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    <p>Ok, I have checked the <strong>c99</strong> standard and can say that this is <strong>undefined behaviour</strong>.</p> <hr> <p><strong>TL;DR:</strong> It is ok to cast/convert one function pointer to another type, but if you call that function, the function types must be compatible, otherwise, you get <strong>undefined behaviour</strong>. You don't get a warning because in each step you are converting between compatible function types, but in the end, <code>helpMe</code> isn't compatible with <code>pFunc</code>.</p> <hr> <p>First of all: </p> <pre><code>char *helpMe(){ return "this is from helpMe"; } </code></pre> <p>Is valid and in C99 is the same as:</p> <pre><code>char *helpMe(void){ return "this is from helpMe"; } </code></pre> <p>From <strong>§6.7.5.3</strong> (emphasis mine):</p> <blockquote> <p>An identifier list declares only the identifiers of the parameters of the function. <strong>An empty list in a function declarator that is part of a definition of that function specifies that the function has no parameters</strong>. The empty list in a function declarator that is not part of a definition of that function specifies that no information about the number or types of the parameters is supplied.</p> </blockquote> <p>I don't know how this behaves in C89, I think that maybe, in C89 that is valid.</p> <p>On the other hand, we have: </p> <pre><code>char * (*fNewFunc)() = helpMe; </code></pre> <p>This cast is valid, because you can cast a function pointer to a function of another type and go back without problem. The compiler won't throw a warning because even their types are compatible! (void against unespecified). </p> <p>Section <strong>§6.3.2.3</strong>, paragraph 8 reads (again, emphasis mine):</p> <blockquote> <p>A pointer to a function of one type may be converted to a pointer to a function of another type and back again; the result shall compare equal to the original pointer. <strong>If a converted pointer is used to call a function whose type is not compatible with the pointed-to type, the behavior is undefined.</strong></p> </blockquote> <p>So, when you pass the pointer to <code>hearThis</code> it is casted again, from <code>char *(*)();</code> to <code>char *(*)(char *);</code> which, again, they are compatible for the compiler (thus, not throwing a warning). </p> <p><strong>BUT</strong> the standard says if the function type you are calling is incompatible with the function type pointed to, then, the <strong>behaviour is undefined</strong>.</p> <p>Te question now is, are they compatible?</p> <p>The answer is <strong>NO</strong>. if you change the type of <code>fNewFunc()</code> to <code>char *(*fNewFunc)(void);</code> you will get a warning: </p> <pre><code>p.c: In function ‘main’: p.c:12:5: warning: passing argument 1 of ‘hearThis’ from incompatible pointer type [enabled by default] p.c:6:6: note: expected ‘char * (*)(char *)’ but argument is of type ‘char * (*)(void)’ </code></pre> <p>In the standard there also says when two functions types are compatible. It is in <strong>§6.7.5.3.15</strong>. It's a bit more complicated than this :P</p> <h3>Why does it work?</h3> <p>Probably it works because the computer doesn't care. It just pushes or passes the parameters as registers, and <code>helpMe</code> just ignores them. But you should know that it is <strong>undefined behaviour</strong>.</p>
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    1. CO+1 for the C99 §6.7.5.3 (C11dr §6.7.6.3 14). LSNED.
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