2 ms·

va_ doesn't change the calling convention. Parameters passed as registers continue to be passed as registers. We could probably argue this some more but I sugg

• mikeash 9y ago
  Good idea! https://gist.github.com/mikeash/ce38d3a77b88734a9e0e9dc3f352fbc7 https://gist.github.com/mikeash/ce38d3a77b88734a9e0e9dc3f352... You'll notice how `normal` takes all of its arguments out of registers `x0` through `x7` and places them on the stack for the call to `printf`. And you'll notice how `vararg` plays a bunch of games with the stack and never touches registers `x1` through `x7`. (It still uses `x0` because the first argument is not variadic.) On the caller side, observe how `call_normal` places its values into `x0` through `x7` sequentially and then invokes the target function, while `call_vararg` places one value into `x0` and places everything else on the stack. So, no, it looks to me like varargs very much change the calling convention.