3 ms·
Close. Applying that to double c = dbg(dbg(a) + dbg(b)); gives a = 1 @ debug.cxx:16 b = 2 @ debug.cxx:16 dbg(a) + dbg(b) = 3 @ debug.cxx:16
by petschge 6y ago
Close. Applying that to
double c = dbg(dbg(a) + dbg(b));
gives
a = 1 @ debug.cxx:16
b = 2 @ debug.cxx:16
dbg(a) + dbg(b) = 3 @ debug.cxx:16
which has extra dbg()s on the third line.
If I could I would probably want to write
double dbg(c) = dbg(a) + dbg(b);
but that of course doesn't work.