# calculating average from the input

Hi, I have this program which you will input a number. It will tell whether which digit is the smallest and which is the biggest and this part works fine. What I can’t get to work is the average part wherein if I input ‘1234’ it will add 1+2+3+4 and divide the sum by 4. if I run this it gives me an output of ‘50’. I also need the ave to be in the form of number and not in ascii form. I can’t seem to find the problem.Please help.Thanks!

``````char output[5];
int sum=0;

void loop()
{
if (Serial.available() > 0)
{
delay(10);
int x = 0;
while (Serial.available()>0)
{
output[x+1] = '\0';
x++;
}

if (x>=5 || x<4)
{
Serial.println("The input must be 4 bytes.");
}
else
{
int input = atoi(output);
Serial.print("The value is: ");
Serial.println(input);
Serial.print("1st byte is: ");
Serial.println(output[0]);
Serial.print("2nd byte is: ");
Serial.println(output[1]);
Serial.print("3rd byte is: ");
Serial.println(output[2]);
Serial.print("4th byte is: ");
Serial.println(output[3]);

if (min(output[0],output[1])<min(output[2],output[3]))
{
Serial.print("Byte ");
Serial.print(min(output[0],output[1]));
Serial.println(" is the smallest byte.");
}
else
{
Serial.print("Byte ");
Serial.print(min(output[2],output[3]));
Serial.println(" is the smallest byte.");
}
if (max(output[0],output[1])>max(output[2],output[3]))
{
Serial.print("Byte ");
Serial.print(max(output[0],output[1]));
Serial.println(" is the biggest byte.");
}
else
{
Serial.print("Byte ");
Serial.print(max(output[2],output[3]));
Serial.println(" is the biggest byte.");
}

for (x=0;x<5;x++)
{
sum+=output[x];
}
int ave=sum/4;
Serial.println(ave);
}
}

}
``````

See my reply to your other question.

You're dealing with numbers and text in the same code, which can be tricky. 1 is one, unless it's the text (ASCII) '1', in which case it's 49. Your average is for the ASCII values, so 50 is almost correct for avg of 1+2+3+4 (it should be 2.5, but integers through away the fraction). One option is to convert ASCII to number as soon as possible, and work with that.