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C# has a set of rules for working out the order in which to evaluate the components of an expression. It does not necessarily work from left to right, because some operators have a higher precedence than others. For example, imagine evaluating this: 1.0 + 3.0 / 4.0 from left to right. Start with 1.0, add 3.0 which gets you to 4.0, and then divide by 4.0 the result would be 1.0. But the conventional rules of arithmetic mean the result should be one and three quarters. And that s just what C# produces the result is 1.75. The division is performed before the addition, because division has higher precedence than division. Some groups of operators have equal precedence. For example, multiplication and division have equal precedence. When expressions contain multiple operations with the same precedence, mathematical operations are evaluated from left to right. So 10.0 / 2.0 * 5.0 evaluates to 25.0. But parentheses trump precedence, so 10.0 / (2.0 * 5.0) evaluates to 1.0. Some programming books go into great depths about all the details of precedence, but it makes for exceptionally tedious reading C# has 15 different levels of precedence. The details are important for compiler writers, but of limited value for developers code that relies heavily on precedence can be hard to read. Using parentheses to make evaluation order explicit can often improve clarity. But if you would like the gory details, you can find them at http://msdn.microsoft.com/en-us/library/aa691323.
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An expression s type matters. The examples we just looked at involve numbers or numeric variables, and are of type double or int. Expressions can be of any type, though. For example, ("Hello, " + "world") is an expression of type string. If you wrote an assignment statement that tried to assign that expression into a variable of type double, the compiler would complain it insists that expressions are either of the same type as the variable, or of a type that is implicitly convertible to the variable s type. Implicit conversions exist for numeric types when the conversion won t lose information for example, a double can represent any value that an int can, so you re allowed to assign an integer expression into a double variable. But attempting the opposite would cause a compiler error, because doubles can be larger than the highest int, and they can also contain fractional parts that would be lost. If you don t mind the loss of information, you can put a cast in front of the expression:
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int approxKmPerHour = (int) kmPerHour;
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This casts the kmPerHour (which we declared earlier as a double) to an int, meaning it ll force the value to fit in an integer, possibly losing information in the process.
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A variable doesn t have to be stuck with its initial value for its whole life. We can assign new values at any time.
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The previous section showed how to assign an expression s value into a newly declared variable:
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double kmPerHour = kmTravelled / (elapsedSeconds / (60 * 60));
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If at some later stage in the program s execution new information becomes available, we could assign a new value into the kmPerHour variable assignment statements aren t required to declare new variables, and can assign into existing ones:
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kmPerHour = updateKmTravelled / (updatedElapsedSeconds / (60 * 60));
This overwrites the existing value in the kmPerHour variable. C# offers some specialized assignment statements that can make for slightly more succinct code. For example, suppose you wanted to add the car s latest lap time to the variable holding the total elapsed time. You could write this:
elapsedSeconds = elapsedSeconds + latestLapTime;
This evaluates the expression on the righthand side, and assigns the result to the variable specified on the lefthand side. However, this process of adding a value to a variable is so common that there s a special syntax for it:
elapsedSeconds += latestLapTime;
This has exactly the same effect as the previous expression. There are equivalents for the other mathematical operators, so -= means to subtract the expression on the right from the variable on the left, *= does the same for multiplication, and so on.
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