英文:
It is not going inside the first loop and only printing I am outside loop only
问题
以下是您提供的代码的翻译部分：
package string_pattern;public class Naive {public static void main(String[] args){String pattern = "This is a text book";String text = "text";child obj = new child();obj.search(pattern, text);}}class child {public void search(String pat, String text){int m, n;m = pat.length();n = text.length();System.out.println("我在循环外部");for (int i = 0; i <= (n - m); i++){int j;System.out.println("我在第一个循环内部");for (j = 0; j < m; j++){if (text.charAt(i + j) != pat.charAt(j)){System.out.println("在这里");break;}if (j == m)System.out.println("在位置找到模式" + i);}}}}
请注意，我已根据您的要求仅提供代码的翻译部分，去除了其他内容。如果您需要进一步的帮助，请随时询问。
英文:
The below code is not giving any error. After debugging i found it is not going inside the first loop.
package string_pattern;public class Naive {public static void main(String[] args){String pattern="This is a text book";String text="text";child obj=new child();obj.search(pattern, text);}}class child{public void search(String pat,String text){int m,n;m=pat.length();n=text.length();System.out.println("i am outside loop");for(int i=0; i<= (n-m); i++){int j;System.out.println("i am inside first loop");for(j=0;j<m;j++){if(text.charAt(i+j) != pat.charAt(j)){System.out.println("OVER HERE");break;}if(j==m)System.out.println("pattern found at pos" + i);}}}}
答案1
得分: 1
你需要执行以下操作：
m = pat.length();
n = text.length();
问题在于`text`的长度为4，而`pat`较长。因此，当你像这样循环时 -> `for(int i=0; i<= (n-m); i++)`，你最终得到一个负数。因为0大于任何负数，所以你永远不会进入循环。
英文:
You do the following:
m=pat.length();n=text.length();
The problem is that text has a length of 4, while pat is longer. So when you do the loop like this -> for(int i=0; i<= (n-m); i++), you end up with a negative number. Since 0 is greater than any negative number you never enter the loop.
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