java - 线程“main”中的异常java.lang.StringIndexOutOfBoundsException

Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 7
    at java.lang.String.charAt(Unknown Source)
    at DNAmatch.findFirstMatchingPosition(DNAmatch.java:100)
    at DNAmatch.run(DNAmatch.java:82)
    at acm.program.Program.runHook(Program.java:1568)
    at acm.program.Program.startRun(Program.java:1557)
    at acm.program.Program.start(Program.java:808)
    at acm.program.Program.start(Program.java:1279)
    at acm.program.Program.main(Program.java:1360)

我收到了这个奇怪的错误，但我不知道是什么原因引起的。我用谷歌搜索了我的问题的解决方案，但没有任何帮助我解决它。
如果您尝试作为chain1 atcg和chain2 tagaagtc的输入，则可以正常工作
但是对于输入chain1 atcg和chain2 tagaagct，则出现此错误。
如果有人可以帮助我将不胜感激。

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import acm.program.*;

public class DNAmatch extends Program
{
public void run()
{
    //The chains must only contain the characters A,T,C or G so we check the string for any irregularities.
    String current, chain2; Boolean flag=true; int errorCount;
    String chain1 = readLine("Please type in the first chain of DNA code: ");
    for (int i=0; i<chain1.length(); i++)
    {
        current = chain1.charAt(i) +"";
        if (!((current.equalsIgnoreCase("a"))||(current.equalsIgnoreCase("t"))||
                (current.equalsIgnoreCase("c"))||(current.equalsIgnoreCase("g"))))
        {
            flag=false; break;
        }
    }
    while (flag==false)
    {
        errorCount=0;
        println("The DNA code you insert must only contain the characters A, T, C and G");
        chain1 = readLine("Please type again in the first chain of DNA code: ");

        for (int i=0; i<chain1.length(); i++)
        {
            current=chain1.charAt(i) +"";
            if (!((current.equalsIgnoreCase("a"))||(current.equalsIgnoreCase("t"))||
                (current.equalsIgnoreCase("c"))||(current.equalsIgnoreCase("g"))))
            {
                errorCount =1; break;
            }
        }
        if (errorCount==0)
        {
            flag=true; break;
        }

    }
    chain2 = readLine("Please type in the second chain of DNA code: ");
    flag=true;
    for (int i=0; i<chain2.length(); i++)
    {
        current = chain2.charAt(i) +"";
        if (!((current.equalsIgnoreCase("a"))||(current.equalsIgnoreCase("t"))||
                (current.equalsIgnoreCase("c"))||(current.equalsIgnoreCase("g")))&&(chain1.length()>chain2.length()))
        {
            flag= false; break;
        }
    }

    while ((flag==false)&&(chain1.length()>chain2.length()))
    {
        errorCount=0;
        if (chain1.length()>chain2.length())
            println("The second DNA chain must be longer or equal to the first one at the most!");
        else
            println("The DNA code you insert must only contain the characters A, T, C and G");
        chain2 = readLine("Please type again the second chain of DNA code: ");
        for (int i=0; i<chain2.length(); i++)
        {
            current = chain2.charAt(i) +"";
            if (!((current.equalsIgnoreCase("a"))||(current.equalsIgnoreCase("t"))||
                    (current.equalsIgnoreCase("c"))||(current.equalsIgnoreCase("g")))&&(chain1.length()>chain2.length()))
            {
                errorCount =1; break;
            }
        }
        if (errorCount==0)
        {
            flag=true;
        }
    }

    int match=findFirstMatchingPosition(chain1,chain2);
    if (match==-1)
        println("Not match found! "+ match);
    else
        println("Match has been found at point "+ match+ "!");

}
public int findFirstMatchingPosition(String shortDNA, String longDNA)
{
    String currentCharShort=""; String currentCharLong="";
    int match=0; int ans=-1; int a;
    for (int i=0; i < longDNA.length(); i++)
    {
        a = i;
        match = 0;

        for (int j=0; j < shortDNA.length(); j++)
        {
            currentCharLong=longDNA.charAt(a)+ "";
            currentCharShort=shortDNA.charAt(j)+ "";

            if ( (currentCharShort.equalsIgnoreCase("a") && currentCharLong.equalsIgnoreCase("t"))||
                (currentCharShort.equalsIgnoreCase("t") && currentCharLong.equalsIgnoreCase("a"))||
                (currentCharShort.equalsIgnoreCase("c") && currentCharLong.equalsIgnoreCase("g"))||
                (currentCharShort.equalsIgnoreCase("g") && currentCharLong.equalsIgnoreCase("c")) )
            {
                match +=1;
                a +=1;
            }
            else break;
        }
        if (match == shortDNA.length())
        {
            ans=i;
            break;
        }

    }
    return ans;
}

}

最佳答案

当您输入此for循环时

for (int j=0; j < shortDNA.length(); j++)
    {
        currentCharLong=longDNA.charAt(a)+ "";
        currentCharShort=shortDNA.charAt(j)+ "";

        if ( (currentCharShort.equalsIgnoreCase("a") && currentCharLong.equalsIgnoreCase("t"))||
            (currentCharShort.equalsIgnoreCase("t") && currentCharLong.equalsIgnoreCase("a"))||
            (currentCharShort.equalsIgnoreCase("c") && currentCharLong.equalsIgnoreCase("g"))||
            (currentCharShort.equalsIgnoreCase("g") && currentCharLong.equalsIgnoreCase("c")) )
        {
            match +=1;
            a +=1;
        }
        else break;
    }

使用“ a”表示您在longDNA中的最后一个字符的索引（如果longDNA有六个字符，则为5），当您输入if分支并放入a + = 1时，您现在拥有的索引大于longDNA的大小。因此，当您再次迭代并搜索charAt（a）时，将引发异常。如果在内部for循环中不同时间执行a + = 1语句，则使用较小的“ a”也可能发生这种情况。