Fix lcd.RGB()
It does not require access to lcd, so let's not check that.
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@@ -834,9 +834,6 @@ Returns a 5/6/5 rgb color code, that can be used with lcd.setColor
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*/
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static int luaRGB(lua_State *L)
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{
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if (!luaLcdAllowed)
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return 0;
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int r = luaL_checkinteger(L, 1);
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int g = luaL_checkinteger(L, 2);
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int b = luaL_checkinteger(L, 3);
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