On 10/11/2013 21:55, Jonas Maebe wrote:
On 10 Nov 2013, at 22:42, Martin wrote:
Why is the
movl %eax,-12(%ebp)
generated at the start of i:=i+1
Why does a statement save an outdated value to memory?
Because the load back from memory that came right after it got removed by a
peephole optimisation. Peephole optimisations are by definition local and have
no clue about what happens next, and hence the store remains.
Ok, i understand, but that would mean hat the save originally was
intended for the "i := 1" statement.
So somewhere in that process, the debug info, and the statements get
out of sync.
I only wonder, if despite the fact that O1 does not guarantee, if in
this case it would be worth to consider it a bug (and intend to fix it)?
Because a lot of users rely on O1 being good for the debugger, and not
everyone might spot such a value as wrong by the debugger. So people may
spent a long time to figure out why there code does something, that it
actually does not do.
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