C#•7mo ago

Float (signed) number comparisons compiled to unsigned version of instruction

Why does

    public bool lteq(float a, float b) {
        return a <= b;
    }

    public bool lteq(float a, float b) {
        return a <= b;
    }

compile to

        IL_0000: ldarg.1
        IL_0001: ldarg.2
        IL_0002: cgt.un
        IL_0004: ldc.i4.0
        IL_0005: ceq
        IL_0007: ret

        IL_0000: ldarg.1
        IL_0001: ldarg.2
        IL_0002: cgt.un
        IL_0004: ldc.i4.0
        IL_0005: ceq
        IL_0007: ret

whereas

    public bool not(float a, float b) {
        return !(a > b);
    }

    public bool not(float a, float b) {
        return !(a > b);
    }

compiles to the same thing but with cgt instead of cgt.un? Are these two expressions not equivalent? I'm also confused why it's using .un for two signed numbers that are ordered sharplab: https://sharplab.io/#v2:EYLgxg9gTgpgtADwGwBYA0AXEBLANgHwAEAmARgFgAoQgZgAIS6BhOgbyrs4fuAglzoA7CBgAUAM1wQAhhjrS0dSTLnAAlGw5dthAOx0AhKOl0AfHXUBuLZwC+N7hb4DcGGAEcJU2fMXKf6pqU2jr6JgA8ALwW1sFc9pS2QA

C#/VB/F# compiler playground.

3 Replies

thegu5OP•7mo ago

i'm new to messing with IL so I probably missed something

Petris•7mo ago

iirc the un actually stands for unordered, not unsigned and signifies how NaN is handled for floats

thegu5OP•7mo ago

thank you!