# # void foo(){ # ifloat f = 1.0fi; # f *= 2.0fi; // illegal but compiles # } # It looks like all arithmetic operators for imaginary types perform insufficient type checks. test cases: http://dstress.kuehne.cn/nocompile/o/opAssign_21_A.d http://dstress.kuehne.cn/nocompile/o/opAssign_21_B.d http://dstress.kuehne.cn/nocompile/o/opAssign_21_C.d
-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 d-bugmail@puremagic.com schrieb am 2006-12-23: > http://d.puremagic.com/issues/show_bug.cgi?id=730 > test cases: > http://dstress.kuehne.cn/nocompile/o/opAssign_21_A.d > http://dstress.kuehne.cn/nocompile/o/opAssign_21_B.d > http://dstress.kuehne.cn/nocompile/o/opAssign_21_C.d Should have been: http://dstress.kuehne.cn/nocompile/o/opMulAssign_21_A.d http://dstress.kuehne.cn/nocompile/o/opMulAssign_21_B.d http://dstress.kuehne.cn/nocompile/o/opMulAssign_21_C.d -----BEGIN PGP SIGNATURE----- iD8DBQFFjT5cLK5blCcjpWoRAtWEAKCOxes9Y4yOPIUHzXdzthB4topa+ACfU6Ho znTa8F9q1YbtXUaIWPGrmkQ= =xHSE -----END PGP SIGNATURE-----
It's not illegal. The mathematical result is 0, which is what is assigned to f.
(In reply to comment #2) > It's not illegal. The mathematical result is 0, which is what is assigned to f. The imaginary part of the expression is indeed 0. But that involves an implicit cast from double to idouble, which is very hard to justify. The type of (1.0i * 2.0i) is double. It's not even a cdouble. The current behaviour violates a *= b <---> a = a*b, since the latter won't compile. idouble *= idouble is a bug in user code, 100% of the time. (I actually got bitten by this, when I wrote generic dot product code. You can't assume that a = a*a compiles).
Fixed in DMD1.048