C Program that prints out the maximum and minimum values from user input involving precision decimals

Question

I am a newbie programmer and just started to teach myself C then decided to tackle some simple problems taken from the internet. Specifically with this one: Problem

And my solution is:

#include <stdio.h>
#include <math.h>
#include <float.h>
#include <limits.h>

int main() {
    double min = DBL_MAX;
    double max = DBL_MIN;
    double n;

    do {
        scanf("%lf", &n);

        if (n < 0.0001) {
            break;
        }

        if (n > max) {
            max = n;
        }

        if (n < min) {
            min = n;
        }
    } while (n > 0.0000);

    printf("Min: %.4f Max: %.4f\n", min, max);
    return 0;
}

However, I need to run my program exactly as the input/output specified by the problem. For example, consider I am inputting in different lines:

1.1 1.2 1.3 3 6 9 11 -2 7
2.2
2.3 12
0

The program should output 12 as the max and -2 as the min, the program ends when the 0 was inputted.

Answer 1

You need to set min and max BOTH to the first value set by scanf and then check for greater and less than on later iterations. The way it works now, min starts at 0 and will never change as nothing can be lower than that without the program exiting.

int main(){
  double min = 0; 
  double max = 0; 
  float n; 
  int count = 0; 

  do{
    scanf("%f", &n);

    if(n < 0.0001){
      break;
    }  
    if( count ) {
        if(n > max){
            max = n; 

        } else if(n < min){
            min = n; 
        }  
    } else {
        max = n; 
        min = n; 
        count = 1; 
    }  

  }while(n > 0); 

  printf("Min: %.4f Max: %.4f\n", min, max);

  return 0; 
}

Also, the proper type for float is %f.

Answer 2

Make sure you read the problem description again, your logic in the code is tainted in comparison to what's being attempted.

According to the problem description, n can be an int as you're using it to define the number of values to be given to the program.

You can then use a for loop (for(i=0; i < n; i++){CODE HERE}) to gather more input from the user. The first number given should be your base value for the min and max values.

After you have a base value you can compare min and max to each input thereafter. (ie. if(max < input) {max = input} and if(min > input) {min = input})

If you have anymore questions feel free to inbox me and I'll try to help you work out the problem :)

Answer 3

1. Using math.h you can use INFINITY and -INFINITY. http://pubs.opengroup.org/onlinepubs/007904975/basedefs/math.h.html
2. Comparing doubles with literals in order to make decisions is tricky at best. Besides it would be better to have positive as well as negative numbers in your program.
3. fflush(stdin) gets rid of any previous input in the buffer.
4. Notice that max starts at -INFINITY and min starts at +INFINITY.
5. Comparing a float with a double will result in the calculation being performed using the double type anyways. Try not to mix these without a good reason.

---

#include <stdio.h>
#include <math.h>

int main()
{
    double min = +INFINITY;
    double max = -INFINITY;
    double n;
    char ans = 'n';

    do
    {
        printf("Enter a number: ");
        scanf("%lf", &n);

        if(n > max)
        {
            max = n;
        }

        if(n < min)
        {
            min = n;
        }
        fflush(stdin);  // gets rid of previous input
        printf("Another number? (y/n): ");
        scanf("%c", &ans);
    }
    while(ans == 'y');

    printf("Min: %.4f Max: %.4f\n", min, max);
    return 0;
}

---

Output:

Enter a number: -45.6
Another number? (y/n): y
Enter a number: 23.0
Another number? (y/n): y
Enter a number: 92.3
Another number? (y/n): y
Enter a number: -100.22
Another number? (y/n): n
Min: -100.2200 Max: 92.3000

Process returned 0 (0x0)   execution time : 15.805 s
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