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if statementif (condition) { statements}• Nested ifif (condition1) { statements1 if (condition2) { statements2 } statements3}
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Nested if• Sometimes possible to simplify nested ifif (condition1) { if (condition2) { statements }}• Same asif ((condition1) && (condition2)) { statements}
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Exampleclass exampleIf { public static void main(String arg[]) { int x=10, y, z; if ((x%2)==0) { System.out.println(x + “ is even.”); if ((x%3)==0) { System.out.println(x + “ is a
multiple of 6.”); y = x/6; } z = x%6; } }}
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if-elseif (condition) { statements1}else { statements2}• if within elseif (condition1) { statements1 }else { if (condition2) { statements2 } statements3}
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if-else if-else if-…-elseif (condition1) { statements1}else if (condition2) { statements2}else if (condition3) { statements3}…else { statementsn}
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Exampleclass greetings { public static void main(String arg[]) { int hour = 3; if ((hour >= 0) && (hour < 12)) { System.out.println(“Good Morning!”); } else if ((hour >= 12) && (hour < 18)) { System.out.println(“Good Afternoon!”); } else if ((hour >=18) && (hour < 24)) { System.out.println(“Good Evening!”); } else { System.out.println(“Bad time!”); } }}
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Announcements
• Wednesday is a holy day– No class– No labs
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Conditional assignmentif ((x%2)==0) { y = x/2;}else { y = (x+1)/2;}• Same asy = ((x%2)==0) ? x/2 : (x+1)/2;
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Integer part and absolute value
class integerPart { public static void main(String arg[]) { double x = -3.7; int ipart; double aval; ipart = ((x >= 0) || ((int)x==x)) ?
(int)x : (int)(x-1); aval = (x >= 0) ? x : -x; System.out.println(“Integer part of ” +
x + “ is ” + ipart + “. Absolute value is ” + aval + “.”);
}}
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Integer part and absolute value
class integerPartAlternate { public static void main(String arg[]) { double x = -3.7; int ipart; double aval; ipart = ((x < 0) && ((int)x!=x)) ? (int)
(x-1) : (int)x; aval = (x < 0) ? -x : x; System.out.println(“Integer part of ” +
x + “ is ” + ipart + “. Absolute value is ” + aval + “.”);
}}
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Sorting three numbersclass sortThree { public static void main(String arg[]) { int x = 2, y = 5, z = 1; int max, mid, min; if ((x > y) && (x > z)) { max = x; if (y > z) { mid = y; min = z; } else { mid = z; min = y; } } // next slide
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Sorting three numbers else { if (y > z) { max = y; if (x > z) { mid = x; min = z; } else { mid = z; min = x; } } else { // the remaining two permutations} } // end else } // end main} // end class
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Loops• Needed in problems that require solving
the same subproblem over and over– Computing the sum of the first 100 natural
numbers and putting the result in y– Algorithm:y = 1;y = y + 2;y = y + 3;…y = y + 100;– Cannot write 99 such additions: use a loop
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whilewhile (condition) { statements}• Can put anything in “statements”
– The entire construct is called a while loop– statements are executed until condition is
true– Even before executing it the first time
condition is evaluated• A while loop may not execute even once
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Exampleclass justAnExample { public static void main(String arg[]) { int x = 5; int y = 0; while (x < 10) { y--; x++; } System.out.println(y); }}
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Exampleclass justAnExample { public static void main(String arg[]) { int x = 15; int y = 0; while (x < 10) { y--; x++; } System.out.println(y); }}
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Sum of natural numbersclass naturalSum { public static void main(String arg[]) { int n = 2; int y = 1; while (n <= 100) { y += n; n++; } System.out.println(“Sum of the first ” +
(n-1) + “ natural numbers is ” + y); }}
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Sum of natural numbersclass naturalSumAnotherWay { public static void main(String arg[]) { int n = 99; int m = n+1; int y = 100; while (n > 0) { y += n; n--; } System.out.println(“Sum of the first ” +
m + “ natural numbers is ” + y) }}
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Integer indexclass integerIndex { public static void main(String arg[]) { int n = 3; double x = 3.14, y = 1.0; int m = n; if (n < 0) { x = 1/x; m = -n; } while (m > 0) { y *= x; m--; } System.out.println(x + “ to the power ” + n + “
is ” + y); }}
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Positive Decimal to Binaryclass positiveDecimalToBinary {
public static void main(String arg[]) { int n = 34, y=0, polyTerm = 1; if (n < 0) { System.out.println(“Sorry, cannot handle
negative integers today!”); } else { while (n > 0) { y += (polyTerm*(n%2)); n /= 2; polyTerm *= 10; } System.out.println(“Required binary: ” + y); } }}
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do-whiledo { statements} while (condition);• “statements” execute at least once
irrespective of condition
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for loopsfor (expression1; condition; expression2)
{ statements}• Same asexpression1while (condition) { statements expression2}
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Sum of natural numbersclass naturalSum { public static void main(String arg[]) { int n; int y = 1; for (n=2; n <=100; n++) { y += n; } System.out.println(“Sum of the first
” + (n-1) + “ natural numbers is ” + y); }}
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Comma operator in for loopfor (expression1a, expression2a, …;
condition; expression1b, expression2b,…) { statements }• Same asexpression1aexpression2a …while (condition) { statements expression1b expression2b …}
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Sum of natural numbersclass naturalSum { public static void main(String arg[]) { int n; int y; for (n=2, y=1; n <=100; y += n,
n++) { } System.out.println(“Sum of the first
” + (n-1) + “ natural numbers is ” + y); }}
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Empty for loop bodyfor (expression1; condition;
expression2) {}• Same asfor (expression1; condition;
expression2);
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Infinite loops• Loops that never terminatewhile (true) { statements}
do { statements} while (true);
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Infinite loopsfor (expression1; ;expression2) { statements}
for (i=0; i > -10; i++) { statements}
for (i=0; i<=100; i--) { statements}
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Perfect squaresclass identifySquareButLessClever { public static void main (String arg[]) { int n = 48; int i; if (n < 0) { System.out.println (n + “ is not a perfect
square.”); } else if ((n==0) || (n==1)) { System.out.println (n + “ is a perfect square.”); } else { for (i=2; i<=n/2; i++) { if ((i*i) == n) { System.out.println(n + “ is square of ” + i); } } } }}
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break• In the last example you may want to come
out of the for loop as soon as you discover that n is a square– The computation done after this is useless– Use break– Breaks out of the loop (while, do-while, or for)
currently you are in
for (i=2; i<=n/2; i++) { if ((i*i)==n) { System.out.println(n + “ is square of ” +
i); break; }}
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break• Another way to achieve the same
effect without using breakfor (i=2; (i<=n/2) && ((i*i) <= n); i++)
{ if ((i*i)==n) { System.out.println(n + “ is square of
” + i); }}
