istream_iterator问题,在循环中使用

Question

为什么这个循环不会终止？程序打印出istream_iterator中的所有元素后冻结.

 /*30*/ int main( int arc, char **arv ) {
 /*31*/     vector<float> numbers( 0 );
 /*32*/     cout << "Please, input even number of floats: ";
 /*33*/     istream_iterator<float> iit (cin);
 /*34*/     istream_iterator<float> eos;
 /*35*/     while( iit != eos ) {
 /*36*/         cout << (*iit) << endl; 
 /*37*/         iit++;
 /*38*/     }   
 /*39*/     if( numbers.size( ) & 1 == 1 ) {
 /*40*/         cerr << "Sorry: you must input"
 /*41*/              << " an even number of inputs" << endl;
 /*42*/     }                                               
 /*43*/     return ( 0 );                                   
 /*44*/ }  

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更新:

所以现在这样做更多.谢谢大家.通过提供的信息,我将简化的内容简化为更少.需要重新获得模数部分,但现在对我来说更清楚了.

 34 int main( int arc, char **arv ) {
 35     vector<float> num;
 36     cout << "Please, input even number of floats: ";
 37     for( istream_iterator<float> iit (cin);
 38          iit!=istream_iterator<float>( );iit++ ) {
 39         num.push_back( (*iit) );
 40     }   
 41 }

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更新2:如果有人还在阅读这个帖子,我可以在"最终"产品上获得社区的内容/风格反馈吗？

/********************************************************************
 * Author: Mattew Hoggan
 * Description: Write a client program that uses the data type point.
 * Read a sequence of points (pairs of floating-point numbers) from
 * standard input, and find the one that is closest to the first.
 * *****************************************************************/

#include <math.h>                                                  
#include <iostream>                                                
#include <vector>                                                  
#include <istream>                                                 
#include <iterator>
#include <algorithm>
#include <limits.h>

using namespace std;                                               

typedef struct point {                                             
    float x;                                                       
    float y;                                                       
} point;                                                           

float calc_distance( const point *a, const point *b ) {            
    return sqrt( pow( a->x-b->x, 2.0 ) + pow( a->y-b->y, 2.0 ) );      
}                                                                  

void print_point( point *a ) {                                 
    cout << "(" << a->x << ", " << a->y << ") ";
}

void delet_point( point *a ) {
    delete a;
}

vector<point*>* form_pairs( vector<float> &num ) {
    vector<point*> *points=NULL;
    if( ( num.size( ) & 1 ) == 1 ) { 
        cerr << "ERROR: You input: " << num.size( ) 
             << " which is odd, failed to build vector "
             << endl;
        return points;
    } else {
        cout << "Going to build points" << endl;
        points = new vector<point*>;
        for( vector<float>::iterator vit=num.begin( );
            vit!=num.end( ); vit+=2 ) {
            point *in = new point( );
            in->x = *(vit);
            in->y = *(vit+1);
            points->push_back( in );     
        }   
        return points;
    }
}

void pop_front( vector<point*> *pairs ) {
    reverse( pairs->begin( ), pairs->end( ) );
    pairs->pop_back( );
    reverse( pairs->begin( ), pairs->end( ) );
}

void min_euclidean_distance( vector<point*> *pairs ) {
    if( pairs->size( ) == 1 ) {
        cerr << "You already know the answer to this" << endl;
        return;
    }
    point *first = pairs->front( );
    pop_front( pairs );
    point *second = pairs->front( );
    pop_front( pairs );

    for_each( pairs->begin( ),pairs->end( ),print_point );
    cout << endl;

    point *p_min = second;
    float f_min = calc_distance( first,second );
    for( vector<point*>::iterator pit = pairs->begin( );
        pit != pairs->end( ); pit++ ) {
        float tmp = calc_distance( first,(*pit) );
        if( tmp < f_min ) {
            f_min = tmp;
            p_min = (*pit);
        }
    }

    cout << "The closest node to "; print_point( first );
    cout << " is "; print_point( p_min );
    cout << " at " << f_min << " units away " << endl;
    delete first;
    delete second;
}

int main( int arc, char **arv ) {                                
    vector<float> num;
    cout << "Please, input even number of floats: ";
    for( istream_iterator<float> iit (cin);
         iit!=istream_iterator<float>( );iit++ ) {
         num.push_back( (*iit) );
    }

    vector<point*>* pairs = form_pairs( num );
    if( pairs ) {
        min_euclidean_distance( pairs );
        for_each( pairs->begin( ),pairs->end( ),delet_point );
        delete pairs;
    }
}  

Answer 1

如果您将istream_iterator一个流连接到一个流,那么istream_iterator它将在它读取的流具有failbit或badbit设置之前有效.只有当您从流中读取格式错误的数据或流用尽数据时,才会发生这种情况.

因为cin从标准输入流(默认情况下在大多数系统上连接到键盘)istream_iterator读取数据,所以总能读取更多数据.也就是说,在输入数字列表后,cin将不会处于错误状态,因为您始终可以在程序中输入更多数字.程序"冻结"的事实是由于它正在等待输入更多输入.您可以在程序处理您输入的所有数据后输入更多数据来验证这一点.

要使cin数据停止读取,您有几种选择.首先,您可以cin通过将数据传输到程序或从文件中读取数据来重定向操作系统.例如,在Linux系统上,您可以像这样运行您的程序:

./my-program < my-input-file

一旦程序读取了所有文本my-input-file,cin将遇到文件结束并将触发failbit.循环然后终止.

或者,您可以明确地导致cin命中文件结尾.在Linux系统上,如果从命令行运行编程,可以按CTRL + D使标准输入发送文件结束信号,这也会导致循环退出.或者,您可以输入无效的浮点值cin,例如字符串STOP!.这会导致failbit触发cin,因为它不能将值视为float,然后会导致循环退出.

希望这可以帮助!

Answer 2

该程序运行正常,只需要在输入中提供文件结尾:

echo 1.1 5.3 | ./myprogram

如果你真的在控制台上打字,发送Ctrl-D信号表示输入结束.

(不相关的,您的条件可能应该是if( numbers.size( ) % 2 != 0 )为了清晰度和语义而阅读;另外,请编译并注意编译器警告.)