Question

为什么这个函数抱怨它应该返回char类型的结果？

public static char getUserResponseToGuess(int guess){
    Scanner input = new Scanner(System.in);     
    char userResponse = 'x';            
    while((userResponse != 'h' && userResponse != 'l' && userResponse != 'c')){
        System.out.print("Is it" + guess + "? (h/l/c)");        
        userResponse = input.next().charAt(0);
        if (userResponse == 'h' || userResponse == 'l' || userResponse =='c'){
            return userResponse;            
        }
        else{
            System.out.print("Is it" + guess + "? (h/l/c)");
        }
    }
}

Answer 1

编译器完成的静态分析无法理解您使用的逻辑，因此根据其分析，所有可能的流程最终都不会返回值

当编译器分析代码时，如果代码中的if条件不满足并且退出循环，则该方法将不返回值。

public static char getUserResponseToGuess(int guess){
    Scanner input = new Scanner(System.in);     
    char userResponse = 'x';            
    while((userResponse != 'h' && userResponse != 'l' && userResponse != 'c')){
        System.out.print("Is it" + guess + "? (h/l/c)");        
        userResponse = input.next().charAt(0);
        if (userResponse == 'h' || userResponse == 'l' || userResponse =='c'){
            return userResponse;            
        }
        else{
            System.out.print("Is it" + guess + "? (h/l/c)");
        }
    }
    return userResponse;  
}

Answer 2

当userResponse不是这三个字符中的一个时会发生什么？它将退出循环并返回什么？每条路径都必须返回char

Answer 3

这应该有效。 Arun P Johny解决方案也有效。

public static char getUserResponseToGuess(int guess){
    Scanner input = new Scanner(System.in);     
    char userResponse = 'x';            
    System.out.print("Is it " + guess + "? (h/l/c): ");
    while((userResponse != 'h' && userResponse != 'l' && userResponse != 'c')){             
        userResponse = input.next().charAt(0);
        if (userResponse == 'h' || userResponse == 'l' || userResponse =='c'){
            break;                          
        }
        else{
            System.out.print("Is it " + guess + "? (h/l/c): ");
        }
    }
    return userResponse;    
}

返回char的Java方法抱怨不返回char。

3 个答案: