Thread: Array average help

Hint: if I declare an int as

Code:

int  x = 3/4;

printf("%d",x);

What value would be printed out?

Originally Posted by internalbalance

3/4?

You have no idea what the value of 3/4 is in C, do you?

Hint: The two best guesses are 0 or 1; what do you think it is?

Tim S.

Code:

for(i=0;i<n;++i);
        sum=sum+i/n;
        printf("Average is %d\n", sum);

The ';' at the end of the "for" loop stops the next line from being ran.

And the indentation of your project is bad - The snip I took from above is a bit too "gotofail" for me. It should look more like this:

Code:

for(i=0;i<n;++i)
{
  sum=sum+i/n; // Another Hint: This does not compute the sum nor average of your ARRAY
}

printf("Average is %d\n", sum);

Originally Posted by internalbalance

Mathmatically no it is not...and it seems like common sense now looking back. However, everything that I thought would make sense hasn't worked very well so far.

Mathematically, assuming the result had to be stored as an integer, what would you expect 3/4 to produce? That is common sense (the only decision needed is whether to round up or down). In C, the literal value 3 has type int, the literal value 4 has type int, so 3/4 is an int.

It is not exactly common sense to expect to produce an integral value that is equal to 0.75.

Originally Posted by Click_here

Code:

for(i=0;i<n;++i);
        sum=sum+i/n;
        printf("Average is %d\n", sum);

The ';' at the end of the "for" loop stops the next line from being ran.

And the indentation of your project is bad - The snip I took from above is a bit too "gotofail" for me. It should look more like this:

Code:

for(i=0;i<n;++i)
{
  sum=sum+i/n; // Another Hint: This does not compute the sum nor average of your ARRAY
}

printf("Average is %d\n", sum);

Agreed. My company's coding standards prohibit any use of loops or if statements without braces - one-line statements always get them, for exactly the reason that it makes the "goto fail" kind of error less likely.

Agreed. My company's coding standards prohibit any use of loops or if statements without braces - one-line statements always get them, for exactly the reason that it makes the "goto fail" kind of error less likely.

[A bit off topic, but...]
One bug that changed the way that I program that was along the lines of this sort of thing -> In embedded programming it is common to see this sort of statement waiting for a register to set/clear

Code:

while (!REGISTER.bitx);

Which is fine - Until you put it into a do/while loop that is very long

Code:

do
{
   /* Lots of sending/receiveing for a few pages */
  /* Lots of indentating... */
  {
    {
    }
  }

  { 
    { 
    }
  }
  while(something);

  /* Lots of indentating... */
  {
    {
    }
  }

  { 
    { 
    }
  }
}
while (somethingelse);

Needless to say that it was heartbreaking to see that the code was... well... confused.

Now I always type out my wait while statements like this:

Code:

while (!REGISTER.bitx)
{
  continue;
}

It is very easy to see that it is not the end of a do/while statement.

Maybe the real problem was that the loop body was too long

Originally Posted by internalbalance

How can I get the values to add up? Example if I enter, 1 2 and 4 my 'n' would be 3, so obviously I would divide by that to get my average correct? However, how can I add the 1 2 and 4??

Let's call 'n' the number of items and lets call 'sum' the current running total. Then to add 1, 2 and 4 into sum, you do something like this in a loop

Code:

sum = 0;
n = 0;

// User enters 1 and you store it in 'x'
sum += x;
n++;

// User enters 2 and you store it in 'x'
sum += x;
n++;

// User enters 4 and you store it in 'x'
sum += x;
n++;