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

Why doesn't the compiler detect the missing semicolon?

There are three things to remember.

1. Line endings in C are just ordinary whitespace.
2. * in C can be both a unary and a binary operator. As a unary operator it means "dereference", as a binary operator it means "multiply".
3. The difference between unary and binary operators is determined from the context in which they are seen.

The result of these two facts is when we parse.

 temp = *a    /* Oops, missing a semicolon here... */
 *a = *b;

The first and last * are interpreted as unary but the second * is interpreted as binary. From a syntax perspective, this looks OK.

It is only after parsing when the compiler tries to interpret the operators in the context of their operand types that an error is seen.