C ++在案例中没有检测到字符串

时间:2014-03-13 23:27:14

标签: c++

C ++初学者。我已经尝试了很多方法来尝试修复此代码,但我认为纠正它的方法远远超出我所知的范围。

switch (num)
{
case 0: enemy == "Ogre";
    enetotal = "250";
    enehealth = "250";
    eneattack = "5";
    break;
case 1: enemy == "Goblin";
    enetotal = "75";
    enehealth = "75";
    eneattack = "10";
    break;
case 2: enemy == "Orc";
    enetotal = "100";
    enehealth = "100";
    eneattack = "25";
    break;
case 3: enemy == "Dark Elf";
    enetotal = "70";
    enehealth = "70";
    eneattack = "15";
    break;
case 4: enemy == "Lizardman";
    enetotal = "200";
    enehealth = "200";
    eneattack = "20";
    break;
}


cout << "You walk past a field when you see a " + enemy +", what do you do? \n";

基本上,应该发生的是,根据随机情况,它会为某个敌人选择一个名字和健康/攻击。当我刚拿到敌人的名字时工作正常,但是一旦我添加了攻击和健康,它就会打印出来

You walk past a field when you see a, what do you do?

有些东西告诉我案件不能处理那么多行,但我已经尝试了许多其他的东西,比如if语句,似乎没什么用。任何人都可以帮我解决这个问题吗?

2 个答案:

答案 0 :(得分:5)

您正在比较名称

enemy == "Orc"

而不是分配

enemy = "Orc"

答案 1 :(得分:0)

您的语法错误:==是一项测试=