Question

int single = 0, doub=0, triple=0, homer=0, atbats=0, totalbase, totalhits;
double slug, battingavg;

Scanner sc = new Scanner(System.in);
System.out.print("Enter singles (-1 to end):  ");
single = sc.nextInt();

while (single != -1)
{
    System.out.print("Enter doubles:  ");
    doub = sc.nextInt();
    System.out.print("Enter triples:  ");
    triple = sc.nextInt();
    System.out.print("Enter home runs:  ");
    homer = sc.nextInt();
    System.out.print("Enter total at bats:  ");
    atbats = sc.nextInt();
    System.out.print("Enter the player's name: ");
    String name = sc.next();
    totalbase = (single + doub * 2 + triple * 3 + homer * 4);
    slug = totalbase / atbats;
    battingavg = (single + doub + triple + homer) / atbats;
    System.out.println("Player's name is " + name);
    System.out.printf("The slugging percentage is %.3f\n", + slug);
    System.out.printf("The batting percentage is %.3f\n", + battingavg);

    System.out.print("Enter singles (-1 to end):  ");
    single = sc.nextInt();
}

该程序在计算后只输出1或0。其他一切都很好，但它似乎没有做计算。

Answer 1

问题是当你进行整数除法时，你会得到整数值。而是使用double或float数据类型。我看到slug和battingavg已经double，但您要为它们分配整数除法的结果。如果将计算中的至少一个值转换为double，则应获得预期的输出。例如：

slug = totalbase / (double) atbats;
battingavg = (single + doub + triple + homer) / (double) atbats;

Answer 2

当你对整数值进行除法并将结果存储在一个加宽转换的double时，但该值被计算为一个整数，因此你可以扩大整数值。

改变这一点，

slug = totalbase / atbats;
battingavg = (single + doub + triple + homer) / atbats;

这样的事情，

slug = ((double) totalbase / atbats);
battingavg = ((double) (single + doub + triple + homer) / atbats);

将double值转换为double变量。

程序不会做计算

2 个答案: