[Bug target/54816] [avr] shift is better than widening mul

["klaus.doldinger64 at googlemail dot com" <gcc-bugzilla@gcc.gnu.org>]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=54816

Wilhelm M <klaus.doldinger64 at googlemail dot com> changed:

           What    |Removed                     |Added
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                 CC|                            |klaus.doldinger64@googlemai
                   |                            |l.com

--- Comment #1 from Wilhelm M <klaus.doldinger64 at googlemail dot com> ---
The following code has the same problem:

#include <avr/io.h>
#include <stdint.h>

uint16_t b;
uint8_t a;

template<typename A, typename B>
B Mul(const A a, const B b) {
    static constexpr uint8_t shift = (sizeof(B) - sizeof(A)) * 8;
    return static_cast<A>(b >> shift) * a ;
}

int main() {
    return Mul(a, b);
}

with 4.6.4. it produces:

main:
        lds r24,a
        lds r25,b+1
        mul r25,r24
        movw r24,r0
        clr r1
        ret

with actual 12.2 it produces missing optimization:

main:
        lds r24,b+1
        ldi r25,0
        lds r18,a
        movw r20,r24
        mul r18,r20
        movw r24,r0
        mul r18,r21
        add r25,r0
        clr __zero_reg__
ret

Interistingly the follwing code produces optimal code also with 12.2:

template<typename A, typename B>
B MulX(const A a, const B b) {
    static const uint8_t shift = (sizeof(B) - sizeof(A)) * 8;
    return static_cast<A>((b >> shift) + 1) * a ;
}