[cfe-users] Why is this block of instructions generated?
Dipanjan Das via cfe-users
cfe-users at lists.llvm.org
Tue Jun 20 09:06:52 PDT 2017
The DataFlowSanitizer pass on LLVM 3.8.0, 64 bit (Ubuntu 16.04.2) generates
the following IR from source:
======================================
test.c
======================================
#include <sanitizer/dfsan_interface.h>
int main(void) {
int i = 1;
dfsan_label i_label = dfsan_create_label("i", 0);
dfsan_set_label(i_label, &i, sizeof(i));
return 0;
}
======================================
clang -c -emit-llvm -fsanitize=dataflow test.c -o test.bc
llvm-dis test.bc
======================================
test.ll
======================================
; Function Attrs: nounwind uwtable
define i32 @main() #0 {
entry:
%0 = alloca i16
%retval = alloca i32, align 4
%i = alloca i32, align 4
%1 = alloca i16
%i_label = alloca i16, align 2
store i16 0, i16* %0
store i32 0, i32* %retval, align 4
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
%2 = ptrtoint i32* %i to i64
%3 = and i64 %2, -123145302310913
%4 = mul i64 %3, 2
%5 = inttoptr i64 %4 to i16*
%6 = bitcast i16* %5 to i64*
store i64 0, i64* %6, align 2
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
store i32 1, i32* %i, align 4
%call = call zeroext i16 @dfsan_create_label(i8* getelementptr inbounds
([2 x i8], [2 x i8]* @.str, i32 0, i32 0), i8* null)
store i16 0, i16* %1
store i16 %call, i16* %i_label, align 2
%7 = load i16, i16* %1
%8 = load i16, i16* %i_label, align 2
%9 = bitcast i32* %i to i8*
call void @dfsan_set_label(i16 zeroext %8, i8* %9, i64 4)
ret i32 0
}
======================================
I don't understand why the block of instruction I separated out is being
generated. Looking at the Transform/Instrumentation/DataFlowsanitizer.cpp,
I can't find the code that inserts the instrumentation above. Can anyone
explain this behavior?
--
Thanks & Regards,
Dipanjan
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