Hi. Andrew Haley wrote:
The clobber list is for physical register names such as "ebx", not for operands.
Yes, that's why my question was not "why it still doesn't work", but rather "what does the "0" mean, when put in the clobber list?" gcc accepts it there, and I wasn't able to find anywhere in the docs, what does it mean.
This is wrong: you're clobbering one of the inputs without telling gcc.
Yes, but the problem is, the way to tell this to gcc looks very strange to me. Here's only what I've been able to find in the docs: --- There is no way for you to specify that an input operand is modified without also specifying it as an output operand. Note that if all the output operands you specify are for this purpose (and hence unused), you will then also need to specify |volatile| for the |asm| construct, as described below, to prevent GCC from deleting the |asm| statement as unused. --- A quick googling shows many people are confused with that, too. I am just curious, why there wasn't a better solution to such a problem? gcc wants me to add the dummy output var for that, and googling reveals some discussion that a better syntax was simply not found. But IMHO, allowing "&" modifier in the input list is the most natural way of expressing that the input is being clobbered after the value is taken.
Assuming you actually want the output, it should be:
No, what I really want, is an input. I.e. I want to put the input in %%ebx. But then, after I already used it, I want to re-use %%ebx in an asm code. And here's the gotcha: re-using %%ebx also alters the input operand, but not always - that depends on an optimization level. I find that counter-intuitive, hence the questions.
