## Section 3.4

TheforStatement

WE TURN IN THIS SECTION to another type of loop, the

forstatement. Anyforloop is equivalent to somewhileloop, so the language doesn't get any additional power by havingforstatements. But for a certain type of problem, aforloop can be easier to construct and easier to read than the correspondingwhileloop. It's quite possible that in real programs,forloops actually outnumberwhileloops.

The

forstatement makes a common type of while loop easier to write. Many while loops have the general form:initializationwhile (continuation-condition) {statementsupdate}For example, consider this example, copied from an example in Section 2:

years = 0; //initializethe variable years while ( years < 5 ) { //conditionfor continuing loop interest = principal * rate; // principal += interest; // do threestatementsSystem.out.println(principal); // years++; //updatethe value of the variable, years }This loop can be written as the following equivalent

forstatement:for ( years = 0; years < 5; years++ ) { interest = principal * rate; principal += interest; System.out.println(principal); }The initialization, continuation condition, and updating have all been combined in the first line of the

The formal syntax of theforloop. This keeps everything involved in the "control" of the loop in one place, which helps makes the loop easier to read and understand. Theforloop is executed in exactly the same way as the original code: The initialization part is executed once, before the loop begins. The continuation condition is executed before each execution of the loop, and the loop ends when this condition isfalse. The update part is executed at the end of each execution of the loop, just before jumping back to check the condition.forstatement is as follows:for (initialization;continuation-condition;update)statementor, using a block statement:

for (initialization;continuation-condition;update) {statements}The

continuation-conditionmust be a boolean-valued expression. Theinitializationcan be any expression, as can theupdate. Any of the three can be empty. If the continuation condition is empty, it is treated as if it were "true," so the loop will be repeated forever or until it ends for some other reason, such as abreakstatement. (Some people like to begin an infinite loop with "for (;;)" instead of "while (true)".)Usually, the initialization part of a

forstatement assigns a value to some variable, and the update changes the value of that variable with an assignment statement or with an increment or decrement operation. The value of the variable is tested in the continuation condition, and the loop ends when this condition evaluates tofalse. A variable used in this way is called a loop control variable. In theforstatement given above, the loop control variable isyears.Certainly, the most common type of

forloop is the counting loop, where a loop control variable takes on all integer values between some minimum and some maximum value. A counting loop has the formfor (variable=min;variable<=max;variable++ ) {statements}where

minandmaxare integer-valued expressions (usually constants). Thevariabletakes on the valuesmin,min+1,min+2, ...,max. The value of the loop control variable is often used in the body of the loop. Theforloop at the beginning of this section is a counting loop in which the loop control variable,years, takes on the values 1, 2, 3, 4, 5. Here is an even simpler example, in which the numbers 1, 2, ..., 10 are displayed on standard output:for ( N = 1 ; N <= 10 ; N++ ) System.out.println( N );For various reasons, Java programmers like to start counting at 0 instead of 1, and they tend to use a "

<" in the condition, rather than a "<=". The following variation of the above loop prints out the ten numbers 0, 1, 2, ..., 9:for ( N = 0 ; N < 10 ; N++ ) System.out.println( N );Using

<instead of<=in the test, or vice versa, is a common source of off-by-one errors in programs. You should always stop and think, do I want the final value to be processed or not?It's easy to count down from 10 to 1 instead of counting up. Just start with 10, decrement the loop control variable instead of incrementing it, and continue as long as the variable is greater than or equal to one.

for ( N = 10 ; N >= 1 ; N-- ) System.out.println( N );Now, in fact, the official syntax of a

forstatemenent actually allows both the initialization part and the update part to consist of several expressions, separated by commas. So we can even count up from 1 to 10 and count down from 10 to 1 at the same time!for ( i=1, j=10; i <= 10; i++, j-- ) { TextIO.put(i,5); // Output i in a 5-character wide column. TextIO.putln(j,5); // Output j in a 5-character column // and end the line. }As a final example, let's say that we want to use a

forloop print out just the even numbers between 2 and 20, that is: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. There are several ways to do this. Just to show how even a very simple problem can be solved in many ways, here are four different solutions (three of which would get full credit):(1) // There are 10 numbers to print. // Use a for loop to count 1, 2, // ..., 10. The numbers we want // to print are 2*1, 2*2, ... 2*10. for (N = 1; N <= 10; N++) { System.out.println( 2*N ); } (2) // Use a for loop that counts // 2, 4, ..., 20 directly by // adding 2 to N each time through // the loop. for (N = 2; N <= 20; N = N + 2) { System.out.println( N ); } (3) // Count off all the numbers // 2, 3, 4, ..., 19, 20, but // only print out the numbers // that are even. for (N = 2; N <= 20; N++) { if ( N % 2 == 0 ) // is N even? System.out.println( N ); } (4) // Irritate the professor with // a solution that follows the // letter of this silly assignment // while making fun of it. for (N = 1; N <= 1; N++) { System.out.print("2 4 6 8 10 12 "); System.out.println("14 16 18 20"); }Perhaps it is worth stressing one more time that a

