关于getline函数的问题 [英] question about getline function

问题描述

#include <iostream.h>
#include <conio.h>
#include <fstream.h>
void main()
{
clrscr();
ifstream g;
char str[100];
g.open("a.txt");
while(str!='\n') problem is here
{
g.getline(str,20);
cout<<str;
}
g.close();
getch();
}

我读到getline函数读取了一个字符序列，直到达到行尾（'\ n'' ）

但我的程序无法正常工作

I read that getline functions reads a seqpence of character untill end of line is reached('\n')
but my program can not work

推荐答案

Quote：

我的程序无法正常工作

不，它不能。

从来没有一个点 str 指向100 char s的指针将等于常数值'\ n'如果我没记错的话会被评估为 13 。

你的，而循环永远不会完成。如果你只想阅读一行但我不确定你要做什你检查你对指针的理解。很多人在C / C ++中发现这个难题。

No it cannot.
There is never going to be a point where str a pointer to 100 chars is going to be equal to the constant value '\n' which if I remember correctly is evaluated as 13.
Your while loop will never complete. Your while loop is also possibly unnecessary if you only want to read one line but I don't know for sure what you're trying to do.

I would recommend you check your understanding of pointers. Many people find this a difficult topic in C/C++.

我已经为你修改了你的程序，这里是代码：

I have modified your program for you, here is the code:

#include <iostream>
#include <fstream>

using namespace std;

int main()
{

	ifstream g;

	char str[100];

	g.open("a.txt");

        /* repeat following actions 
           until end of file is reached */

	while( ! g.eof() ) 
	{
                /* clear our string, so we can get
                   new data */

		memset( &str, '\0', sizeof(str) );
		
                // get 100 characters into str
	
		g.getline(str,100);

                // write the result, and go to new line

		cout<<str<<endl;
		
	}

	g.close();

	return 0;
}

我已经处理了这个问题已经有一段时间了，但我希望这会对你有帮助。

祝您好运！

It has been a while since I have dealt with this, but I hope that this will help you.
Best regards and good luck!

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