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Description of the program code

Apart from the line in red above, everything else should be straight forward : two int variables are declared ( inches and feet ). The value 5 is stored in the feet variable when it is declared, whereas the value of the inches variable is undefined when it is declared.

The line in red multiplies the value of the feet variable by 12 and stores the result in the inches variable. The sign for multiplication in programming languages is not x like in standard maths, but * If you read the line in your mind it says "inches equals feet times 12" which makes sense to me. With calculations anything on the left of the equals sign (=) is where the result is stored, anything on the right is the calculation to create that result. The output from the program is shown below :

There are 60 inches in 5 feet

Mathematical operators in C

Operator	Symbol	Eg	Desc
Addition	+	a + 1	Adding two values together
Subtraction	-	a - 1	Subtract right value from left
Multiplication	*	a * 1	Multiply two values together
Division	/	a / 2	Divide the left value by the right
Modulus	%	a % 2	The remainder left after a / 2
Increment	++	a++	Add 1 to the value of a
Decrement	--	a--	Subtrace 1 from the value of a

Most of the above table is straight forward apart from the multiplication and division symbols. The last three entries ( modulus, increment and decrement ) may need explaining.

The modulus of a division is the remainder left over and is represented in C by the percentage sign. For example 5 % 2 = 1 ( 5 / 2 = 2 remainder 1 ).. or .. 11 % 3 = 2

Note : When programming and using division or modulus you have to always be aware that dividing a number by 0 ( zero ) will cause a runtime error .. This is the worse kind of error, as the compiler will compile the code without any problems at all, but when the program is running it will crash with a divide overflow error. If you think about it, it is impossible to divide any number by zero, eg 10 / 0 = ? ( how many 0's in 10 ? ) impossible.

The increment and decrement operators are used as an efficient way of adding or subtracting 1 from a variable. The following simple program and output shows the increment operator in action :

#include <stdio.h>

void main(void)
{

int num=25;
num++;
printf("The value of the variable num is %d",num);

}

The value of the variable num is 26

A quick question

What will be the value of the variable num after this line of code ?

num=1+2*3;

The answer is not 9, it is actually 7. If you don't believe me write a small program to prove it, or use a calculator and enter 1 + 2 x 3 = . The answer both the program and the calculator throw out will be 7 not 9. Why ? The answer lies with BODMAS ( Brackets, Ordinals, Division, Multiplication, Addition, Subtraction ) This order is the order that is used to find the value of equations both in computer programs and maths. First you work out the sums of anything in brackets, then you calculate the values of ordinals ( powers ), then divisions, then multiplications, then additions, and finally last of all subtractions.

If you thought the answer was 9 then what you have done is work from left to right through the equation ( 1+2 = 3 * 3 = 9 ) If you follow the BODMAS rule then the 2 x 3 is worked out first ( because multiplication comes before addition BODMAS ). This gives the result 6, then finally the 1 is added giving the answer 7, which is correct.

You can force the answer 9 by using brackets because they are worked out before everything else according to the BODMAS rule. eg

num=(1+2)*3;

This forces the 1+2 calculation to be worked out first because it's in brackets, then that value is multiplied by the final 3, this time giving the answer 9.

Program 5	Simple conversion	Source code - prog5.c
From now on things start to get a bit more interesting, the programs are still small, but after these next few you should start to see some "use" for learning how to program. You are probably thinking by now that it seems like a lot of taxing stuff to do very little, which would be true considering what programs have been done so far, but the best is yet to come. It's just one of those things that the boring bits have to be understood first :-( The program here shows how to multiply two numbers together and store the result. It would be far more useful if the user of the program could enter the numbers and the program produce the result, but input does not come in until program 8, so it will have to make do by setting the values in the program code for now :-( This is known as "hard-coding"