How to find the highest and lowest number of user input.

[td0g03]

Hello again guys. I have a question. I'm working on a program for my class and I'm stuck on how to find the lowest and highest number of a user input. We aren't at arrays yet so that's out of the question, but what would be another way of finding the highest and lowest. I remember hearing something in the lines of "cout <<" in class, but forgot. Also I could try the if else statement, but would seem like alot of work. I hope I am making this clear, but bottom line is how to find the lowest and highest number of a user input ( 5 inputs). Thanks in advance.

[kvbreddy]

Hi,
we can do it without need of array. Here i am trying to explain the logic...
You can implement it.

1) min and max are two unsigned integer variables.
min is initialized with possible maximum value,
max is initialized with zero.
2) Loop the following three statements for n times.
Read a number .
if the number is < min, then min = number.
if the number is > max, then max = number.

3) Finally print min, max

[td0g03]

Hi,
we can do it without need of array. Here i am trying to explain the logic...
You can implement it.

1) min and max are two unsigned integer variables.
min is initialized with possible maximum value,
max is initialized with zero.
2) Loop the following three statements for n times.
Read a number .
if the number is < min, then min = number.
if the number is > max, then max = number.

3) Finally print min, max

Hello, let me see if I am getting this right. So the user input 5 integers and that will be initialized into a,b,c,d, and e all int. Then how would I would do something like... to find the min

float findLowest (int a, int b,int c, int d, int e)
{
//    Local Definitions
    int min;
 
//    Statements
     if (a < b && a < c && a < d && a < e)
    return a;
    else
    if (b < a && b < c && b < d && a < e);
    return b;
    else
    if (c < a && c < b && c < d && c < e);
    return c;
    else
    if (d < a && d < b && d < c && d < e);
    return d;
    else
    if (e < a && e < b && e < c && e < d);
    return e;
 
 

I'm pretty sure that I have the else, if and return wrongs. Sorry, I haven't mastered or really understand it yet. Perhaps you can correct for me so I can see my mistakes and learn from them and understand why I got it why.

[rajesh6695]

#include <stdio.h>
#include <conio.h>
int main()
{
    unsigned int min,max,in,buf;
    min=0,max=0;
 
    printf("\nEnter The Numbers or press Zero to exit....");
    do
    {
        scanf("%u",&in);
        if(in!=0)
        {
            if(in>=max)
            {
                max=in;
            }
            if(in<=buf)
            {
                min=in;
            }
            buf=in;
        }
        else
        {
            printf("The Max & Min value Entered are : %u %u",max,min);
            break;
        }
    }while(1);
    return 0;
}

Try this code......

[td0g03]

#include <stdio.h>
#include <conio.h>
int main()
{
unsigned int min,max,in,buf;
min=0,max=0;

printf("\nEnter The Numbers or press Zero to exit....");
do
{
scanf("%u",&in);
if(in!=0)
{
if(in>=max)
{
max=in;
}
if(in<=buf)
{
min=in;
}
buf=in;
}
else
{
printf("The Max & Min value Entered are : %u %u",max,min);
break;
}
}while(1);
return 0;
}

Try this code......

Hmm, all that codes seems to do for me is let me input unlimited amount of numbers and never actually gives me the min/max

[rajesh6695]

Yes for the above code you can give n number of inputs but to stop getting input you have enter 0

[kvbreddy]

rajesh,
what is there in the buf initially????? It contains garbase value. So, in the first checking of in with buf, will give unexpected result.
In ur code we don't need buf. Simply use min in place of buf and make sure min is initialized with 32767 or 65535. (Maximum value that we can give)

Yes for the above code you can give n number of inputs but to stop getting input you have enter 0

[Ganon11]

I hope I am making this clear, but bottom line is how to find the lowest and highest number of a user input ( 5 inputs).

You will have to have 3 int variables - call them min, max, and num. You will have to ask the user for input into num before anything else - you will then initialize min and max to num, because after one input, num is the min AND the max.

You will then have to create a loop that will execute 4 more times (a for loop will do nicely). In each execution of the loop, you will 1) ask the user for input into num, 2) check if num > max (and if so, max = num), and 3) check if num < min (and if so, min = num). Once outside of the loop, print the min/max.

If you cannot use loops, you will have to manually ask the user for input 4 more times and write the checks after each input statement.

The function you wrote will work, but it is a bit tedious - the method described here will likely be much faster.

[td0g03]

You will have to have 3 int variables - call them min, max, and num. You will have to ask the user for input into num before anything else - you will then initialize min and max to num, because after one input, num is the min AND the max.

You will then have to create a loop that will execute 4 more times (a for loop will do nicely). In each execution of the loop, you will 1) ask the user for input into num, 2) check if num > max (and if so, max = num), and 3) check if num < min (and if so, min = num). Once outside of the loop, print the min/max.

If you cannot use loops, you will have to manually ask the user for input 4 more times and write the checks after each input statement.

The function you wrote will work, but it is a bit tedious - the method described here will likely be much faster.

Hello, something like this. I am doing two different functions for just for min and one for max. I already asked the user for 5 inputs which are a,b,c,d,e. I'm sorta confused on how to do a loop. Anyway you can give me a actual code example?

[Ganon11]

I assume you have a section of code like

int num1, num2, num3, num4, num5;
cout << "Enter the first number: ";
cin >> num1;
cout << "Enter the second number: ";
cin >> num2;
cout << "Enter the third number: ";
cin >> num3;
cout << "Enter the fourth number: ";
cin >> num4;
cout << "Enter the fifth number: ";
cin >> num5;

Then you will display your min and max by calling your min/max functions. This is perfectly fine.

