| Issue |
141863
|
| Summary |
Bad codegen for reversing bits in bytes without swapping bytes
|
| Labels |
new issue
|
| Assignees |
|
| Reporter |
tom-rein
|
The following code on RISC-V target with Zbkb extension ([godbolt](https://godbolt.org/z/cTqEYsbaz)):
```C++
#include <cstdint>
uint8_t brev8(uint8_t x) {
return __builtin_bitreverse8(x);
}
```
Results in:
```
brev8(unsigned char):
rev8 a0, a0
brev8 a0, a0
srli a0, a0, 56
ret
```
The expected result would be just `brev8 a0, a0` since it doesn't swap bytes.
On armv8 target (and similarly on armv7) bit reversing bytes in `uint16_t` uses `rev16` which results in an extra shift:
```C++
uint16_t brev8(uint16_t x) {
return __builtin_bswap16(__builtin_bitreverse16(x));
}
```
```
brev8(unsigned short):
rbit w8, w0
lsr w8, w8, #16
rev16 w0, w8
ret
```
This could instead be:
```
brev8(unsigned short):
rbit w0, w0
rev w0, w0
ret
```
(Also the register allocation somehow uses w8 instead of reusing w0, which I find weird, since that is done on other targets, including armv7)
When doing `bswap16` first and then `bitreverse16`, it doesn't even use `rev16`, but adds two shifts:
```C++
uint16_t brev8_2(uint16_t x) {
return __builtin_bitreverse16(__builtin_bswap16(x));
}
```
```
brev8_2(unsigned short):
rev w8, w0
lsr w8, w8, #16
rbit w8, w8
lsr w0, w8, #16
ret
```
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