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// Demonstrate the % operator using System; class ModDemo { static void Main() { int iresult, irem; double dresult, drem; iresult = 10 / 3; irem = 10 % 3; dresult = 100 / 30; drem = 100 % 30; ConsoleWriteLine("Result and remainder of 10 / 3: " + iresult + " " + irem); ConsoleWriteLine("Result and remainder of 100 / 30: " + dresult + " " + drem); } }

Substituting Eq (64) in Eq (61) and eliminating N1 0 = - L0 + F cosa - mF Substituting gives the external load 2A + B L0 = F cosa - m sin a B A sin a - mN2 B (66)

The output from the program is shown here:

Result and remainder of 10 / 3: 3 1 Result and remainder of 100 / 30: 333333333333333 1

x = 2/2

Solving for the force normal to the cam F= L0 2A + B cosa - m B sin a (67)

Introduced in 2, the ++ and the are the increment and decrement operators As you will see, they have some special properties that make them quite interesting Let s begin by reviewing precisely what the increment and decrement operators do The increment operator adds 1 to its operand, and the decrement operator subtracts 1 Therefore,

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x = x + 1;

x++;

The normal force F is a maximum (equals in nity) which means that the follower will jam in its guide when the denominator of Eq (67) equals zero Therefore, cosa m - m 2A + B sin am = 0 B

Sin 1 (x 2 + x)

x = x - 1;

x--;

The maximum pressure angle without locking the follower in its guide is am = tan -1 B m ( 2 A + B) (68)

4:

Understand, however, that in the increment or decrement forms, x is evaluated only once, not twice This can improve efficiency in some cases Both the increment and decrement operators can either precede (prefix) or follow (postfix) the operand For example

x =1/3

Let us substitute some trial values to compare the magnitude of the results If we let A = B and assume the values for the coef cient of friction of bronze on steel to be m ( kinetic) = 010 and m (static) = 015 Substituting in Eq (68), we nd the maximum pressure angle for each condition: am = tan -1 am = tan -1 B = 73 degrees for m = 010 ( 010)(2 B + B) B = 66 degrees for m = 015 ( 015)(2 B + B)

x = x + 1;

In the foregoing example, there is no difference whether the increment is applied as a prefix or a postfix However, when an increment or decrement is used as part of a larger expression, there is an important difference When an increment or decrement operator precedes its operand, the result of the operation is the value of the operand after the increment If the operator follows its operand, the result of the operation is the value of the operand before the increment Consider the following:

.

x = 10; y = ++x;

x = 3

In this case, y will be set to 11 This is because x is first incremented and then its value is returned However, if the code is written as

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