The indef plural code in Adjudicate differs from the 6/11 version.

From Vince at http://www.intfiction.org/forum/viewtopic.php?f=7&t=19868&start=40#p108572

Here's 6/11:

for (j=BestGuess() : j~=-1 && i<indef_wanted && i+offset<63 : j=BestGuess()) {
    flag = 0;
    if (j hasnt concealed && j hasnt worn) flag = 1;

Here's 6/12:

for (j=BestGuess() : j~=-1 && i<indef_wanted && i+offset<63 : j=BestGuess()) {
    flag = 1;
    if (j has concealed && j has worn) flag = 0;

In both cases, the meaning of the flag variable is the same. flag == 1 means accept an object, and flag == 0 means reject it.

In 6/11 the flag is initialized to 0 (reject), and various tests set it to accept object j if it's a good choice in context. In 6/12, the flag is initialized to 1 (accept) and various tests clear it to reject object j if it's not a good choice.

In 6/11, we see that j is provisionally accepted if it's not concealed and it's not worn. This makes sense in the context of commands like TAKE ALL (where we wouldn't want to include concealed objects) or DROP ALL (where we wouldn't want to drop things that the actor is wearing), and the excerpt from the DM4 that we quoted above is consistent with this: "mark each as accept unless (i) it has worn or concealed".

In 6/12, because the default value of the flag was reversed, this test also had to be reversed. Instead of accepting if j isn't concealed and isn't worn, we want to reject j if it is concealed or is worn. However, the code, as written, rejects it only if it's both concealed and worn. This is a bug.

The code should read:

if (j has concealed or worn) flag = 0;

DavidGriffith / inform6lib

The indef plural code in Adjudicate differs from the 6/11 version. #37