How the parameter of this 2 type of pointer is different?

Question

In this post ( Why the result of this code is the same when the arg is different? ), the parameter of chg is yay *lol and the inside of it is lol (notice there's no asterisk in front of it). But why in this code, it shows up an error?

void chg (int *lol) {lol=9;}

int main ()
{
    int a=5;
    int *boi=&a;
    printf ("%d\n", *boi);
    chg (boi);
    printf ("%d\n", *boi);

    return 0;
}

[Error] invalid conversion from 'int' to 'int*' [-fpermissive]

So, different data types in parameter means it works differently?

Answer 1

In function you should use like this

void chg (int *lol) {
    *lol=9;
}

Because it is pointer, it keeps an address. With * sign, you say that. Go to this address and assign this value.

Also you can use your function like this

int main ()
{
    int a=5;
    int *boi=&a;
    printf ("%d\n", *boi);
    chg (&a); //send address of a.
    printf ("%d\n", *boi);

    return 0;
}

Like this. It means same thing