printf("%f", 20); results in the output 0.000000, not 20.000000. I'm guessing this has to do with how an int is represented in memory and how a double is represented in memory. Surprisingly for me, no matter how I alter 20, for example by making the number larger, the output is still 0.000000. Could someone please explain the underlying mechanics of this?
Most probably you are compiling your code on a platform/ABI where even for varargs functions data is passed into registers, and in particular different registers for integer/floating point values. x86_64 on Linux/OS X behaves like that.
The caller has an integer to pass, so it puts it into rsi; on the other side, printf expects a floating point value, so it tries to read it from xmm0. No matter how you change your integer argument to any other value printf will be unaffected - if will just print whatever happens to stay into xmm0 at the moment of the call.
You can actually check if this is the case by changing your call to:
printf("%f", 20, 123.45);
if it's working as I described, you should see 123.45 printed (the caller here puts 123.45 into xmm0, as it is the first floating point parameter passed to the function; printf behaves as before, but this time finds another value into xmm0).