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copied!<h2>Don't ever write code like that.</h2> <hr> <p>For <code>j<1000</code>, <code>j/1000</code> is zero (integer division). So:</p> <pre><code>(&main + (&exit - &main)*(j/1000))(j+1); </code></pre> <p>is equivalent to:</p> <pre><code>(&main + (&exit - &main)*0)(j+1); </code></pre> <p>Which is:</p> <pre><code>(&main)(j+1); </code></pre> <p>Which calls <code>main</code> with <code>j+1</code>.</p> <p>If <code>j == 1000</code>, then the same lines comes out as:</p> <pre><code>(&main + (&exit - &main)*1)(j+1); </code></pre> <p>Which boils down to</p> <pre><code>(&exit)(j+1); </code></pre> <p>Which is <code>exit(j+1)</code> and leaves the program.</p> <hr> <p><code>(&exit)(j+1)</code> and <code>exit(j+1)</code> are essentially the same thing - quoting C99 §6.3.2.1/4:</p> <blockquote> <p>A function designator is an expression that has function type. Except when it is the operand of the sizeof operator <em>or the unary & operator</em>, a function designator with type "<strong>function returning type</strong>" is converted to an expression that has type "<strong>pointer to function returning type</strong>".</p> </blockquote> <p><code>exit</code> is a function designator. Even without the unary <code>&</code> address-of operator, it is treated as a pointer to function. (The <code>&</code> just makes it explicit.)</p> <p>And function calls are described in §6.5.2.2/1 and following:</p> <blockquote> <p>The expression that denotes the called function shall have type <strong>pointer to function</strong> returning void or returning an object type other than an array type.</p> </blockquote> <p>So <code>exit(j+1)</code> works because of the automatic conversion of the function type to a pointer-to-function type, and <code>(&exit)(j+1)</code> works as well with an explicit conversion to a pointer-to-function type.</p> <p>That being said, the above code is not conforming (<code>main</code> takes either two arguments or none at all), and <code>&exit - &main</code> is, I believe, undefined according to §6.5.6/9:</p> <blockquote> <p>When two pointers are subtracted, <strong>both shall point to elements of the same array object</strong>, or one past the last element of the array object; ...</p> </blockquote> <p>The addition <code>(&main + ...)</code> would be valid in itself, and could be used, <em>if</em> the quantity added was zero, since §6.5.6/7 says:</p> <blockquote> <p>For the purposes of these operators, a pointer to an object that is not an element of an array behaves the same as a pointer to the first element of an array of length one with the type of the object as its element type.</p> </blockquote> <p>So adding zero to <code>&main</code> would be ok (but not much use).</p>

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