argument type mismatch ?? -
b0b - 22.07.2016
error 035: argument type mismatch (argument 2) I get this all time
All lines,.
PHP код:
COMMAND:help(playerid, params[])
{
if(PlayerLoggedIn[playerid] == true)
{
SCM(playerid, COL_WHITE, "__________________HELP__________________");
SCM(playerid, COL_GREEN, "CMDS");
SCM(playerid, COL_GREEN, "CMDS");
SCM(playerid, COL_WHITE, "_______________________________________");
return 0;
}
Re: argument type mismatch ?? -
AndySedeyn - 22.07.2016
COL_WHITE is by convention probably defined as an embedded color code:
PHP код:
{FFFFFF} // white
Embedded color codes can only be used in the middle of strings, hence them being embedded:
PHP код:
""COL_WHITE"This is white\n"COL_GREEN"While this is green."
Colors as function arguments must be in their hexadecimal form:
PHP код:
0xFFFFFFFF //white
In your case, you should define the color with its hexadecimal value:
PHP код:
#define COLOR_WHITE 0xFFFFFFFF
Re: argument type mismatch ?? -
Stinged - 22.07.2016
You have a missing bracket ( } ) above return 0;
Also, it should be return 1; or else it will show "SERVER: Unknown command."
Re: argument type mismatch ?? -
b0b - 22.07.2016
Okey now i get this fixey !! But now : error 030: compound statement not closed at the end of file (started at line 671)
Re: argument type mismatch ?? -
K0P - 22.07.2016
Код:
COMMAND:help(playerid, params[])
{
if(PlayerLoggedIn[playerid] == true)
{
SCM(playerid, 0xFFFFFFFF, "__________________HELP__________________");
SCM(playerid, 0x33AA33AA, "CMDS");
SCM(playerid, 0x33AA33AA, "CMDS");
SCM(playerid, 0xFFFFFFFF, "_______________________________________");
return 0;
}
}
Re: argument type mismatch ?? -
FuNkYTheGreat - 22.07.2016
@K0P ur command wont work because ur returning 0
Try this b0b
Код:
COMMAND:help(playerid, params[])
{
if(PlayerLoggedIn[playerid] == true)
{
SCM(playerid, 0xFFFFFFFF, "__________________HELP__________________");
SCM(playerid, 0x33AA33AA, "CMDS");
SCM(playerid, 0x33AA33AA, "CMDS");
SCM(playerid, 0xFFFFFFFF, "_______________________________________");
return 1;
}
}
Re: argument type mismatch ?? -
Stinged - 22.07.2016
If you return something inside an if statement, it will ask give you a warning for not returning anything outside.
You should return 1; before adding the last bracket.