如何防止scanf失败的无限循环 [英] How to prevent an infinite loop with scanf failure

问题描述

该准则应该能为美元带来零钱，并且效果很好.但是这位教授说，他将在输入随机数和字母的同时输入.它可以很好地处理数字，但是当输入字母时，会出现无限循环吗?

The Code is supposed to make change for a dollar and works fine. but the professor says that he will be enter random numbers along with letters. It works fine with numbers but when a letter is entered an infinite loop will occur any suggestions?

#include <stdio.h>
#include <stdlib.h>
#define amtPaid 1
#define SENTINAL -1
int quarters(int numChange);
int dimes(int numChange);
int nickels(int numChange);
int pennies(int numChange);

int main(void)
{
    double amtDue = 0; // how much is paid


while(1){
    printf("\nPlease enter the price less than 1 dollar: ");
    scanf(" %lg", &amtDue);
    int changeReturn = (amtPaid - amtDue) * 100 + 0.5; // convert decimal to whole number
    int updated = 0;    // remaining change after amt of change
    int numberQuarters = quarters(changeReturn); // number of quarters needed

   if(changeReturn >= 0 && changeReturn <= 100){ // if changereturn  is between 0 and 100 execute code

        printf("\nNice!");
        printf("\nWe owe you %i cents" , changeReturn);
        if(numberQuarters >= 0){    // get and print number of quarters
            printf("\nQuarters: %i", numberQuarters);
            updated = changeReturn % 25;
        }
        if(dimes(updated) >= 0){ // get and print number of dimes
            printf("\nDimes: %i", dimes(updated));
            updated = updated % 10;
        }
        if(nickels(updated)>= 0){ // get and print number of nickels
            printf("\nNickels: %i", nickels(updated));
            updated = updated % 5;
        }
        if(pennies(updated) >= 0){ // get and print number pennies
            printf("\nPennies: %i", pennies(updated));
        }

    }

    else if(amtDue == SENTINAL){
        break;
    }

    else {
        printf("That does not make sense to me. please type a valid number");
    }

    printf("\n %g", amtDue);

}
return 0;
}

int quarters(int numChange){
    int numQuarters = 0;

    numQuarters = numChange / 25;

    return numQuarters;
}
int dimes(int numChange){
    int numDimes = 0;

    numDimes = numChange / 10;

    return numDimes;
}
int nickels(numChange){
    int numNickels = 0;

    numNickels = numChange / 5;

    return numNickels;
}
int pennies(numChange){

    return numChange;
}

推荐答案

如果提供了除scanf()格式说明符的预期值以外的不适当值，则scanf()将失败并且不当值保留在输入缓冲区中，将 feed 提供给下一个scanf()，只会导致连续失败.在这种情况下，您需要清理输入缓冲区，然后再进行下一个输入.您可以使用类似的

In case an inappropriate value is supplied other than the expected value of the format specifier with scanf(), the scanf() will fail and the inappropriate value will remain in the input buffer, providing the feed to next scanf(), only to cause successive failures. In that case, you need to clean up the input buffer before going for next input. You can use something like

1. 检查scanf()
的返回值
2. 万一发生故障，请使用while( getchar() != '\n' );清理输入缓冲区.

1. check the return value of scanf()
2. In case of failure, use while( getchar() != '\n' ); to clean the input buffer.

也就是说，int nickels(numChange)现在在c中(C99起)无效.您必须明确地将其设置为int.

That said, int nickels(numChange) is now invalid in c (C99 onwards). You have to make it as int explicitly.

这篇关于如何防止scanf失败的无限循环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IT屋！