如何将第二行读入两个不同的数组 [英] How do I read the second line into two different arrays

问题描述

好吧，我有第一行下来

这就是我正在读取的文件



BECDCBAADEBACBAEDDBEDCBAAECDCB

1234 BECDXACDXAXDXEDBXCABCDEXXECDCB



但问题是我无法正确读取第二行，内部while循环完全正常，因为它正在读取第一行，正确的答案，成阵列。



我在尝试读取数字时得到的输出是234234而不是1234.



谢谢！



我的尝试：

okay I have the first line down
This is what I'm reading from a file

BECDCBAADEBACBAEDDBEDCBAAECDCB
1234 BECDXACDXAXDXEDBXCABCDEXXECDCB

But the problem is I can't read the second line properly, the inner while loop is perfectly fine, as it's reading the first line, the right answers, into an array.

The output I'm getting when trying to read the numbers are 234234 instead of 1234.

Thank you!

What I have tried:

    infile>>r;

while  (!infile.eof()){
    while (r != '\n' && d<30){
        right [d] = r;
        cout<<right [d];
        infile>>r;
        d++;
    }
    infile>>i;
    cout<<i;
}

推荐答案

如需更多帮助，请尝试使用可执行的代码改进您的问题。



如果你不明白你的代码在做什么或为什么它做了什么，答案是调试器。

使用调试器查看代码正在执行的操作。只需设置断点并查看代码执行情况，调试器允许您逐行执行第1行并在执行时检查变量，这是一个令人难以置信的学习工具。



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调试器在这里向您展示您的代码在做什么你的任务是与它应该做的事情进行比较。

调试器中没有魔法，它没有发现错误，它只是帮助你。当代码没有达到预期效果时，你就接近一个bug。

For more help, try to improve your question with code that can be executed.

When you don't understand what your code is doing or why it does what it does, the answer is debugger.
Use the debugger to see what your code is doing. Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute, it is an incredible learning tool.

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The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't find bugs, it just help you to. When the code don't do what is expected, you are close to a bug.

你好会员13068214 ，



请使用 Peek（）查看换行符。由于某种原因，它不能在循环条件下识别换行符。

Hi Member 13068214,

Please use Peek() to check newline. For some reason it is not recognizing newline at the loop condition.

while (!infile.eof()) {
	while (d < 30) {
		right[d] = r;
		cout << right[d];
		if (infile.peek() == '\n')
		{
			d = 0;
			break;				
		}
		infile >> r;
		d++;
	}
	infile >> r;
	/*cout << r;*/
}

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