如何在 C++ 中实现 Java“扫描仪”？
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How to implement Java "Scanner" in C++?
提问by Lemon Juice
Please have a look at the following java code
请看下面的java代码
import java.util.Scanner;
public class Main
{
static int mul=1;
static String convert;
static char[] convertChar ;
static StringBuffer buffer = new StringBuffer("");
public static void main(String[]args)
{
Scanner scan = new Scanner(System.in);
int number=0;
int loopValue = scan.nextInt();
//System.out.println("print: "+loopValue);
for(int i=0;i<loopValue;i++)
{
number = scan.nextInt();
// System.out.println("print: "+number);
for(int a=1;a<=number/2;a++)
{
if(number%a==0)
{
//System.out.println(a);
mul = mul*a;
//System.out.println(mul);
}
}
convert = String.valueOf(mul);
convertChar = convert.toCharArray();
if(convertChar.length>4)
{
/*System.out.print(convertChar[convertChar.length-4]);
System.out.print(convertChar[convertChar.length-3]);
System.out.print(convertChar[convertChar.length-2]);
System.out.print(convertChar[convertChar.length-1]);
System.out.println();*/
buffer.append(convertChar[convertChar.length-4]);
buffer.append(convertChar[convertChar.length-3]);
buffer.append(convertChar[convertChar.length-2]);
buffer.append(convertChar[convertChar.length-1]);
System.out.println(buffer);
}
else
{
System.out.println(mul);
}
//System.out.println(mul);
mul = 1;
}
}
}
This code is built to compute the product of positive divisors of a given number. I have used scanner here because I don't know how many inputs will be entered. That is why I can't go something like
此代码用于计算给定数字的正除数的乘积。我在这里使用了扫描仪，因为我不知道将输入多少个输入。这就是为什么我不能去像
int a, b;
cin >> a >> b
in C++. All the inputs will be inserted by a test engine, into one single like like following
在 C++ 中。所有输入都将由测试引擎插入，如下所示
6 2 4 7 8 90 3456
6 2 4 7 8 90 3456
How can I implement the Java "Scanner" using C++ ? Is there a header file for that? Please help!
如何使用 C++ 实现 Java“扫描仪”？有没有头文件？请帮忙！
回答by Rob?
You seem to be using Scannerto read one integer at a time from the standard input stream. This is easily accomplished with the extraction operator, operator>>.
您似乎正在使用Scanner从标准输入流中一次读取一个整数。这可以通过提取运算符轻松完成operator>>。
Replace this code:
替换此代码：
Scanner scan = new Scanner(System.in);
int number=0;
int loopValue = scan.nextInt();
//System.out.println("print: "+loopValue);
for(int i=0;i<loopValue;i++)
{
number = scan.nextInt();
// System.out.println("print: "+number);
With this:
有了这个：
int number=0;
int loopvalue=0;
std::cin >> loopvalue;
for(int i = 0; i < loopValue; i++)
{
std::cin >> number;
You should check the value of std::cinafter the >>operations to ensure that they succeeded.
您应该检查std::cinafter>>操作的值以确保它们成功。
Refs:
参考：
回答by Simon
If you use std::cin >> value;to read the value then you can only process the entire line once a new-line has been detected.
如果您使用std::cin >> value;读取值，那么您只能在检测到换行后处理整行。
If you want to process each number as it is typed then you could use a function like:
如果您想按输入的方式处理每个数字，则可以使用如下函数：
int nextInt()
{
std::stringstream s;
while (true)
{
int c = getch();
if (c == EOF) break;
putch(c); // remove if you don't want echo
if ((c >= '0' && c <= '9') || (c == '-' && s.str().length() == 0))
s << (char)c;
else if (s.str().length() > 0)
break;
}
int value;
s >> value;
return value;
}
OK, there are probably more efficient ways to write that but it will read the input character by character until a number is encountered and will return whatever number is read when anything other than a number is encountered.
好吧，可能有更有效的方法来编写它，但它会逐个字符地读取输入，直到遇到一个数字，并在遇到除数字以外的任何数字时返回读取的任何数字。
E.g. 1 2 3 4 would return 1 on the first call, 2 on the second etc.
例如 1 2 3 4 将在第一次调用时返回 1，在第二次调用时返回 2，以此类推。