forstatement, like any statement, never occurs on its own in a real program. A statement must be inside themainroutine of a program or inside some other subroutine. And that subroutine must be defined inside a class. I should also remind you that every variable must be declared before it can be used, and that includes the loop control variable in aforstatement. In all the examples that you have seen so far in this section, the loop control variables should be declared to be of typeint. It is not required that a loop control variable be an integer. Here, for example, is aforloop in which the variable,ch, is of typechar:// Print out the alphabet on one line of output. char ch; // The loop control variable; // one of the letters to be printed. for ( char ch = 'A'; ch <= 'Z'; ch++ ) System.out.print(ch); System.out.println();

Let's look at a less trivial problem that can be solved with a

forloop. IfNandDare positive integers, we say thatDis a divisor ofNif the remainder whenDis divided intoNis zero. (Equivalently, we could say thatNis an even multiple ofD.) In terms of Java programming,Dis a divisor ofNifD % Nis zero.Let's write a program that inputs a positive integer,

N, from the user and computes how many different divisorsNhas. The numbers that could possibly be divisors ofNare 1, 2, ...,N. To compute the number of divisors ofN, we can just test each possible divisor ofNand count the ones that actually do divideNevenly. In pseudocode, the algorithm takes the formGet a positive integer, N, from the user Let divisorCount = 0 for each number, testDivisor, in the range from 1 to N: if testDivisor is a divisor of N: Count it by adding 1 to divisorCount Output the countThis algorithm displays a common programming pattern that is used when some, but not all, of a sequence of items are to be processed. The general pattern is

for each item in the sequence: if the item passes the test: process itThe

forloop in our divisor-counting algorithm can be translated into Java code asfor (testDivisor = 1; testDivisor <= N; testDivisor++) { if ( N % testDivisor == 0 ) divisorCount++; }On a modern computer, this loop can be executed very quickly. It is not impossible to run it even for the largest legal

intvalue, 2147483647. (If you wanted to run it for even larger values, you could use variables of typelongrather thanint.) However, it does take a noticeable amount of time for very large numbers. So when I implemented this algorithm, I decided to output a period every time the computer has tested one million possible divisors. In the improved version of the program, there are two types of counting going on. We have to count the number of divisors and we also have to count the number of possible divisors that have been tested. So the program needs two counters. When the second counter reaches 1000000, we output a '.' and reset the counter to zero so that we can start counting the next group of one million. Reverting to pseudocode, the algorithm now looks likeGet a positive integer, N, from the user Let divisorCount = 0 // Number of divisors found. Let numberTested = 0 // Number of possible divisors tested // since the last period was output. for each number, testDivisor, in the range from 1 to N: if testDivisor is a divisor of N: Count it by adding 1 to divisorCount Add 1 to numberTested if numberTested is 1000000: print out a '.' Let numberTested = 0 Output the countFinally, we can translate the algorithm into a complete Java program. Here it is, followed by an applet that simulates it:

public class CountDivisors { /* This program reads a positive integer from the user. It counts how many divisors that number has, and then it prints the result. */ public static void main(String[] args) { int N; // A positive integer entered by the user. // Divisors of this number will be counted. int testDivisor; // A number between 1 and N that is a // possible divisor of N. int divisorCount; // Number of divisors of N that have been found. int numberTested; // Used to count how many possible divisors // of N have been tested. When the number // reaches 1000000, a period is output and // the value of numberTested is reset to zero. /* Get a positive integer from the user. */ while (true) { TextIO.put("Enter a positive integer: "); N = TextIO.getlnInt(); if (N > 0) break; TextIO.putln("That number is not positive. Please try again."); } /* Count the divisors, printing a "." after every 1000000 tests. */ divisorCount = 0; numberTested = 0; for (testDivisor = 1; testDivisor <= N; testDivisor++) { if ( N % testDivisor == 0 ) divisorCount++; numberTested++; if (numberTested == 1000000) { TextIO.put('.'); numberTested = 0; } } /* Display the result. */ TextIO.putln(); TextIO.putln("The number of divisors of " + N + " is " + divisorCount); } // end main() } // end class CountDivisors

