Support for different platforms. Platform_desktop_sdl.
This commit is contained in:
50
src/core.c
50
src/core.c
@@ -1887,56 +1887,6 @@ int lcoreSetExitKey( lua_State *L ) {
|
||||
return 0;
|
||||
}
|
||||
|
||||
/*
|
||||
> keyName = RL.GetKeyName( int key, int scancode )
|
||||
|
||||
This function returns the name of the specified printable key, encoded as UTF-8.
|
||||
This is typically the character that key would produce without any modifier keys,
|
||||
intended for displaying key bindings to the user. For dead keys, it is typically
|
||||
the diacritic it would add to a character.
|
||||
|
||||
Do not use this function for text input. You will break text input for many
|
||||
languages even if it happens to work for yours.
|
||||
|
||||
If the key is KEY_UNKNOWN, the scancode is used to identify the key,
|
||||
otherwise the scancode is ignored. If you specify a non-printable key,
|
||||
or KEY_UNKNOWN and a scancode that maps to a non-printable key,
|
||||
this function returns nil but does not emit an error.
|
||||
|
||||
- Success return string or nil
|
||||
*/
|
||||
int lcoreGetKeyName( lua_State *L ) {
|
||||
int key = luaL_checkinteger( L, 1 );
|
||||
int scancode = luaL_checkinteger( L, 2 );
|
||||
|
||||
const char *keyName = glfwGetKeyName( key, scancode );
|
||||
|
||||
if ( keyName != NULL ) {
|
||||
lua_pushstring( L, keyName );
|
||||
}
|
||||
else {
|
||||
lua_pushnil( L );
|
||||
}
|
||||
|
||||
return 1;
|
||||
}
|
||||
|
||||
/*
|
||||
> scancode = RL.GetKeyScancode( int key )
|
||||
|
||||
This function returns the platform-specific scancode of the specified key.
|
||||
If the key is KEY_UNKNOWN or does not exist on the keyboard this method will return -1.
|
||||
|
||||
- Success return int
|
||||
*/
|
||||
int lcoreGetKeyScancode( lua_State *L ) {
|
||||
int key = luaL_checkinteger( L, 1 );
|
||||
|
||||
lua_pushinteger( L, glfwGetKeyScancode( key ) );
|
||||
|
||||
return 1;
|
||||
}
|
||||
|
||||
/*
|
||||
## Core - Input-related functions: gamepads
|
||||
*/
|
||||
|
||||
Reference in New Issue
Block a user