Why doesn't the compiler report a missing semicolon?

Question

I have this simple program:

#include 

struct S
{
    int i;
};

void swap(struct S *a, struct S *b)
{
    struct S temp;
    temp = *a    /*          


        

Answer 1

Some good answers above, but I will elaborate.

temp = *a *a = *b;

This is actually a case of x = y = z; where both x and y are assigned the value of z.

What you are saying is the contents of address (a times a) become equal to the contents of b, as does temp.

In short, *a *a = is a valid statement. As previously pointed out, the first * dereferences a pointer, while the second multiplies two values.