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    <p>The relevant part of the C standard is section 6.3.2.1 "Lvalues, arrays, and function designators". Here's what it says about functions:</p> <blockquote> <p>A <em>function designator</em> is an expression that has function type. Except when it is the operand of the <code>sizeof</code> operator<sup>65</sup> or the unary <code>&amp;</code> operator, a function designator with type ‘‘function returning <em>type</em>’’ is converted to an expression that has type ‘‘pointer to function returning <em>type</em>’’.</p> <p>[footnote 65] Because this conversion does not occur, the operand of the <code>sizeof</code> operator remains a function designator and violates the constraint in 6.5.3.4 [ed: the constraint in 6.5.3.4 says that you may not apply <code>sizeof</code> to a function designator - it's a semantic error].</p> </blockquote> <p>An identifier that names a function is the simplest sort of "expression that has function type". So what this means is, if <code>foo</code> has been declared as a function, the identifier <code>foo</code> evaluates as a pointer to that function, <em>except</em> when it's the operand of <code>&amp;</code> (in which case the larger expression <code>&amp;foo</code> evaluates as a pointer to that function) or the operand of <code>sizeof</code> (in which case the larger expression, <code>sizeof(foo)</code>, provokes a compile error).</p> <p>tl,dr: When <code>foo</code> is a function, <code>foo</code> and <code>&amp;foo</code> are equivalent by definition. This is a special rule for functions. It's not entirely unlike the special rule for arrays, which also "decay" to pointers in many contexts (that rule is one paragraph up from the one I quoted).</p> <p>Aside: Yes, this means that the function-call operator <em>always</em> operates on a pointer-to-function. When <code>pfunc</code> is a pointer-to-function variable, <code>(*pfunc)()</code> is processed as if it read <code>(&amp;(*pfunc))()</code> ... or just <code>pfunc()</code>.</p>
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    1. COSo you're saying `BrandNewFunc` is the same as `&BrandNewFunc`? And how are arrays like that? `BrandNewIdentifier` is the address of the first byte; `&BrandNewIdentifier` is the address of the address of the first byte.
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    2. COAll I mean is, arrays do "decay" to pointers, and so do functions; you can't pass either of them by value. It's true that the type of `BrandNewIdentifier` is not the same as the type of `&BrandNewIdentifier` if `BrandNewIdentifier` is an array; that's not so for functions, the types of `BrandNewFunc` and `&BrandNewFunc` are the same if used in a context that expects a pointer-to-function.
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    3. COCan you edit your question to explain what you mean by "context"? As far as I know, expressions in C don't change meaning based on context, that's for crazy languages like Perl.
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