08.06.2014, 09:43
sampgdk
08.06.2014, 10:04
Maybe that's because you didn't initialize logprintf.
08.06.2014, 10:35
I added
And still the same
Код:
typedef void (*logprintf_t)(char* format, ...); logprintf_t logprintf;
08.06.2014, 11:14
That's not initialization - initializing a variable means assigning something to it, and logprintf can be initialized like this:
(and you would have known that if you looked at any plugin's source code...)
Код:
logprintf = (logprintf_t)ppData[PLUGIN_DATA_LOGPRINTF];
08.06.2014, 11:24
RCON1, can I give you an advice?
Learn C/++ first, then yes, try to do a plugin.
That's basically what we had done, and that's what you have to do, since the moment you say(had done) initializing is the same thing as declaring a variable...
Learn C/++ first, then yes, try to do a plugin.
That's basically what we had done, and that's what you have to do, since the moment you say(had done) initializing is the same thing as declaring a variable...
08.06.2014, 11:30
Quote:
That's not initialization - initializing a variable means assigning something to it, and logprintf can be initialized like this:
Код:
logprintf = (logprintf_t)ppData[PLUGIN_DATA_LOGPRINTF]; |
08.06.2014, 12:03
You can know why such a thing?
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