What I was suggesting is something like this:

int max, min, num;
 
cout << "Enter number 1: ";
cin >> num;
max = num;
min = num;
 
for (int i = 1; i < 5; i++) {
   cout << "Enter number " << i + 1 << ": ";
   cin >> num;
   //Check if num > max
   //Check if num < min
}
// max and min are now the largest/smallest values entered

However, you might not have been taught loops yet, which means the original code you have will be perfectly fine.

[td0g03]

I assume you have a section of code like

int num1, num2, num3, num4, num5;
cout << "Enter the first number: ";
cin >> num1;
cout << "Enter the second number: ";
cin >> num2;
cout << "Enter the third number: ";
cin >> num3;
cout << "Enter the fourth number: ";
cin >> num4;
cout << "Enter the fifth number: ";
cin >> num5;

Then you will display your min and max by calling your min/max functions. This is perfectly fine.

What I was suggesting is something like this:

int max, min, num;
 
cout << "Enter number 1: ";
cin >> num;
max = num;
min = num;
 
for (int i = 1; i < 5; i++) {
   cout << "Enter number " << i + 1 << ": ";
   cin >> num;
   //Check if num > max
   //Check if num < min
}
// max and min are now the largest/smallest values entered

However, you might not have been taught loops yet, which means the original code you have will be perfectly fine.

Well, the code I have is let me just go ahead and post the whole thing. It isn't finish yet.

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
using namespace std;
 
//    Function Declarations 
char getOption (void);
void getData (int* a, int* b, int* c, int* d, int e);
float calc   (int option, int a, int b, int c, int d, int e);
 
int main (void)
{
//    Local Declarations 
    char toLower;
    int a,b,c,d,e;
    float result;
 
//    Statements 
    toLower = getOption ();
 
 
    return 0;
}    // main 
 
/*    ================= getOption ================= 
    This asks what option the user will like to display
    Such as smallest, largest, sum, average.
       Pre  getOption
       Post Tax income, total tax, and tax due
            calculated
*/
char getOption (void)
{
//   Local Declarations
     char toLower;
 
//   Statement
    printf  ("\t**********************************");
    printf("\n\t*              MENU              *");
    printf("\n\t*                                *");
    printf("\n\t*  L. Find Lowest                *");
    printf("\n\t*  H. Find Highest               *");
    printf("\n\t*  S. Calculate Sum              *");
    printf("\n\t*  A. Calculate Average          *");
    printf("\n\t*                                *");
    printf("\n\t**********************************");
 
    printf("\n\nPlease type your choice ");
    printf("and key Enter: ");
    scanf ("%c", &toLower);
    return toLower;
}    // getOption 
 
/* ====================== getData ==================== 
    This function reads two integers from the keyboard.
       Pre   Parameters a,b,c,d, and e are addresses
       Post  Data read into parameter addresses
*/
void getData (int* a, int* b, int* c, int* d, int e)
{
    printf("Please enter five integer numbers: ");
    scanf("%d %d %d %d %d", a, b, c, d, e);
    return;
}    // getData 
 
/*    ==================== doCalc ====================
    This function determines the type of operation
    and calls a function to perform it. 
       Pre   option contains the operation
             num1 & num2 contains data
       Post  returns the results
*/
 
float calc (int toLower, int a, int b, int c, int d, int e) 
{
// Local Declarations 
float result;
 
//    Statements 
    switch(toLower)
        {
         case 1:  result = findLowest (a, b, c, d, e);
                  break;
         case 2:  result = findHighest(a, b, c, d, e);
                  break;
         case 3:  result = 3.0;              // Multiply
                  break;
         case 4:  if (num2 == 0.0)           // Divide
                     {
                      printf("\n\a\aError: ");
                      printf("division by zero ***\n");
                      exit (100);
                     } // if 
                  else
                      result = 4.0;
                  break;
         /* Better structured programming would validate
            option in getOption. However, we have not  
            yet learned the technique to code it there.
         */
         default: printf("\aOption not available\n");
                  exit (101);
        } // switch 
printf("**In calc result is: %6.2f\n", result);
    return result;
}    // calc 
 
/* ====================== findLowest =================
    This function adds two numbers and returns the sum.
       Pre   a and b contain values to be added
       Post  Returns the lowest number
*/
float findLowest (int a, int b,int c, int d, int e)
{
//    Local Definitions
    int min;
 
//    Statements
    if (a < b && a < c && a < d && a < e)
    return a;
    else
    if (b < a && b < c && b < d && a < e);
    return b;
    else
    if (c < a && c < b && c < d && c < e);
    return c;
    else
    if (d < a && d < b && d < c && d < e);
    return d;
    else
    if (e < a && e < b && e < c && e < d);
    return e;
 
}    // findLowest 
 

Well, I have to use a function that will call getData (The user inputed numbers) then use findLowest to find the min number. I haven't learned looped yet, but the first code you posted will only work *I think* if it isn't in functions. Perhaps I am wrong.

 
 
 
 
    Expand|Select|Wrap|Line Numbers
 
 
 
    
 
 int toLowest (int a,int b,int c,int d,int e;)
 
int num1,num2,num3,num4,num5
  
 
cout << a;
 
cin >> num1;
 
cout << b;
 
cin >> num2;
 
cout << c;
 
cin >> num3;
 
cout << d;
 
cin >> num4;
 
cout << e;
 
cin >> num5;