In my article "C: Hello World on Linux", we saw how to write a very simple C program, and how to compile and run it on Linux. In this article, we'll learn how to work with numbers in C, and also how to accept input and print output. I like to use Linux when writing or teaching C/C++, but you can just as well follow along if you're on Windows (e.g. by using Microsoft Visual C++). If you're going to use Linux, you can use your distribution's text editor, or else something like vi (check out my articles, "Linux: vi Essentials" and "Linux: Navigating Text with vi", if you feel you need a refresher on vi).
So, we'll begin with the following basic template:
#include <stdio.h>
int main(int argc, char ** argv)
{
return 0;
}
Including the stdio.h header file gives us access to the input/output (I/O) functions that we'll need. The first of these is printf(), which we've already seen in my original article, "C: Hello World on Linux".
In our program, we're first going to ask the user how many numbers he's going to enter. Let's say he enters 5. In that case, we'll ask him for five numbers, and then add them up, and then compute the average.
Before the return statement, add the following:
printf("How many numbers do you want to average?\n");
The \n translates to a newline.
We then declare a variable called count, of type int (integer), so that we can store how many numbers the user needs to enter:
int count = 0;
Just a note... in C/C++ it is important to always initialise your variables with a value. If you don't, they will contain random garbage data, and using them before they're assigned might lead to unpredictable results.
Now, we can then use the scanf() function to accept input, as follows:
scanf("%d", &count);
This looks pretty scary, doesn't it? The first parameter is a format string; it indicates what kind of data you're trying to accept. In this case, %d means we are expecting an integer. The second parameter indicates the variable where we'll store the input. By preceding the count variable with the & symbol, we're telling scanf() that the input should go to the location in memory where count resides. This might sound a bit weird until we talk about pointers and references (because it is), but just remember to include that & when accepting input, and you'll be fine.
Next, we need a few new variables:
int i = 0;
int number = 0;
int total = 0;
We'll accept each number into the number variable, after which we'll add it to total. The i variable will be used to track how many numbers have been accepted, so that we know when to stop.
Next, we can write a for loop to repeatedly accept numbers as input, until we have reached the number of numbers we were expecting:
for (i = 0; i < count; i++)
{
}
The code inside the curly brackets (which we will write shortly) will execute as many times as the count variable is set to. If count is 5, then it will execute 5 times. If you look at the first line of the for loop, we're starting with i set to zero, and we're telling it to continue executing as long as i is less than count. Each time the loop's code executes, i is increased by 1 (that's the i++ bit), until the i < count condition becomes true and the loop ends.
Sidenote: if you're coming from some other language, you might be wondering why I didn't declare i in the for loop. That's because some versions of C don't allow it (although it's fine in C++), so you may end up with an error like this:
error: ‘for’ loop initial declarations are only allowed in C99 mode
Let's add some code inside the for loop to accept numbers as input:
printf("Enter number %d of %d: ", i, count);
scanf("%d", &number);
total = total + number;
In the first line, we can see the use of the %d (integer) placeholder within printf(). When you run the program, the placeholders are replaced with the other parameters, i.e. the first %d is replaced with the value of i, and the second one is replaced with the value of count.
In the second line, we're accepting a number as input, and storing it in the number variable (remember the &). Finally, in the third line, we add number to total. There's actually a more concise and efficient way of adding a value to a variable, which is the following:
total += number;
After the end of the for loop, we can now calculate the average by dividing total by the number of items (count):
float average = total / count;
printf("Average: %f\n", average);
The float data type allows us to store numbers that are not integers (whole numbers), and which might have some decimal value. When using floats with printf(), we need to use the %f placeholder, instead of the %d that is used for integers.
Here's the code we have so far, with syntax highlighting courtesy of vim:
From the Linux terminal, the following command will compile your code to an executable called average:
gcc average.c -o average
...and the following command executes it:
./average
Let's test the program to see that it works correctly:
Hmmm, wait just a minute. 1 + 2 + 4 gives us 7, and divided by 3, that doesn't give us exactly 2 (it's actually 2.333333.....). The problem is in the following line of code:
float average = total / count;
We're dividing an integer by another integer, and that gives us back an integer - losing any digits after the decimal point. To get this to work correctly, we can convert count to a float as follows:
float average = total / (float) count;
This is called type casting: you put a data type in brackets before a variable to indicate that it should be interpreted as the specified data type. When we recompile and run the program, we can see that it now behaves as we originally expected:
Nice! :)
This article showed how to write a simple program - one that calculates the average of a set of numbers - in C. This simple program revealed several features of the C language, including input/output (via printf() and scanf()), placeholders for format strings (%d and %f), int and float variables, for loops, and type casting.
I hope you found this interesting. Stick around for more articles! :)
