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Thread: Return smallest value
 06052011, 06:55 PM #1Member
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Return smallest value
Hi I have an error with my code but I do not know what it is, here are my codes:
Java Code:public class Q2 { public static void main(String[] args) { int[] numbers={1,2,4,5,10,100,2,22}; int mostSmall= find(numbers); System.out.println(mostSmall); } public static int find(int[] numbers){ int smallest=0; for(int i=0;i<numbers.length;i++){ if(numbers[i]<=numbers[i+1]){ smallest= numbers[i]; } else{ smallest= numbers[i+1]; } } return smallest; } }
 06052011, 07:16 PM #2
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Most readers here aren't psychic; I am so I knew an ArrayIndexOutOfBounds error was thrown. Suppose your array length is 10; variable i goes from 0 to 9 (inclusive) but you are trying to access numbers[9+1] which goes kaboom. You have to fix that. b.t.w. you don't have to compare two consecutive numbers in the array to find the smallest number in the array.
kind regards,
Joscenosillicaphobia: the fear for an empty beer glass
 06052011, 07:49 PM #3Member
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Java Code:import java.math.*; public class Q2 { public static void main(String[] args) { int[] numbers={1,2,4,5,10,100,2,22}; int mostSmall= find(numbers); System.out.println(mostSmall); } public static int find(int[] numbers){ int smallest=0; for(int i=0;i<numbers.length;i++){ smallest= Math.min(numbers[7], numbers[i]); } return smallest; } }
 06052011, 07:51 PM #4Member
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Thanks for the tips.
 06052011, 08:25 PM #5
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 06062011, 03:29 PM #6Member
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It means comparing all the numbers in the array with the last number of the array to get the smallest value. Why do you ask?
 06062011, 03:52 PM #7
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 06062011, 03:57 PM #8Member
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Well based on my question paper(college), yeah this particular question contains only eight numbers. But I get what you mean :)
 06062011, 06:11 PM #9
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