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# help getting average

 P: n/a Hi, I have a temperature conversion program down pat, but I was told to add an average, meaning, i need to get the average temperature for as many times as it was entered. i do not know where to start, and my stpid book doesnt seem to help. here is my program #include void main( ) { int num1, num2, count; char choice, again; count=1; do {cout<<"This Program will Convert Temperature"<>choice; switch (choice) { case '1': cout<>num1; cout<>num2; cout<>again; } while (again!= 'N' && again!='n'); } Feb 15 '06 #1
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 P: n/a Gary wrote: Hi, I have a temperature conversion program down pat, but I was told to add an average, meaning, i need to get the average temperature for as many times as it was entered. i do not know where to start, and my stpid book doesnt seem to help. Consider this: To print the average you need to calculate it first. For this you need to divide the sum of all values inputted by the amount of values inputted. The sum is variable, increasing by the inputted value every time the loop runs. The amount is variable as well, increasing by one every time the loop runs. Small fragment example: int sum, count; sum=0; count=0; do { count++; int number=readNumber(); // Or use cin or whatever sum=sum+number; // IMPORTANT! If your number is integer, you need to convert // to double first to avoid getting integer division cout << "The average is " << sum/(double)count << endl; } while (runLoop); Feb 15 '06 #2

 P: n/a Gary wrote: Hi, I have a temperature conversion program down pat, but I was told to add an average, meaning, i need to get the average temperature for as many times as it was entered. i do not know where to start, and my stpid book doesnt seem to help. here is my program #include I take it, you meant: #include using std::cin; using std::cout; using std::endl; void main( ) That should be int main ( ) { int num1, num2, count; char choice, again; count=1; do {cout<<"This Program will Convert Temperature"<>choice; switch (choice) { case '1': cout<>num1; cout<>num2; cout<>again; } while (again!= 'N' && again!='n'); } In order to compute an average of the temperatures, you need to a) express all the different temperature using the same units, e.g, Celsius. b) at the end, compute and output the average. The most straight forward way to do that, would be to store the temperatures in a std::vector<> while they are entered and use them at the end to form an average. Best Kai-Uwe Bux Feb 15 '06 #3

 P: n/a Thank you so much. Got it to work!! Feb 15 '06 #4

 P: n/a Kai-Uwe Bux wrote: Gary wrote: Hi, I have a temperature conversion program down pat, but I was told to add an average, meaning, i need to get the average temperature for as many times as it was entered. i do not know where to start, and my stpid book doesnt seem to help. here is my program #include I take it, you meant: #include using std::cin; using std::cout; using std::endl; void main( ) That should be int main ( ) { int num1, num2, count; char choice, again; count=1; do {cout<<"This Program will Convert Temperature"<>choice; switch (choice) { case '1': cout<>num1; cout<>num2; cout<>again; } while (again!= 'N' && again!='n'); } In order to compute an average of the temperatures, you need to a) express all the different temperature using the same units, e.g, Celsius. b) at the end, compute and output the average. The most straight forward way to do that, would be to store the temperatures in a std::vector<> while they are entered and use them at the end to form an average. One drawback with retaining all of the previous temperatures though is that both the amount of memory needed by the vector to hold the previous results and the amount of time needed to update their average will keep increasing as each new data point is added. In other words, our program would not be able to run indefinitely since at some point it would run out of memory or become too slow for anyone to want to use it. A more efficient solution would be to maintain a running average of the temperature conversions. For a running average the program needs to keep only a running total of the number of previous conversions performed. Then each new conversion performed adds 1.0/runningTotal * conversionValue to the running average (which was initialized to 0 before the first conversion). An actual C++ implementation of this technique has been left as an exercise for the reader. Greg Feb 16 '06 #5

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