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    <p>The type of <code>addition</code> is <code>int (&amp;)(int,int)</code> which can decay into a pointer of type <code>int (*)(int,int)</code> which is same as that of <code>operation</code> function's third parameter. So you can pass <code>addition</code> as third argument to the function <code>operation</code>. </p> <p>The type of <code>subtraction</code> is also the same as that of <code>addition</code>. In your code, the address of <code>subtraction</code> is first stored as local variable of the compatible type, and then that variable is passed as argument to <code>operation</code> function. </p> <p>In case of <code>addition</code>, it's address is not stored as local variable, instead its passed as such to <code>operation</code>. Its initializing the function's third parameter directly with the function's address, without using any local variable. </p> <p>A conversion from <code>int (&amp;)(int,int)</code> to <code>int (*)(int,int)</code> occurs in both cases. Its just that with <code>substration</code>, the conversion occurs when initializing the local variable, and with <code>addition</code>, the conversion occurs when initializing the function parameter.</p> <p>An analogy would be this:</p> <pre><code>void f(double a, double b) {} int main() { double x = 100;//first store 100 in a local variable f(x, 100); //pass the local variable as first arg, //and pass 100 as second arg without using any local variable. } </code></pre> <p>Note the type of <code>100</code> is <code>int</code>, so it first converts to <code>double</code> type, which is then stored as local variable <code>x</code>, which in turn is passed to the function <code>f</code> as first argument. And the second argument <code>100</code> is passed directly to the function, so even now it first converts to <code>double</code> and then it initializes <code>b</code> (the second parameter of the function). </p> <p>Again, a conversion from <code>int</code> to <code>double</code> occurs in both cases. Its just that first argument conversion occurs when initializing the local variable <code>x</code>, and second argument conversion occurs when initializing the <em>second</em> parameter of the function. </p>
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    1. COIn fact, `int (&)(int, int)` would not decay into function pointer - `int(int,int)`, on the other hand, will. But other than that, very precise answer, so +1. :)
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    2. CO@Vitus: `int(int,int)` is same as `int(*)(int,int)`, as far as I know, there is no difference. BTW, `int(&)(int,int)` can decay into function pointer : http://www.ideone.com/YcLVj
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    3. COI don't believe it _decays_ - more likely there are some other conversions going on. There are three types of decay - **lvalue to rvalue**, **array of T to pointer to T** and **function F to pointer to function F** (§4.1 through §4.3, and also §20.9.7.6, `decay` struct - in latest draft, at least). Hmmm, but on the other hand, lvalue reference behaves as lvalue...
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