Question regarding the syntax of someting related to macro in C++ - c++

Question

#define minimum(mn, b) mn = min(b)
#define maximum(mx, c) mx = max(c)
#define INF 10000000
#define toStr(a) #a
#define io(v) v.push_back()
#define FUNCTION(a,b) #
#include <iostream>
#include <vector>
using namespace std;
#if !defined toStr || !defined io || !defined FUNCTION || !defined INF
#error Missing preprocessor definitions
#endif
FUNCTION(minimum, <)
FUNCTION(maximum, >)
int main(){
int n; cin >> n;
vector<int> v(n);
foreach(v, i) {
io(v)[i];
}
int mn = INF;
int mx = -INF;
foreach(v, i) {
minimum(mn, v[i]);
maximum(mx, v[i]);
}
int ans = mx - mn;
cout << toStr(Result =) <<' '<< ans;
return 0;
}
This is a sample question from a hackerrank question, so there would be some mistakes in it. I want to know what does the part
FUNCTION(minimum, <)
FUNCTION(maximum, >)
mean ? It doesn't look like a macro as it doesn't start with #define. The discussion forums too didn't have a good explanation about this part, just that it relates to the macro defined as #define minimum(mn, b) mn = min(b) and #define maximum(mx, c) mx = max(c).
'
I want to know what's it called and some advice on good resource to read about it.
The title is poorly written as I have no clue what to refer to that doubtful part of code as.

Answer 1

Let's break this answer into parts:
First, the thing you show here isn't valid code. #define FUNCTION(a, b) # doesn't compile on GCC, it correctly says:
error: '#' is not followed by a macro parameter
# is supposed to work like shown in your toStr example, where #a causes it to emit a string with a's value as contents.
Second, I assume that the poster of this "solution" wanted FUNCTION to just do nothing.
It seems that some compilers do treat a sole # that way, but it's not correct. It would have worked to write #define FUNCTION(a, b) without anything afterwards (or
#define FUNCTION(a, b) // like I saw in some other solutions for this question).
So, the answer to "what does this part do" is literally "nothing". (At least, it was intended this way.)
Third, even with that fixed, the code doesn't look right, because unless I'm misreading this, it doesn't really calculate any minimum or maximum, because min doesn't get the previous minimum as argument (same for max)! I'd expect this...
#define minimum(mn, b) mn = min(mn, b)
#define maximum(mx, c) mx = max(mx, c)
...instead of the current min(b) and max(c) there.
Forth, it seems that most solutions on this hackerrank question turn FUNCTION into a no-op and instead create minimum and maximum as macros.
But, I feel like the intention behind FUNCTION was something else, something a bit more clever. You could define it as macro that creates functions (hence the name) named after the first argument, that compare the minimum/maximum based on the second argument used as comparison operator:
#define FUNCTION(_FN_NAME_, _COMPARISON_OP_) \
void _FN_NAME_(int& limit, int newValue) { \
if (newValue _COMPARISON_OP_ limit) { \
limit = newValue; \
} \
}
This way, FUNCTION(minimum, <) and FUNCTION(maximum, >) would expand to:
void minimum(int& limit, int newValue) {
if (newValue < limit) {
limit = newValue;
}
}
void maximum(int& limit, int newValue) {
if (newValue > limit) {
limit = newValue;
}
}