Unnecessary code is generated when using 8 bit and 16 bit variables

Using gcc 7.2.1 for ARM thumb with -O3, I see unnecessary instructions being emitted. I tried it on gcc 8.2 and see the same output.

Code is shown below and can be found here: https://godbolt.org/z/AgFwNE

void shortMult()
{
    volatile short int a;
    volatile short int b;

    a *= b;
}

void charMult()
{
    volatile char x;
    volatile char y;

    x *= y;
}

void shortIntMult(void)
{
    volatile short int a;
    volatile int b;

    b *= a;
}

This generates:
shortMult():
        sub sp, sp, #8
        ldrh r1, [sp, #6]
        ldrh r2, [sp, #4]
        mul r3, r2, r1
        lsl r3, r3, #16
        asr r3, r3, #16
        strh r3, [sp, #4] @ movhi
        add sp, sp, #8
        bx lr
charMult():
        sub sp, sp, #8
        ldrb r2, [sp, #7] @ zero_extendqisi2
        ldrb r1, [sp, #6] @ zero_extendqisi2
        mul r3, r2, r1
        and r3, r3, #255
        strb r3, [sp, #6]
        add sp, sp, #8
        bx lr
shortIntMult():
        sub sp, sp, #8
        ldrh r3, [sp, #2]
        ldr r2, [sp, #4]
        lsl r1, r3, #16
        asr r1, r1, #16
        mul r3, r2, r1
        str r3, [sp, #4]
        add sp, sp, #8
        bx lr

Looking at shortMult():
The strh r3, [sp, #4] writes out 16 bits. There is no need to do the lsl and asr - they don't modify the bits we write out.

Looking at charMult():
Similar issue as above - strb r3, [sp, #6] writes out 8 bits. There is no need to do the and r3, r3, #255 - it doesn't modify the bits we write out.

Looking at shortIntMult():
a is loaded into r3 using ldrh and then sign extended using lsl r1, r3, #16 and asr r1, r1, #16
a can be loaded with sign extension using ldrsh

Thank you.