The value bits differ, memset will not help here. You can watch the values in the gdb: (gdb) n 43 printBits(sizeof(r1), &r1); (gdb) info locals c = -1 r1 = 0.36787944117144232158305751367866065 r2 = 0.36787944117144232161016256799079827 пт, 27 нояб. 2020 г. в 00:54, Stefan Ring <stefanrin@xxxxxxxxx>: > On Thu, Nov 26, 2020 at 9:21 PM Alex Markin via Gcc-help > <gcc-help@xxxxxxxxxxx> wrote: > > > > Hello. > > > > I don't know if it is a gcc or binutils problem. I have different results > > for `expl' function depending on the argument was variable or literal: > > > > long double c, r1, r2; > > r1 = expl(-1); > > > > c = -1; > > r2 = expl(c); > > > > In this example r1 and r2 differ in the lower bits. The entire example is > > here: https://godbolt.org/z/xqn4bd. The llvm does not have such a > problem. > > The reason is that the Intel-specific 80 bit long double format is > used. So the first 80 bits of r1 and r2 contain the value, and the > rest remains uninitialized. You could memset both to 0 first, then the > comparison would check out. I'm sure there is a good reason for sizeof > producing 16, not 10. >
