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// A promotion surprise! using System; class PromDemo { static void Main() { byte b; b = 10; b = (byte) (b * b); // cast needed!! ConsoleWriteLine("b: "+ b); } }

Determine the equation of the line with slope 3 that passes through the point (2, 1)

(v ) (v ) k - 1 ( s2 - y j ) M j , k2 -1 ( s2 ) ( y j + k2 - s2 ) M j +1, k2 -1 ( s2 ) ) M (j vk2 ( s2 ) = 2 + , k2 - v - 1 y j + k2 -1 - yi y j + k2 - y j +1

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Somewhat counterintuitively, a cast to byte is needed when assigning b * b back to b! The reason is because in b * b, the value of b is promoted to int when the expression is evaluated Thus, b * b results in an int value, which cannot be assigned to a byte variable without a cast Keep this in mind if you get unexpected type-incompatibility error messages on expressions that would otherwise seem perfectly correct This same sort of situation also occurs when performing operations on chars For example, in the following fragment, the cast back to char is needed because of the promotion of ch1 and ch2 to int within the expression

char ch1 = 'a', ch2 = 'b'; ch1 = (char) (ch1 + ch2);

(546)

Without the cast, the result of adding ch1 to ch2 would be int, which can t be assigned to a char Type promotions also occur when a unary operation, such as the unary , takes place For the unary operations, operands smaller than int (byte, sbyte, short, and ushort) are promoted to int Also, a char operand is converted to int Furthermore, if a uint value is negated, it is promoted to long

SOLUTION Let (x, y) be a variable point on the line Then we can use that variable point together with (2, 1) to calculate the slope: m= y 1 x 2

A cast can be applied to a specific portion of a larger expression This gives you fine-grained control over the way type conversions occur when an expression is evaluated For example, consider the following program It displays the square roots of the numbers from 1 to 10 It also displays the whole number portion and the fractional part of each result, separately To do so, it uses a cast to convert the result of MathSqrt( ) to int

[ S1 (f2 , s2 )] = [ N (f2 )] [ p] [ M ( s2 )]

// Using casts in an expression using System; class CastExpr { static void Main() { double n; for(n = 10; n <= 10; n++) {

3:

where S1 (f 21 , s21 ) S f , s ( ) [ S1 (f2 , s2 )] = 1 22 21 M S1 (f 2 n , s21 ) S1 (f 21 , s22 ) L S1 (f 21 , s2 m ) S1 (f 22 , s22 ) L S1 (f 22 , s2 m ) M O M S1 (f 2 n , s22 ) L S1 (f 2 n , s2 m ) n m

On the other hand, we are given that the slope is m = 3 We may equate the two expressions for slope to obtain 3= y 1 x 2 ( )

ConsoleWriteLine("The square root of {0} is {1}", n, MathSqrt(n));

ConsoleWriteLine("Whole number part: {0}" , (int) MathSqrt(n)); ConsoleWriteLine("Fractional part: {0}", MathSqrt(n) - (int) MathSqrt(n) ); ConsoleWriteLine(); } } }

(547)

Here is the output from the program:

This may be simpli ed to y = 3x 5 Math Note: The form y = 3x 5 for the equation of a line is called the slopeintercept form The slope is 3 and the line passes through (0, 5) (its y-intercept) Math Note: Equation ( ) may be rewritten as y 1 = 3(x 2) In general, the line with slope m that passes through the point (x0 , y0 ) can be written as y y0 = m(x x0 ) This is called the point-slope form of the equation of a line

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