Operator precedence is confusing in this case

Lets say I have a struct

struct s
{
    int a;
}

and a main function like this

int main()
{
    struct s s1 = {.a = 10};
    
    struct s *sptr = &s1;
    
    sptr->a--;
    printf("%d\n", sptr->a);
    
    s1.a = 10;
    
    --sptr->a;
    printf("%d\n", sptr->a);
    
    return 0;
}

Output :

9
9

As per operator precedence, -> and -- have same precedence and left->right associativity.

In case of sptr->a--, I can understand sptr->a was done first and then -- was applied.

But in case of --sptr->a, -- should be applied to sptr first. This may lead to undefined behaviour but it should be the case. Why does this still work same as sptr->a-- ?

As per operator precedence, -> and -- have same precedence and left->right associativity.

-> and postfix -- have the same precedence and left-to-right associativity, so sptr->a-- is (sptr->a)--.

-> and prefix -- don't have the same precedence, -- is lower. Thus --sptr->a is --(sptr->a).