2, 3, 4}
        I1 = 2
        R0 = 0.
        Result = R1

1   R1 = R0
2   I0 = Length[ T(I1)0]
3   R2 = I0
4   R1 = R2
5   R2 = I1
6   R3 = Reciprocal[ R2]
7   R2 = R1 * R3
8   R1 = R2
9   Return
"
*)

In this output, the integers start with "I" and the reals start with "R".

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.