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LinkedList迭代异常(LinkedList Iteration Exception)

 我正在努力解决这个错误。 我觉得它很简单，但无法理解为什么会出错。  
 当我遍历链表中的值时，我不断收到NullPointerException 。  
 代码片段：  
private void updateBuyBook(LimitOrder im) {

    LimitOrder lm = null;

    Iterator itr = buyBook.entrySet().iterator();

    boolean modify = false;

    while (itr.hasNext() && !modify) {

        Map.Entry pairs = (Map.Entry) itr.next();
        if ((((LinkedList<IMessage>) pairs.getValue()).size() > 0)) {
            LinkedList<ILimitOrder> orders = (LinkedList<ILimitOrder>) pairs
                .getValue();
            ListIterator listIterator = orders.listIterator();
            while (listIterator.hasNext() && !modify) {
                LimitOrder order = (LimitOrder) listIterator.next();

                if (order.getOrderID().equalsIgnoreCase(im.getOrderID())) { // error at this line
                    lm = order;
                    addToBuyMap(im);
                    modify = true;
                    break;
                }
            }
            if (modify = true) {
                orders.remove(lm);
                break;
            }
        }
    }
}
 
 错误在这一行：  
Exception in thread "main" java.lang.NullPointerException
order.getOrderID().equalsIgnoreCase(im.getOrderID()));
 
 请帮忙。 我的作业是否有任何错误？ 请帮忙！！！ 谢谢 

I am struggling with this error. I feel its really simple but cannot understand why am getting the error. 
I keep getting a NullPointerException when I iterate through values in my linked list. 
Code Snippet: 
private void updateBuyBook(LimitOrder im) {

    LimitOrder lm = null;

    Iterator itr = buyBook.entrySet().iterator();

    boolean modify = false;

    while (itr.hasNext() && !modify) {

        Map.Entry pairs = (Map.Entry) itr.next();
        if ((((LinkedList<IMessage>) pairs.getValue()).size() > 0)) {
            LinkedList<ILimitOrder> orders = (LinkedList<ILimitOrder>) pairs
                .getValue();
            ListIterator listIterator = orders.listIterator();
            while (listIterator.hasNext() && !modify) {
                LimitOrder order = (LimitOrder) listIterator.next();

                if (order.getOrderID().equalsIgnoreCase(im.getOrderID())) { // error at this line
                    lm = order;
                    addToBuyMap(im);
                    modify = true;
                    break;
                }
            }
            if (modify = true) {
                orders.remove(lm);
                break;
            }
        }
    }
}
 
Error is at this line:  
Exception in thread "main" java.lang.NullPointerException
order.getOrderID().equalsIgnoreCase(im.getOrderID()));
 
Please help. Is my assignment wrong in any way??? Please help!!! Thanks

原文：https://stackoverflow.com/questions/5255285

更新时间：2023-12-17 11:12

最满意答案

 这里的问题是即使在-n 1模式下读取仍然是读取分隔线。 所以当它看到换行符时，它仍然认为是一个行分隔符并将其移除（并留下一个空变量）。  
$ find_newlines() {
    while IFS= read -r -n 1 c; do
        declare -p c;
    done
}
$ printf 'a\nb' | find_newlines
declare -- c="a"
declare -- c=""
declare -- c="b"
 
 正如Cyrus用bash 4.1+ ^{（ ref ）}所回答的，你可以使用-N标志来read以避免这个问题。  
 -N选项记录为（从Bash参考手册条目中read ）：  
 
  -N nchars  
  在读取精确的nchars字符后读取返回值，而不是等待完整的输入行，除非遇到EOF或读取超时。 在输入中遇到的分隔符字符不会被专门处理，并且不会导致读取返回，直到读取nchars字符。  
 
 对于非4.1 +版本的bash（2.04+，看起来像），可以使用-d标志进行read以指定一个备用分隔符来解决此问题。 输入中不存在的任何分隔符都可以使用。 对于许多输入流而言，最可能的值可能是NUL / \0字符，您可以将其指定为-d ''来read 。  
$ find_newlines() {
    while IFS= read -d '' -r -n 1 c; do
        declare -p c;
    done
}
$ printf 'a\nb' | find_newlines
declare -- c="a"
declare -- c="
"
declare -- c="b"

The problem here is that read even in -n 1 mode is still reading delimited lines. So when it sees the newline it still considers that a line delimiter and removes it (and leaves you with an empty variable). 
$ find_newlines() {
    while IFS= read -r -n 1 c; do
        declare -p c;
    done
}
$ printf 'a\nb' | find_newlines
declare -- c="a"
declare -- c=""
declare -- c="b"
 
As indicated in the answer by Cyrus with bash 4.1+^(ref) you can use the -N flag to read to avoid this problem. 
The -N option is documented as (from the Bash Reference Manual entry for read): 
 
 -N nchars 
 read returns after reading exactly nchars characters rather than waiting for a complete line of input, unless EOF is encountered or read times out. Delimiter characters encountered in the input are not treated specially and do not cause read to return until nchars characters are read. 
 
For non-4.1+ versions of bash (2.04+ it looks like) you can use the -d flag to read to specify an alternate delimiter to work around this problem. Any delimiter that isn't going to exist in your input will work. The most likely value for that for many input streams is likely to be the NUL/\0 character which you can specify to read as -d ''. 
$ find_newlines() {
    while IFS= read -d '' -r -n 1 c; do
        declare -p c;
    done
}
$ printf 'a\nb' | find_newlines
declare -- c="a"
declare -- c="
"
declare -- c="b"

LinkedList迭代异常(LinkedList Iteration Exception)

最满意答案

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