在C中强制输入仅为正数并进行错误处理

Question

我有一个微不足道的问题要问。我的程序应该只接受正整数。如果有非法内容，应提示用户重新输入号码。

我现在的代码是:

#include<stdio.h>
int main(){
int reads;
int num=0;
char a;
while(num<=0){
  printf("Please Enter positive integer: ");
  while(((reads = scanf("%d%c", &num, &a)) != 2 && reads != EOF) || a != 'n' ){
    do {
     printf("Please Enter positive integer: ");
     reads = scanf("%c", &a);
    }while(reads != EOF && a != 'n');
  }
}
printf("Num is: %d", num);
}

上面的代码几乎完成了我想要的;但是，当输入是多个字母时，输出提示将被打印多次，这让我很困扰。

   Please Enter positive integer: pdesf
    Enter positive numbers only: Enter positive numbers only: Enter positive numbers only: Enter positive numbers only: Enter positive numbers only: dfwerasdfwe
    Enter positive numbers only: Enter positive numbers only: Enter positive numbers only: Enter positive numbers only: Enter positive numbers only: Enter positive numbers only: Enter positive numbers only: Enter positive numbers only: Enter positive numbers only: Enter positive numbers only: Enter positive numbers only: 

如果你能帮我解决这个问题，或者为这个看似微不足道的问题提供更好的解决方案，我将不胜感激。谢谢吧!

Answer 1

使用fgets将整行读入缓冲区。如果您只想处理第一个字符，则可以忽略其余字符。类似以下语句:

char buf[MAX_LINE_LEN];
if (fgets(buf, MAX_LINE_LEN, stdin))
{
    char a = buf[0];
    /* Do handling... */
}
else
{
    /* error */
}

在浏览器中编码，可能包含错误痕迹

Answer 2

while(num<=0){
  printf("Please Enter positive integer: ");
  while(((reads = scanf("%d%c", &num, &a)) != 2 && reads != EOF) || a != 'n' ){
    printf("Please Enter positive integer: ");
    while(getchar() != 'n');
  }
}

Answer 3

我已经解决了你的程序。在Turbo c++中试试下面的程序:

#include<stdio.h>
#include<conio.h>
main()
{
    char n[2],ni;
    clrscr();
    GET:
    printf("Enter positive number: ");
    scanf("%s",&n);
    if(atoi(n)>0)
    {
        printf("You have entered ");
        ni=atoi(n);
        printf("%d",ni);
    }
    else if(atoi(n)<=0)
    {
        printf("Wrong choicen");
        goto GET;
    }
    else
    {
        printf("Wrong choicen");
        goto GET;
    }
    getch();
}