如何在不使用某种数组的情况下输入多个值

Question

So basically I'm taking an intro tp programming class and we've been taught the basics (loops, if statements, variable types etc.) I'm solving a program where I have to ask the user to enter 6 different temperature values, and then print out the maximum, average and range of the 6 values.

How and where should I store these 6 numbers?

cout<< "Enter 6 diff numbers" << endl;
float numbers;
cin >> numbers;

for ( .... i_++)

max = ;
min = ;


cout << .. << .... << endl;

//This should not help because float can only keep one number stored and not 6. How should I do this without using any sort of arrays, functions etc. ?

I was thinking using substrings and declaring it like a string or something??

Thanks for all your help.

Answer 1

I'm putting up the basic algorithm you can use without using arrays. Assuming everything in Kelvin.

float max = 0; // Minimum Value Set for comparing with larger values
float min;
float sum = 0;
float avg = 0;
float tmp;
string number, alltheNumbers;
for( int i = 0; i < 6; i++ ){
   cin>>number;
   tmp = <float> number;
   if( tmp > max ){
      max = tmp;
   }
   sum += tmp;
   alltheNumbers += ',' + number; // Save all the numbers in comma seperated Strings
}
min = max;    // Maximum Found value set for finding minimum
std::string delimiter = ",";
size_t pos = 0;
while ((pos = alltheNumbers.find(delimiter)) != std::string::npos) {
    number = alltheNumbers.substr(0, pos); // Use the comma to retrieve all those numbers
    tmp = <float> number;
    if( tmp < min ){
      min = tmp;
    }
    alltheNumbers.erase(0, pos + delimiter.length());
}
avg = sum / 6;

So, You have the following variables with the required data.

max <- will have the maximum value
min <- will have the minimum value
avg <- will have the average value.

Answer 2

This is one way to do it without arrays without spoiling the rest of your homework.

cout<< "Enter 6 different numbers" << endl;

float num1, num2, num3, num4, num5, num6, max, min, sum, avg;

cin >> num1 >> num2 >> num3 >> 
       num4 >> num5 >> num6;

Goodluck!

Answer 3

An improvement on @Zion's the first answer, I think initializing the min and max to the first digit should suffice. Before the for loop input number, then set max = number and min = number and then start for loop from i=1.