Thread: Casting a type of function to another type of function pointer

The behaviour is probably undefined.

I also think its undefined. But if I look at K&R example they do something which "looks" like they doing the same undefined action - but I stand corrected. They define two functions which will be used as a parameter to another function:

Code:

int numcmp(char *s1, char *s2){
//do stuff
}

Code:

int strcmp(char *s, char *t){
//do other stuff...
}

The below function is one in which either one of the above 2 functions are pointed to through the use of a function pointer as a fourth argument. Note however that the pointer is void and not of the same type of the functions above - this is so it can accommodate any type of function pointing.

Code:

void qsortcustom(/*arg 1*/, /*arg 2*/, /*arg 3*/, int (*comp)(void *, void *)){
//Do yet more stuff
}

Everything up until now makes sense. However:

Finally, when we call the above function, the code is written like this:

Code:

qsortcustom((void**) lineptr, 0, nlines-1, (int (*)(void*,void*))(numeric ? numcmp : strcmp));

It makes no sense - numcmp and strcmp are functions and now its been cast to a pointer? So my question is, is the above undefined behaviour? Its in the K&R textbook

Okay, I'll stop being lazy and dig up the actual quote from the standard:

Originally Posted by C11 Clause 6.3.2.3 Paragraph 8

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. If a converted pointer is used to call a function whose type is not compatible with the referenced type, the behavior is undefined.

Originally Posted by Vespasian_2

So is the K&R quote an example of undefined behaviour?

Probably not. The undefined behaviour happens "if a converted pointer is used to call a function whose type is not compatible with the referenced type", so presumably this is with reference to say, casting a pointer to a function with 2 arguments to be a pointer to a function with 3 arguments, or where the return/parameter types simply are not compatible. In the case of int (*)(char*, char*) versus int (*)(void*, void*), I would say that they are compatible since a char* can be implicitly converted to a void* and vice versa.

Originally Posted by algorism

The function names are already pointers (to the start of the function code). They're just being cast to a different kind of pointer.

Effectively, though the standard does distinguish between functions and function pointers:

Originally Posted by C11 Clause 6.3.2.1 Paragraph 4

A function designator is an expression that has function type. Except when it is the operand of the sizeof operator, or the unary & operator, a function designator with type "function returning type" is converted to an expression that has type "pointer to function returning type".