Question

How does this code find a non-positive value?

#include <map>
#include <cstdio>

#define until(cond) while(!(cond))

using namespace std;

int main() {
  // initialization
  map<int,int> second;
  int i=10;
  int testme[10]={4,3,1,-3,7,-10,33,8,4,14};
  while (i --> 0) second[i]=testme[i];

  // find the first non-positive value in second
  map<int,int>::reverse_iterator p = --second.rend();
  do {
    printf("Is %d non-positive?\n",second[++i]);
  } until(-- p --->   second < 0);
    // "second < 0" to check if the value in second is non-positive

  printf("Yes it is!\n");
}

The output is:

Is 4 non-positive?
Is 3 non-positive?
Is 1 non-positive?
Is -3 non-positive?
Yes it is!

So how does the “second < 0” string check for a non-positive value?

Answer 1

Some hints for parsing --p--->second. It is evaluated as --((p--)->second). (Thanks to @AProgrammer for fixing my blatant error!)

• p is a pointer or iterator.

• p-- decrements p, but returns its previous value as an rvalue

• (p--)->second accesses that value’s member second.

• --((p--)->second) decrements that value (i.e. the mapped value) and returns the new, decremented value

• That new value is compared against 0

Notes:

• The p-- takes care of iterating over the container. Note that the loop doesn’t otherwise have any explicit change of p.

• The outer -- makes 0 count as a negative number. As a side effect, the loop decrements every value in the map.

• The second use of i is somewhat redundant. You could have written p->second inside the loop rather than second[++i], since you already have an iterator. In fact, second[++i] necessitates a whole tree search.

The code is equivalent to:

do { /* ... */
    auto q = p;
    --p;
    --(q->second);
} until (q->second < 0);