Why the use of the macro #define FREE_ARG char* for free memory alllocated by malloc (Numerical recipes)

Question

I am studying some routines from Numerical Recipes and I did not understand the use of the macro #define FREE_ARG char* in the implementations to deallocating memory. Here goes one piece of the code:

#include <stdio.h>
#include <stddef.h>
#include <stdlib.h>
#define NR_END 1
#define FREE_ARG char*

float *vector(long nl, long nh)
/* allocate a float vector with subscript range v[nl..nh] */
{
    float *v;

    v=(float *)malloc((size_t) ((nh-nl+1+NR_END)*sizeof(float)));
    if (!v) nrerror("allocation failure in vector()");
    return v-nl+NR_END;
}


void free_vector(float *v, long nl, long nh)
/* free a float vector allocated with vector() */
{
    free((FREE_ARG) (v+nl-NR_END));
\\           ^
\\           '----- why?
}

My question is: Why the memory is being released in this manner free((FREE_ARG) (v+nl-NR_END));?

Answer 1

That is probably a very bad code.^note

free() takes the argument type void *. Any pointer type can be implicitly converted to a void *, no cast needed.

---

Note:

In this line

 v=(float *)malloc((size_t) ((nh-nl+1+NR_END)*sizeof(float)));

the cast is also not required and should be avoided. See this discussion on why not to cast the return value of malloc() and family in C.. This is also one of the reasons behind the usage of bad earlier.