三元和强制转换为qsort参数

Question

I'm trying to practice pointer to function but I don't understand the technique used.

Context:

char *lineptr[5000];
int strcmp (char *, char *);
int numcmp (char *, char *);`

The function prototype is:

void qsort ( void *lineptr[], int left, int right, int (*comp)(void *, void *));

I don't understand how the following call works:

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

Specifically this part:

(int (*)(void*, void*))(numeric ? numcmp : strcmp)

The function prototype has only 4 argument whereas in the function call, there are 5 parameters and the 5th one is an equation. And this compiles without error.

At one point I assumed that it would assign either numcomp or strcmp to the function pointer (*comp) based the value of numeric . But what I don't understand is how will this assignment happens.

Answer 1

Only 4 arguments are passed, the 4th argument

(int (*)(void*, void*))(numeric ? numcmp : strcmp)

will either pass a pointer to the numcmp (if numeric is truthy) or a pointer to the strcmp function (if numeric is falsy).

(int (*)(void*, void*)) is just a cast to the function signature to help the compiler, because the compiler is expecting (int (*)(void*, void*)) as function pointer (see qsort function prototype) and you're passing in a pointer to numcmp/strcmp which is (int (*)(char*, char*)) .

Answer 2

There are only 4 arguments in:

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

Broken down to one-per-line, it looks like:

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

The last one seems to be the one that's confusing you. The ?: expression is selecting one of two functions with the same declared type:

numeric?numcmp:strcmp

This is then parenthesized (for precedence) and cast to a function pointer type that takes two void * arguments:

(int(*)(void*,void*))funcptr

The type in the cast operator is a pointer to a function that takes two void * arguments and returns an int result.

Answer 3

The last argument to qsort is a bit complex.

qsort((void**)lineptr,0,nlines-1,(int(*)(void*,void*))(numeric?numcmp:strcmp));
//                               |<--        one argument                 -->

At the cost of more verbose code, you can simplify that line to

int(*compare_function)(void*,void*) = (int(*)(void*,void*))numcmp;
if ( !numeric )
{
   compare_function = (int(*)(void*,void*))strcmp;
}

qsort((void**)lineptr,0,nlines-1, compare_function);