For example, this program
process = no.noise <: *(mem);
outputs zeroes due to integer overflow.
process = no.noise <: *;
the same. This one works correctly
process = no.noise : ^(2);
-------------------------------------------------------------------------------
I think this is because something like
(float1 * int1) * (float2 * int2)
can be normalized to, say,
float1 * (float2 * (int1 * (int2)))
mterm::operator*=(Tree t) simply does
if (isSigBinOp(t, &op, x, y) && (op == kMul)) {
*this *= x; // float2
*this *= y; // int2
}
and in general after that nothing can tell that int2 should be promoted to real.
generateBinOp() does promote2real() if (t1 == kReal) || (t2 == kReal), but in
the case above we will have
float(int1 * int2)
and this can overflow.
Consider this trivial program:
sig1 = int : *(2.0);
sig2 = int : *(3.0);
process = sig1 * sig2;
it compiles into
output0[i] = FAUSTFLOAT((6.0f * float((int(float(input1[i])) *
int(float(input0[i]))))));
and this is obviously wrong.
sig1 = int : *(2.0);
process = sig1 <: *;
results in
output0[i] = FAUSTFLOAT((4.0f *
float(mydsp_faustpower2_i(int(float(input0[i]))))));
which is equally wrong, the same problem.
Just in case... this is not because of mterm.fCoef logic. Say,
process = int * ffunction(float f(), <math.h>, "") <: *;
results in
output0[i] = FAUSTFLOAT((mydsp_faustpower2_f(float(f())) *
float(mydsp_faustpower2_i(int(float(input0[i]))))));
-------------------------------------------------------------------------------
At the same,
sig1 = int : *(2.0) : float;
sig2 = int : *(3.0) : float;
process = sig1 * sig2;
fixes the problem but
output0[i] = FAUSTFLOAT(((3.0f * float(int(float(input1[i])))) * (2.0f
* float(int(float(input0[i]))))));
is suboptimal. In fact sigFloatCast() can "confuse" faust very easily,
sig = *(2.0) : float;
process = sig * 3.0;
leads to
output0[i] = FAUSTFLOAT((3.0f * (2.0f * float(input0[i]))));
perhaps operator*() could me more clever if isSigXxxCast(t) ...
Oleg.
------------------------------------------------------------------------------
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