How to shoot yourself in the foot with struct pointers

#include <stdio.h>

struct thing {
    unsigned int a;
    int b;
};

void do_2(struct thing);
void do_3(struct thing*);
void do_4(struct thing*);
void do_5(struct thing*);

void do_2(struct thing t){
    printf( "do_2(): t.a: %u. Good.\n"
            "        setting t.a to 17\n\n", t.a);
    t.a = 17;
}
void do_3(struct thing *t){
    printf( "do_3(): t->a: %u. Good.\n"
            "        setting t->a to 17\n\n", t->a);
    t->a = 17;
}
void do_4(struct thing *t){
    printf( "do_4(): t->a: %u. Good.\n"
            "        outsourcing complex operation to do_5()\n\n", t->a);
    do_5(&t); // this is wrong. the t in this function is _already_ a pointer. The result of this error is quite interesting.
    printf("do_4(): t->a: %u. What the hell!? Hey do_5, that was wrong! W.R.O.N.G. Oh wait...sorry. Let's do it again.\n\n", t->a);
    do_5(t); // This is the proper way to do it...
    printf("do_4(): Alright! Here you go, main!\n\n");
}
void do_5(struct thing *t){
    printf("do_5(): t->a: %u. ", t->a);
    printf("Wow, do_4! That is a large number. Okay... I'm going to increment it.\n\n");
    t->a ++;
}
int main(){
    struct thing t = {5,4};
    printf("main(): t.a: %u. Good.\n\n", t.a);
    do_2(t);
    printf("main(): t.a: %u. No good. I wanted it to change *t.a*, not a copy of t.a.\n\n", t.a);
    do_3(&t);
    printf("main(): t.a: %u. Good.\n\n", t.a);
    do_4(&t);
    printf("main(): t.a: %u. What th--? do_4, you're fired!\n", t.a); // ...however, the damage has already been done.
    printf("-- Ya shot yourself in the foot. --\n");
}