Displaying 4 results from an estimated 4 matches for "fin0".
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2018 Apr 04
2
llvm.localsescape/recover
...ign 4
>> store i8* %0, i8** %1
>> %2 = alloca i32
>> store i32 0, i32* %2
>> call void (...) @llvm.localescape(i8** %1, i32* %2)
>>
>> call my finaly like:
>>
>> %loctmp = call i8* @llvm.localaddress()
>> call void @"TEST$Fin0"(i8 0, i8* %loctmp)
>>
>> and recover it with:
>>
>> %2 = call i8* @llvm.localrecover(i8* bitcast (void (i8*)* @TEST to i8*), i8* %1, i32 0)
>> %3 = bitcast i8* %2 to i8**
>> %4 = load i8*, i8** %3
>>
>> I get the right value passed...
2018 Apr 04
0
llvm.localsescape/recover
On Wed, Apr 4, 2018, at 08:58, Carlo Kok via llvm-dev wrote:
> That's quite likely. I can reproduce this issue with llvm 6.0 though.
> I'll try with svn HEAD.
>
> On Tue, Apr 3, 2018, at 19:22, Reid Kleckner wrote:
> > I would guess that %threedoubles has a large alignment and that is making things go wrong. I thought we fixed this bug, though. I can't find it in
2018 Apr 03
2
llvm.localsescape/recover
...using locals recover to have a seh finally in a separate function:
%1 = alloca i8*, align 4
store i8* %0, i8** %1
%2 = alloca i32
store i32 0, i32* %2
call void (...) @llvm.localescape(i8** %1, i32* %2)
call my finaly like:
%loctmp = call i8* @llvm.localaddress()
call void @"TEST$Fin0"(i8 0, i8* %loctmp)
and recover it with:
%2 = call i8* @llvm.localrecover(i8* bitcast (void (i8*)* @TEST to i8*), i8* %1, i32 0)
%3 = bitcast i8* %2 to i8**
%4 = load i8*, i8** %3
I get the right value passed to my original function (ie %1)
If I however *just* add a struct with 3 double...
2018 Apr 03
0
llvm.localsescape/recover
...arate function:
>
> %1 = alloca i8*, align 4
> store i8* %0, i8** %1
> %2 = alloca i32
> store i32 0, i32* %2
> call void (...) @llvm.localescape(i8** %1, i32* %2)
>
> call my finaly like:
>
> %loctmp = call i8* @llvm.localaddress()
> call void @"TEST$Fin0"(i8 0, i8* %loctmp)
>
> and recover it with:
>
> %2 = call i8* @llvm.localrecover(i8* bitcast (void (i8*)* @TEST to i8*),
> i8* %1, i32 0)
> %3 = bitcast i8* %2 to i8**
> %4 = load i8*, i8** %3
>
> I get the right value passed to my original function (ie %1)
>...