英文:
Java compiler just takes into account the first digit
问题
如果输入为13，则编译器只会考虑第一个数字1。
我会给你Java代码。
class Max {
static void checkFever(double choice) {
// 如果温度低于37度，打印出"你发烧了"
if (choice > 37) {
System.out.println("你发烧了。");
}
else {
System.out.println("你没发烧。");
}
}
public static void main(String[] args) throws IOException {
char choice;
choice = (char) System.in.read();
checkFever(choice);
System.out.println(choice); // 这会显示结果
}
}
输出如下：
38
你发烧了。
3
英文:
If the input is 13 the compiler will just take into account the first digit that is 1.
I'll let you the java code.
class Max {
static void checkFever(double choice) {
// if temperature is less than 37, print "You have fever"
if (choice > 37) {
System.out.println("You have fever.");
}
else {
System.out.println("You don't have fever.");
}
}
public static void main(String[] args) throws IOException {
char choice;
choice = (char) System.in.read();
checkFever(choice);
System.out.println(choice); // this will make appear the result
}
}
the output is the following:
38
You have fever.
3
答案1
得分: 2
使用包装了System.in（这是一个InputStream）的Scanner来读取整行输入，并将其解析为适当的基本数据类型或对象数据类型。在你的情况下，对于温度，int 应该已经足够，但是推荐使用十进制的 float 或者 double。
Scanner scanner = new Scanner(System.in);
double choice = scanner.nextDouble();
checkFever(choice);
System.out.println(choice);
你的代码不起作用，因为你使用了 InputStream::read 方法来读取单个字节。
> 从输入流中读取下一个字节的数据。返回的值是一个范围在 0 到 255 之间的 int 值。
英文:
Use a Scanner that wraps the System.in (which is an InputStream) to read the whole line parsed into an appropriate primitive or object data type. In your case for the temperature int would be sufficient but a decimal float or double is preferred.
Scanner scanner = new Scanner(System.in);
double choice = scanner.nextDouble();
checkFever(choice);
System.out.println(choice);
Your code doesn't work because you read a single byte using method InputStream::read
> Reads the next byte of data from the input stream. The value byte is returned as an int in the range 0 to 255.
答案2
得分: 1
System.in.read()只读取一个单独字符。你可以通过java.util.Scanner辅助类来读取整行，因此：
import java.io.IOException;
import java.util.Scanner;
class Max {
static void checkFever(double choice) {
// 如果体温低于37度，打印“你发烧了”
if (choice > 37) {
System.out.println("你发烧了。");
}
else {
System.out.println("你没有发烧。");
}
}
public static void main(String[] args) throws IOException {
Scanner input = new Scanner(System.in);
String answer = input.nextLine();
int temp = Integer.parseInt(answer);
checkFever(temp);
System.out.println(answer); // 这将显示结果
}
}
如果用户未输入有效的整数，这会产生NumberFormatException。随意改进代码。 
（帮助：try{...} catch(...)）
英文:
System.in.read() reads only a single character. You can read a whole line by the java.util.Scanner helper class, so:
import java.io.IOException;
import java.util.Scanner;
class Max {
static void checkFever(double choice) {
// if temperature is less than 37, print "You have fever"
if (choice > 37) {
System.out.println("You have fever.");
}
else {
System.out.println("You don't have fever.");
}
}
public static void main(String[] args) throws IOException {
Scanner input = new Scanner(System.in);
String answer = input.nextLine();
int temp = Integer.parseInt(answer);
checkFever(temp);
System.out.println(answer); // this will make appear the result
}
}
This will give a NumberFormatException if the user did not enter a valid integer. Feel free to improve the code.
<sub>(Help: try{...} catch(...))</sub>
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论