## Nested Loops

Control structures in Java are statements that contain statements. In particular, control structures can contain control structures. You've already seen several examples of

ifstatements inside loops, but any combination of one control structure inside another is possible. We say that one structure is nested inside another. You can even have multiple levels of nesting, such as awhileloop inside anifstatement inside anotherwhileloop. The syntax of Java does not set a limit on the number of levels of nesting. As a practical matter, though, it's difficult to understand a program that has more than a few levels of nesting.Nested

forloops arise naturally in many algorithms, and it is important to understand how they work. Let's look at a couple of examples. First, consider the problem of printing out a multiplication table like this one:1 2 3 4 5 6 7 8 9 10 11 12 2 4 6 8 10 12 14 16 18 20 22 24 3 6 9 12 15 18 21 24 27 30 33 36 4 8 12 16 20 24 28 32 36 40 44 48 5 10 15 20 25 30 35 40 45 50 55 60 6 12 18 24 30 36 42 48 54 60 66 72 7 14 21 28 35 42 49 56 63 70 77 84 8 16 24 32 40 48 56 64 72 80 88 96 9 18 27 36 45 54 63 72 81 90 99 108 10 20 30 40 50 60 70 80 90 100 110 120 11 22 33 44 55 66 77 88 99 110 121 132 12 24 36 48 60 72 84 96 108 120 132 144The data in the table are arranged into 12 rows and 12 columns. The process of printing them out can be expressed in a pseudocode algorithm as

for each rowNumber = 1, 2, 3, ..., 12: Print the first twelve multiples of rowNumber on one line Output a carriage returnThe first step in the

forloop can itself be expressed as aforloop:for N = 1, 2, 3, ..., 12: Print N * rowNumberso a refined algorithm for printing the table has one

forloop nested inside another:for each rowNumber = 1, 2, 3, ..., 12: for N = 1, 2, 3, ..., 12: Print N * rowNumber Output a carriage returnAssuming that

rowNumberandNhave been declared to be variables of typeint, this can be expressed in Java asfor ( rowNumber = 1; rowNumber <= 12; rowNumber++ ) { for ( N = 1; N <= 12; N++ ) { // print in 4-character columns TextIO.put( N * rowNumber, 4 ); } TextIO.putln(); }This section has been weighed down with lots of examples of numerical processing. For our final example, let's do some text processing. Consider the problem of finding which of the 26 letters of the alphabet occur in a given string. For example, the letters that occur in "Hello World" are D, E, H, L, O, R, and W. More specifically, we will write a program that will list all the letters contained in a string and will also count the number of different letters. The string will be input by the user. Let's start with a pseudocode algorithm for the program.

Ask the user to input a string Read the response into a variable, str Let count = 0 (for counting the number of different letters) for each letter of the alphabet: if the letter occurs in str: Print the letter Add 1 to count Output the countSince we want to process the entire line of text that is entered by the user, we'll use

TextIO.getln()to read it. The line that reads "for each letter of the alphabet" can be expressed as "for (letter='A'; letter<='Z'; letter++)". But the body of thisforloop needs more thought. How do we check whether the given letter,letter, occurs instr? One idea is to look at each letter in the string in turn, and check whether that letter is equal toletter. We can get thei-th character ofstrwith the function callstr.charAt(i), whereiranges from 0 tostr.length() - 1. One more difficulty: A letter such as 'A' can occur instrin either upper or lower case, 'A' or 'a'. We have to check for both of these. But we can avoid this difficulty by convertingstrto upper case before processing it. Then, we only have to check for the upper case letter. We can now flesh out the algorithm fully. Note the use ofbreakin the nestedforloop. It is required to avoid printing or counting a given letter more than once. Thebreakstatement breaks out of the innerforloop, but not the outerforloop. Upon executing thebreak, the computer continues the outer loop with the next value ofletter.Ask the user to input a string Read the response into a variable, str Convert str to upper case Let count = 0 for letter = 'A', 'B', ..., 'Z': for i = 0, 1, ..., str.length()-1: if letter == str.charAt(i): Print letter Add 1 to count break // jump out of the loop Output the countHere is the complete program and an applet to simulate it:

public class ListLetters { /* This program reads a line of text entered by the user. It prints a list of the letters that occur in the text, and it reports how many different letters were found. */ public static void main(String[] args) { String str; // Line of text entered by the user. int count; // Number of different letters found in str. char letter; // A letter of the alphabet. TextIO.putln("Please type in a line of text."); str = TextIO.getln(); str = str.toUpperCase(); count = 0; TextIO.putln("Your input contains the following letters:"); TextIO.putln(); TextIO.put(" "); for ( letter = 'A'; letter <= 'Z'; letter++ ) { int i; // Position of a character in str. for ( i = 0; i < str.length(); i++ ) { if ( letter == str.charAt(i) ) { TextIO.put(letter); TextIO.put(' '); count++; break; } } } TextIO.putln(); TextIO.putln(); TextIO.putln("There were " + count + " different letters."); } // end main() } // end class ListLetters

In fact, there is an easier way to determine whether a given letter occurs in a string,

str. The built-in functionstr.indexOf(letter)will return-1ifletterdoesnotoccur in the string. It returns a number greater than or equal to zero if it does occur. So, we could check whetherletteroccurs instrsimply by checking "if (str.indexOf(letter) >= 0)". If we used this technique in the above program, we wouldn't need a nestedforloop. This gives you preview of how subroutines can be used to deal with complexity.

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