在我的游戏中,我正在尝试制作一个菜单,当您按“q”时可以访问该菜单,但目前我遇到了一些问题。我认为它会切换到CompScreen
视图,然后快速返回currentroom
视图,我可能错了。我得到了cout CompMenu,HELP和你好读数,所以我知道它正在运行程序,但是当我按q时我仍然在同一个地方,没有任何事情发生。
EventManager.h
#ifndef EventManager_h
#define EventManager_h
#endif /* EventManager_h */
int windowWidth = 5000;//width of window
int windowHeight = 5000;//height of window
sf::View leveltwo(sf::FloatRect(x, y, 5000, 5000));
sf::View start(sf::FloatRect(0, 0, 2500, 1500));
sf::View ComputerScreen(sf::FloatRect(50000, 50000, 5000, 5000));
sf::RenderWindow window(sf::VideoMode(windowWidth, windowHeight ), "Awesome Game" );
Character player("/Users/danielrailic/Desktop/Xcode /NewGame/ExternalLibs/Sprites/Player.png");
bool InMenu = false;
enum Levels{
StartRoom, LevelTwo
};
Levels room = StartRoom;
int currentroom;
void WhatRoom(int TheLevel){
switch (room){
case StartRoom:
currentroom = 1;
window.setView(start);
if (TheLevel == 2){
room = LevelTwo;
}
break;
case LevelTwo:
currentroom = 2;
window.setView(leveltwo);
break;
}
};
enum States{
compmenu, mainmenu, NoMenu
};
States menu = NoMenu;
void CompMenu(){
window.setView(ComputerScreen);
cout << "HELP";
InMenu = true;
}
void WhatMenu(int TheMenu){
switch (menu){
case compmenu:
cout << "CompMenu";
CompMenu();
break;
case mainmenu:
break;
case NoMenu:
if (TheMenu == 2){
menu = compmenu;
}
break;
if (TheMenu == 3){
menu = mainmenu;
}
break;
}
}
main.cpp(在int main中)
if (sf::Keyboard::isKeyPressed(sf::Keyboard::Q) and InMenu == false){
WhatMenu(2);
}
if (sf::Keyboard::isKeyPressed(sf::Keyboard::Q) and InMenu == true){
InMenu = false;
WhatRoom(currentroom);
cout << "hello";
}
如果您有任何疑问或需要查看更多代码,请告知我们。感谢。
答案 0 :(得分:0)
我认为您在第一次else
阻止后错过if
因此可能会立即执行,因为此时InMenu
可能会更改为true
:< / p>
else if (sf::Keyboard::isKeyPressed(sf::Keyboard::Q) and InMenu == true){
此外,您应该处理按键长时间(仅在进入按下状态时作出反应)并摆脱所有全局变量的情况,因为它们已经造成了相当混乱。