Summary:
This comes with `Implicit Conversion Sanitizer - integer sign change` (D50250):
```
signed char test(unsigned int x) { return x; }
```
`clang++ -fsanitize=implicit-conversion -S -emit-llvm -o - /tmp/test.cpp -O3`
* Old: {F6904292}
* With this patch: {F6904294}
General pattern:
X & Y
Where `Y` is checking that all the high bits (covered by a mask `4294967168`)
are uniform, i.e. `%arg & 4294967168` can be either `4294967168` or `0`
Pattern can be one of:
%t = add i32 %arg, 128
%r = icmp ult i32 %t, 256
Or
%t0 = shl i32 %arg, 24
%t1 = ashr i32 %t0, 24
%r = icmp eq i32 %t1, %arg
Or
%t0 = trunc i32 %arg to i8
%t1 = sext i8 %t0 to i32
%r = icmp eq i32 %t1, %arg
This pattern is a signed truncation check.
And `X` is checking that some bit in that same mask is zero.
I.e. can be one of:
%r = icmp sgt i32 %arg, -1
Or
%t = and i32 %arg, 2147483648
%r = icmp eq i32 %t, 0
Since we are checking that all the bits in that mask are the same,
and a particular bit is zero, what we are really checking is that all the
masked bits are zero.
So this should be transformed to:
%r = icmp ult i32 %arg, 128
The transform itself ended up being rather horrible, even though i omitted some cases.
Surely there is some infrastructure that can help clean this up that i missed?
https://rise4fun.com/Alive/3Ou
The initial commit (rL339610)
was reverted, since the first assert was being triggered.
The @positive_with_extra_and test now has coverage for that case.
Reviewers: spatel, craig.topper
Reviewed By: spatel
Subscribers: RKSimon, erichkeane, vsk, llvm-commits
Differential Revision: https://reviews.llvm.org/D50465
llvm-svn: 339621
At least one buildbot was able to actually trigger that assert
on the top of the function. Will investigate.
This reverts commit r339610.
llvm-svn: 339612
Summary:
This comes with `Implicit Conversion Sanitizer - integer sign change` (D50250):
```
signed char test(unsigned int x) { return x; }
```
`clang++ -fsanitize=implicit-conversion -S -emit-llvm -o - /tmp/test.cpp -O3`
* Old: {F6904292}
* With this patch: {F6904294}
General pattern:
X & Y
Where `Y` is checking that all the high bits (covered by a mask `4294967168`)
are uniform, i.e. `%arg & 4294967168` can be either `4294967168` or `0`
Pattern can be one of:
%t = add i32 %arg, 128
%r = icmp ult i32 %t, 256
Or
%t0 = shl i32 %arg, 24
%t1 = ashr i32 %t0, 24
%r = icmp eq i32 %t1, %arg
Or
%t0 = trunc i32 %arg to i8
%t1 = sext i8 %t0 to i32
%r = icmp eq i32 %t1, %arg
This pattern is a signed truncation check.
And `X` is checking that some bit in that same mask is zero.
I.e. can be one of:
%r = icmp sgt i32 %arg, -1
Or
%t = and i32 %arg, 2147483648
%r = icmp eq i32 %t, 0
Since we are checking that all the bits in that mask are the same,
and a particular bit is zero, what we are really checking is that all the
masked bits are zero.
So this should be transformed to:
%r = icmp ult i32 %arg, 128
https://rise4fun.com/Alive/3Ou
Reviewers: spatel, craig.topper
Reviewed By: spatel
Subscribers: RKSimon, erichkeane, vsk, llvm-commits
Differential Revision: https://reviews.llvm.org/D50465
llvm-svn: 339610
Roman Lebedev