如何只整数scanf函数重复，如果用户输入非数字字符阅读？ [英] How to scanf only integer and repeat reading if the user enter non numeric characters?

问题描述

下面是一个小青年初学者与C code题从输入一个字符或整数只是想prevent用户小于0或超过23。

Here is a little young tyro's problem with C code trying simply to prevent the user from typing a character or an integer less than 0 or more than 23.

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    const char *input;
    char *iPtr;
    int count = 0;
    int rows;

    printf("Enter an integer: ");
    scanf("%s", input);
    rows = strtol(input, &iPtr, 0);
    while( *iPtr != '\0') // Check if any character has been inserted
    {
        printf("Enter an integer between 1 and 23: ");
        scanf("%s", input);
    }
    while(0 < rows && rows < 24) // check if the user input is within the boundaries
    {
        printf("Select an integer from 1 to 23: ");
        scanf("%s", input);
    }  
    while (count != rows)  
    {  
        /* Do some stuff */  
    }  
    return 0;  
}

我成功了一半和一个小的推升将AP preciated。

I made it halfway through and a small push up will be appreciated.

推荐答案

使用 scanf函数（％d个，＆安培;行）而不是 scanf函数（％S，输入）

这允许你从标准输入获得direcly整数值，而不需要转换为int。

This allow you to get direcly the integer value from stdin without need to convert to int.

如果用户输入包含非数字字符，那么你有一个 scanf函数之前清理你的标准输入一个字符串（％d个，＆安培;行）。

If the user enter a string containing a non numeric characters then you have to clean your stdin before the next scanf("%d",&rows).

您code可能是这样的：

your code could look like this:

#include <stdio.h>  
#include <stdlib.h> 

int clean_stdin()
{
    while (getchar()!='\n');
    return 1;
}

int main(void)  
{  
    int count = 0;  
    int rows =0;  
    char c;
    do
    {  
        printf("\nEnter an integer from 1 to 23: ");

    } while (((scanf("%d%c", &rows, &c)!=2 || c!='\n') && clean_stdin()) || rows<1 || rows>23);

    return 0;  
}

说明

1）

scanf("%d%c", &rows, &c)

此装置从用户输入期望的整数，接近它的非数字字符。

This means expecting from the user input an integer and close to it a non numeric character.

例1：的如果用户输入 aaddk 然后 ENTER ，scanf函数将返回0没什么capted

Example1: If the user enter aaddk and then ENTER, the scanf will return 0. Nothing capted

例2：的如果用户输入 45 然后 ENTER ，scanf函数将返回2（2元素capted）。在这里，％d个是capting 45 和％C 是capting \\ n

Example2: If the user enter 45 and then ENTER, the scanf will return 2 (2 elements are capted). Here %d is capting 45 and %c is capting \n

例3：的如果用户输入 45aaadd 然后 ENTER ，scanf函数将返回2（2元素capted）。在这里，％d个是capting 45 和％C 是capting A

Example3: If the user enter 45aaadd and then ENTER, the scanf will return 2 (2 elements are capted). Here %d is capting 45 and %c is capting a

2）

(scanf("%d%c", &rows, &c)!=2 || c!='\n')

在例1：的这个条件 TRUE 因为scanf函数返回 0 （！= 2 ）

In the example1: this condition is TRUE because scanf return 0 (!=2)

在例2：的这个条件 FALSE 因为scanf函数返回 2 和ç=='\\ n'

In the example2: this condition is FALSE because scanf return 2 and c == '\n'

在例3：的这个条件 TRUE 因为scanf函数返回 2 和ç=='A'（！='\\ n'）

In the example3: this condition is TRUE because scanf return 2 and c == 'a' (!='\n')

3）

((scanf("%d%c", &rows, &c)!=2 || c!='\n') && clean_stdin())

clean_stdin（）总是 TRUE ，因为函数的返回总是 1

clean_stdin() is always TRUE because the function return always 1

在例1：的的（scanf函数（％D％C，＆安培;行，和C）= 2 | ！| c ='\\ n'）是 TRUE 等以后条件的＆放大器;＆安培; 应进行检查，在 clean_stdin（）将被执行，整个条件是 TRUE

In the example1: The (scanf("%d%c", &rows, &c)!=2 || c!='\n') is TRUE so the condition after the && should be checked so the clean_stdin() will be executed and the whole condition is TRUE

在例2：的的（scanf函数（％D％C，＆安培;行，和C）= 2 | ！| c ='\\ n'）是 FALSE 等以后条件的＆放大器;＆安培; 将不会检查（因为曾经它的结果是什么，整个状况会 FALSE ），使 clean_stdin（）不会被执行，整个条件是 FALSE

In the example2: The (scanf("%d%c", &rows, &c)!=2 || c!='\n') is FALSE so the condition after the && will not checked (because what ever its result is the whole condition will be FALSE ) so the clean_stdin() will not be executed and the whole condition is FALSE

在例3：的的（scanf函数（％D％C，＆安培;行，和C）= 2 | ！| c ='\\ n'）是 TRUE 等以后条件的＆放大器;＆安培; 应进行检查，在 clean_stdin（）将被执行，整个条件是 TRUE

In the example3: The (scanf("%d%c", &rows, &c)!=2 || c!='\n') is TRUE so the condition after the && should be checked so the clean_stdin() will be executed and the whole condition is TRUE

所以，你可以说此话 clean_stdin（）将被执行，只有当用户输入包含非数字字符的字符串。

So you can remark that clean_stdin() will be executed only if the user enter a string containing non numeric character.

和这个条件（（scanf函数（％D％C，＆安培;行，和放大器;！C）= 2 || C ='\\ n'）及！＆安培; clean_stdin（ ））将返回 FALSE 仅当用户输入整数，没有别的

And this condition ((scanf("%d%c", &rows, &c)!=2 || c!='\n') && clean_stdin()) will return FALSE only if the user enter an integer and nothing else

如果条件（（scanf函数（％D％C，＆安培;行，和放大器;！C）= 2 || C ='\\ n'）及和放大器; clean_stdin （））是 FALSE 和整数之间和 1 和 23 那么，而循环将打破别人的而循环将继续

And if the condition ((scanf("%d%c", &rows, &c)!=2 || c!='\n') && clean_stdin()) is FALSE and the integer is between and 1 and 23 then the while loop will break else the while loop will continue

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