C ++切换语句案例错误
我正在使用游戏循环的switch语句编写一个简单的基于文本的RPG。 该程序工作正常，直到我试图添加另一个case语句，在这一点上它给了我以下三个错误：“跳转到案例标签”（错误发生在新增案例的行），和两个“十字架初始化” ClassName * objectName'“（在情况2中创建新对象时发生错误）。 我会粘贴重要的代码，如果有人需要更多，请告诉我。
int main(void)
{
// add weapons to array
Weapon *weaponList[12];
// Rusty Sword
weaponList[0] = new Weapon(0,0,0);
weaponList[0]->SetAll(0,2,3);
// Bronze Sword
weaponList[1] = new Weapon(0,0,0);
weaponList[1]->SetAll(1,5,10);
// Bronze Battle Axe
weaponList[2] = new Weapon(0,0,0);
weaponList[2]->SetAll(2,15,30);
// Iron Sword
weaponList[3] = new Weapon(0,0,0);
weaponList[3]->SetAll(3,25,70);
// add armor to array
Armor *armorList[12];
// Worn Platemail
armorList[0] = new Armor(0,0,0);
armorList[0]->SetAll(0,2,3);
// Bronze Chainmail
armorList[1] = new Armor(0,0,0);
armorList[1]->SetAll(1,5,8);
// Bronze Platemail
armorList[2] = new Armor(0,0,0);
armorList[2]->SetAll(2,7,20);
// Iron Chainmail
armorList[3] = new Armor(0,0,0);
armorList[3]->SetAll(3,15,60);
while(gamestate != 8)
{
switch(gamestate)
{
case 0:
cout << " /| Welcome!n"
<< " || n"
<< " || n"
<< " || n"
<< "_||_ n"
<< " 88 n"
<< " 88 Name: ";
cin >> heroName;
gamestate = GAME_STATE_MENU;
break;
case 1:
cout << "n"
<< "'/stats' will show you your statsn"
<< "'/shop' will let you visit the weapon shopn"
<< "secret commands: /setweapon # /setarmor # /setheroexp #n"
<< "n";
cout << "Command: ";
cin >> command;
if (strcmp(command, "/stats") == 0)
{
gamestate = 2;
break;
}
else if (strcmp(command, "/shop") == 0)
{
gamestate = 3;
break;
}
else if (strcmp(command, "/fight") == 0)
{
gamestate = 4;
break;
}
else if (strcmp(command, "/setweapon") == 0)
{
cin >> testNum;
heroWeapon = testNum;
break;
}
else if (strcmp(command, "/setarmor") == 0)
{
cin >> testNum;
heroArmor = testNum;
break;
}
else if (strcmp(command, "/setheroexp") == 0)
{
cin >> testNum;
heroExp = testNum;
LevelUp();
break;
}
else if (strcmp(command, "/exit") == 0)
{
gamestate = 8;
break;
}
else
{
cout << "Please enter a valid command.n";
gamestate = 2;
break;
}
case 2:
Weapon *wCurrent = weaponList[heroWeapon];
Armor *aCurrent = armorList[heroArmor];
heroWeaponPower = wCurrent->GetWeaponAttack();
heroArmorDefense = aCurrent->GetArmorDefense();
heroPowerDefault = ((heroLevel - 1) * 10) + 10;
heroPower = heroPowerDefault + (heroStrength * 2) + heroWeaponPower;
heroDefenseDefault = ((heroLevel - 1) * 2) + 5;
heroDefense = heroDefenseDefault + (heroAgility / 5) + heroArmorDefense;
heroHealthDefault = (heroLevel * 5) + 20;
heroHealth = heroHealthDefault + (heroStamina * 10);
cout << "nS T A T SnName: "
<< heroName
<< "nLevel: "
<< heroLevel
<< "nExp: "
<< heroExp << "/" << expForLevel[heroLevel]
<< "nGold: "
<< heroGold
<< "nHealth: "
<< heroHealth
<< "nPower: "
<< heroPower
<< "nDefense: "
<< heroDefense
<< "nWeapon: "
<< weaponNameList[heroWeapon]
<< "nArmor: "
<< armorNameList[heroArmor]
<< "nn";
system("PAUSE");
gamestate = 2;
break;
case 3:
break;
}
}
return 0;
}
通过它的声音，你有：
case 2:
Type somevar = ...;
...
break;
case 3:
为了达到somevar情况，编译器生成一个跳过somevar初始化的somevar 。
为了解决这个问题，使用大括号来创建围绕变量声明的块：
case 2:
{
Type somevar = ...;
...
}
break;
将声明封装在堆栈中...呃... 本地范围 ... :)
switch(gamestate)
{
case 0:
{
Apple a;
a.DoSomething();
}
break;
case 1: /* etc. */ break;
case 2: /* etc. */ break;
}
...或将它们移动到开关外部：
Apple A;
switch(gamestate)
{
case 0: a.DoSomething(); break;
考虑以下：
switch (x)
{
case 0:
int i = 0;
case 1:
i = 5;
}
如果x是1呢？ 然后我们跳过i的初始化并开始使用它。 这就是你所得到的： case 3可以访问case 2变量，但是如果你使用它们，你已经开始使用它们而不需要运行它们的初始化。
常见的解决办法是引入范围：
switch (x)
{
case 0:
{
int i = 0;
}
case 1:
{
i = 5; // not possible, no i in this scope
}
}