C无限做while循环

Question

This is a simple program for a class that prompts the user for the length of his or her shower in minutes (as a positive integer, re-prompting as needed) and then prints the equivalent number of bottles of water (as an integer).

It assumes the shower uses 1.5 gallons of water per minute (192 oz) and a plastic bottle size of 16 oz

My do-while loop successfully rejects negative numbers and 0, however, if I input text such as "foo" when prompted for the length of a shower in minutes, the program runs into an infinite loop, forever running the loop and printing "How long is your shower(in minutes)?:"

Any ideas how to refine the while condition to avoid this?

#include <stdio.h>

int min_to_bottles(int *n);

int main(void)
{
    int minutes;
    int bottles;
    do
    {
        printf("How long is your shower(in minutes)?:");
        scanf("%i", &minutes);
    }
    while (minutes < 1);

    bottles = min_to_bottles(&minutes);

    printf("Equivalent of bottles used per shower is: %i\n", bottles);

}

int min_to_bottles(int *n)
{
    int bottles;
    int oz = 192 * *n;
    bottles = oz/16;
    return bottles;
}

Answer 1

Always check the return value of scanf() :

int result;
do {
    printf("How long is your shower(in minutes)?:");
    result = scanf("%d", &minutes);
    if(result != 1) 
        break;
} while (minutes < 1);

A shorter version (if only one scan is needed):

printf("How long is your shower(in minutes)?:");

while ((scanf("%d", &minutes) == 1) && (minutes < 1))
    ;

---

There is no need to use a pointer as parameter in int min_to_bottles(int *n); :

#include <stdio.h>

int min_to_bottles(int n)
{
    return (192 * n) / 16;
}

int main(void)
{
    int minutes = 0;
    int bottles = 0;

    printf("How long is your shower(in minutes)?: ");

    while ((scanf("%d", &minutes) == 1) && (minutes < 1 || minutes > 100))
        printf("Enter a number between 1 and 100 : ");

    // if(minutes == 0) here means invalid data was entered.
    //    so a check could be done before continuing.

    bottles = min_to_bottles(minutes);

    printf("Equivalent of bottles used per shower is: %d\n", bottles);

    return 0;
}

Initializing minutes to 0 will avoid calculating the bottles with an undefined value in case scanf() failed (by entering text for example).

Answer 2

When you enter in text, it doesn't match the %i format specifier, so the text gets stuck in the input buffer and it keeps trying to read the same thing.

You need to flush the buffer if you didn't get a good match. You'll know if that the case by checking the return value of scanf , which returns the number of patters successfully matched.

int minutes = 0;
while (minutes < 1)
{
    printf("How long is your shower(in minutes)?:");
    int count = scanf("%i", &minutes);
    if (count < 1) {
        scanf("%*s");   // The * tells scanf to read the input but not assign it to anything
    }
}

Answer 3

Do not use scanf("%i",...) .

The main problem is that bad input that did not convert to a number remains in stdin until another function reads it. As code did not check the return value of scanf() , the value of minutes is not known to be good and do ... while (minutes < 1); can easily repeated loop.

Solution: Read a line of input, convert to number, valid number:

---

To handle unexpected user input begin by reading a line of input

char buf[80];
if (fgets(buf, sizeof buf, stdin) == NULL) Handle_EOF();

Then parse the buffer for the number.

errno = 0;
char *endptr;
long num == strtol(buf, &endptr, 0);

// detect overflow and no conversiosn
if (errno || buf == endptr) Handle_BadInput();

// Ignore trailing white-space
while (isspace((unsigned char) *endptr) endptr++;

if (*endptr) Handle_BadInput();

Validate the number.

#define SHOWER_TIME_MIN (1 /* minute */)
#define SHOWER_TIME_MAX (60 /* minute */)
if (num < SHOWER_TIME_MIN || num > SHOWER_TIME_MAX) Handle_BadInput();

Put this all in a helper function

Example How to test input is sane