如果用户输入非数字字符，如何仅扫描整数并重复阅读? [英] How to scanf only integer and repeat reading if the user enters non-numeric characters?

问题描述

这里有一些 C 代码只是为了防止用户输入字符或小于 0 或大于 23 的整数.

#include #include int main(void){常量字符*输入；字符 * iPtr;整数计数 = 0;整数行；printf("请输入一个整数:");scanf("%s", 输入);行 = strtol(输入, & iPtr, 0);while( *iPtr != '')//检查是否插入了任何字符{printf("请输入一个 1 到 23 之间的整数:");scanf("%s", 输入);}while(0 < rows && rows < 24)//检查用户输入是否在边界内{printf("从1到23中选择一个整数:");scanf("%s", 输入);}而(计数！=行){/* 做一些事情 */}返回0；}

我已经完成了一半，将不胜感激.

解决方案

使用 scanf("%d",&rows) 而不是 scanf("%s",input)

这允许您直接从 stdin 获取整数值，而无需转换为 int.

如果用户输入包含非数字字符的字符串，那么您必须在下一个 scanf("%d",&rows) 之前清理标准输入.

您的代码可能如下所示:

#include #include int clean_stdin(){而 (getchar()!='
');返回 1；}int main(void){整数行 =0;字符 c;做{printf("
请输入一个 1 到 23 之间的整数:");} while (((scanf("%d%c", &rows, &c)!=2 || c!='
') && clean_stdin()) || rows<1 ||行＞23)；返回0；}

说明

1)

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

这意味着期望用户输入一个整数并在其附近输入一个非数字字符.

Example1: 如果用户输入 aaddk 然后ENTER，scanf 将返回 0.没有被捕获

Example2:如果用户输入45然后ENTER，scanf将返回2(2个元素被俘).这里 %d 正在捕获 45 而 %c 正在捕获

示例 3: 如果用户输入 45aaadd 然后 ENTER，scanf 将返回 2(2 个元素被俘).这里 %d 正在捕获 45 而 %c 正在捕获 a

2)

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

在示例 1 中: 这个条件是 TRUE 因为 scanf 返回 0 (!=2)

在示例 2 中: 这个条件是 FALSE 因为 scanf 返回 2 和 c ==' '

在示例 3 中: 这个条件是 TRUE 因为 scanf 返回 2 和 c =='a' (!=' ')

3)

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

clean_stdin() 总是 TRUE 因为函数返回总是 1

在示例 1 中: (scanf("%d%c", &rows, &c)!=2 || c!=' ') 是 TRUE 所以应该检查 && 之后的条件，以便 clean_stdin()将被执行并且整个条件为 TRUE

在示例 2 中: (scanf("%d%c", &rows, &c)!=2 || c!=' ') 是 FALSE 所以 && 之后的条件不会被检查(因为它的结果是整个条件将是FALSE ) 所以 clean_stdin() 不会被执行，整个条件是 FALSE

在示例 3 中: (scanf("%d%c", &rows, &c)!=2 || c!=' ') 是 TRUE 所以应该检查 && 之后的条件，以便 clean_stdin()将被执行并且整个条件为 TRUE

所以你可以注意到 clean_stdin() 只有当用户输入包含非数字字符的字符串时才会执行.

还有这个条件 ((scanf("%d%c", &rows, &c)!=2 || c!=' ') && clean_stdin()) 只会在用户输入 integer 而没有其他内容时返回 FALSE

如果条件 ((scanf("%d%c", &rows, &c)!=2 || c!=' ') && clean_stdin()) 是 FALSE 并且 integer 介于 1 和 23 之间，然后 while 循环将中断，否则 while 循环将继续

Here is some 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 != '') // 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;  
}

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

解决方案

Use scanf("%d",&rows) instead of scanf("%s",input)

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

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

your code could look like this:

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

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

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

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

    return 0;  
}

Explanation

1)

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

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

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

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

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!='
')

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

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

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

3)

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

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

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

In the example2: The (scanf("%d%c", &rows, &c)!=2 || c!=' ') 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

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

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

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

And if the condition ((scanf("%d%c", &rows, &c)!=2 || c!=' ') && 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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