为什么 scanf 会跳过输入?

Why does scanf appear to skip input?(为什么 scanf 会跳过输入?)

问题描述

我对以下程序中 scanf 的行为感到困惑.scanf 似乎输入一次，然后不再输入，直到打印出一个字符流.

I am confused about scanf's behaviour in the following program. scanf appears to input once, and then not input again, until a stream of characters is printed.

下面是一个C程序

#include<stdio.h>
int main()
{
    int i, j=0;

    do
    {
        ++j;
        scanf("%d", &i);
        printf("

%d %d

", i, j);
    }
    while((i!=8) && (j<10));  

    printf("
J = %d
", j);
    return 0;
}

在这里，直到我输入任何整数程序都可以正常工作，但是当输入一个字符时，它会继续打印 i 的最后输入值并且永远不会停止(直到循环退出时 j 为 10)以便 scanf 接受下一个输入.

here, Till i am inputting any integer program works perfectly fine, but when a character is inputted it goes on printing the last inputed value of i and never stops(untill j is 10 when loop exits) for scanf to take next input.

output::  
1    <-----1st input
1 1
2    <---- 2nd input
2 2
a    <---- character input
2 3  
2 4
2 5
2 6
2 7
2 8
2 9
2 10

J = 10  

同样的事情也在 C++ 中发生.

same thing is happening in c++ also.

#include<iostream>
using namespace std;
int main()
{
    int i, j=0;

    do
    {
        ++j;
        cin>>i;
        cout<<i<<" "<<j<<"
";
    }
    while((i!=8) && (j<10));

    cout<<"
j = "<<j<<"
";
}   


output of c++ program ::  
1     <-----1st input
1 1
2     <-----2nd input
2 2
a    <------ character input
0 3
0 4
0 5
0 6
0 7
0 8
0 9
0 10

j = 10  

c++ 中唯一的变化是打印 0 而不是最后一个值.

only change in c++ is that 0 is being printed instead of last value.

我知道这里程序需要整数值，但我想知道当输入字符代替整数时会发生什么?以上所有发生的原因是什么?

I know here integer values are expected by the program, but i want to know what happens when character is inputted in place of an integer? what is the reason of all happening above?

推荐答案

当你输入 a 时，然后 cin >>i 无法读取，因为 i 的类型是 int 无法读取字符.这意味着，a 永远留在流中.

When you enter a, then cin >> i fails to read it because the type of i is int to which a character cannot be read. That means, a remains in the stream forever.

现在为什么 i 打印 0 是另一回事.实际上它可以打印任何东西.一旦尝试读取失败，就不会定义 i 的内容.scanf 也会发生类似的事情.

Now why i prints 0 is a different story. Actually it can print anything. The content of i is not defined once the attempt to read fails. Similar thing happens with scanf as well.

正确的写法:

do
{
    ++j;
    if (!(cin>>i)) 
    {
       //handle error, maybe you want to break the loop here?
    }
    cout<<i<<" "<<j<<"
";
}
while((i!=8) && (j<10));

或者简单地这个(如果你想在发生错误时退出循环):

Or simply this (if you want to exit loop if error occurs):

int i = 0, j = 0;
while((i!=8) && (j<10) && ( cin >> i) )
{
    ++j;
    cout<<i<<" "<<j<<"
";
}

这篇关于为什么 scanf 会跳过输入?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！