AN INTRODUCTION TO C AND GUI PROGRAMMING16
…you have:
float a;
This tells the compiler that a is now a oating-point value, rather than an integer. Compile
and run your program. What happens?
Oops! That doesn’t look right, does it? What has happened is that, while the maths is still
all correct, the printf statement is now wrong; you’re telling it to print a, which is a oating-
point value, as a decimal integer. To x that, change the rst %d in the printf function to %f,
which is the format specier for a oating-point number, like this:
printf ("The sum of adding %f and %d is %d\n", a, b, c);
That should produce something a lot more sensible when you run it. This is an important
lesson about C: it will do exactly what you tell it to, even if it makes no sense. You told it to
show you a oating-point number as if it were a decimal integer, and the compiler assumed
that was what you wanted, even though the result was nonsense.
When you’re working with variables, always keep track of what values you’re putting in what
types, as it’s easy to introduce errors by assuming a variable is of one type when it’s actually
another. One common error is to put the results of a calculation on oating-point values into
an integer.
Try this: make
b a oat as well (not forgetting to change its format specier in the printf),
but leave c as an int, and set the two oats to values with decimal points, like this:
float a;
float b = 3.641;
int c;
a = 2.897;
c = a + b;
printf ("The sum of adding %f and %f is %d\n", a, b, c);
You’ll see a result like:
The sum of adding 2.897000 and 3.641000 is 6
6? That’s not right! But it’s exactly what you’ve asked for. What the compiler did was to add
the two oating-point values together, and got the answer 6.538, but you then told the compiler
to put that into c, an integer variable. So the compiler just threw away everything after the
decimal point! If you change
c to a oat, and change the nal %d to %f, you’ll nd it gives the
correct answer.
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