I have this simple program:

    #include &lt;stdio.h&gt;

    struct S
    {
        int i;
    };

    void swap(struct S *a, struct S *b)
    {
        struct S temp;
        temp = *a    /* Oops, missing a semicolon here... */
        *a = *b;
        *b = temp;
    }

    int main(void)
    {
        struct S a = { 1 };
        struct S b = { 2 };

        swap(&amp;a, &amp;b);
    }

As seen [on e.g. ideone.com](http://ideone.com/jaNv9B) this gives an error:

&gt;     prog.c: In function &#39;swap&#39;:
&gt;     prog.c:12:5: error: invalid operands to binary * (have &#39;struct S&#39; and &#39;struct S *&#39;)
&gt;          *a = *b;
&gt;          ^

Why doesn&#39;t the compiler detect the missing semicolon?

---

Note: This question and its answer is motivated by [this question](https://stackoverflow.com/q/40134786). While there are [other questions](https://stackoverflow.com/q/7071168) similar to this, I didn&#39;t find anything mentioning the free-form capacity of the C language which is what is causing this and related errors.

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Total of 1 Freeform Solutions

Why doesn't the compiler report a missing semicolon?