Talk:Use familiar: Difference between revisions
imported>StDoodle No edit summary |
imported>RogerMexico No edit summary |
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I'm also seeing expected behavior; true most of the time, false if I try a familiar I don't have. | I'm also seeing expected behavior; true most of the time, false if I try a familiar I don't have. | ||
--[[User:StDoodle|StDoodle (#1059825)]] 14:52, 10 April 2010 (UTC) | --[[User:StDoodle|StDoodle (#1059825)]] 14:52, 10 April 2010 (UTC) | ||
I find the syntax as described on this page confusing. With other functions I do not have to specify the datatype, for example:<br/> | |||
outfit($string[Backup]);<br/> | |||
outfit("Backup");<br/> | |||
both work identically. However, use_familiar("hobo");<br/> | |||
returns the error "Function 'use_familiar( string )' undefined..."<br/> | |||
This just happened to be the first function I looked up on this wiki, and if the discussion with examples had not been here, I don't think I would have worked it out at all.--[[User:RogerMexico|RogerMexico]] 12:04, 25 September 2010 (UTC) |
Revision as of 12:04, 25 September 2010
This always seems to return false for me, can anyone confirm?
--Slyz 20:56, 9 April 2010 (UTC)
- > ash use_familiar($familiar[sandworm])
- Putting Mexicana the Hovering Sombrero back into terrarium...
- Taking Hecho en Mexico the Baby Sandworm out of terrarium...
- Returned: true
I assume that the ash command should be enough to check return value; correct me if I'm wrong. --Heeheehee 21:03, 9 April 2010 (UTC)
I'll have to recheck after RO, but here is what I was getting (doing things slightly differently):
- > ash print( use_familiar($familiar[slimeling]) )
- Putting Dusty the Baby Sandworm back into terrarium...
- Taking Yuk the Slimeling out of terrarium...
- false
- Returned: void
--Slyz 03:33, 10 April 2010 (UTC)
Huh. Using that method, I get:
- > ash print( use_familiar($familiar[sombrero]) )
- Putting Hecho en Mexico the Baby Sandworm back into terrarium...
- Taking Mexicana the Hovering Sombrero out of terrarium...
- true
- Returned: void
--Heeheehee 07:51, 10 April 2010 (UTC)
I'm also seeing expected behavior; true most of the time, false if I try a familiar I don't have. --StDoodle (#1059825) 14:52, 10 April 2010 (UTC)
I find the syntax as described on this page confusing. With other functions I do not have to specify the datatype, for example:
outfit($string[Backup]);
outfit("Backup");
both work identically. However, use_familiar("hobo");
returns the error "Function 'use_familiar( string )' undefined..."
This just happened to be the first function I looked up on this wiki, and if the discussion with examples had not been here, I don't think I would have worked it out at all.--RogerMexico 12:04, 25 September 2010 (UTC)