[英]Char* Array Memory Leak
I am having issues de-allocating memory that I used in my char*
array. 我在char*
在char*
数组中使用的内存时遇到问题。 In my code snippet below, I am creating a char*
array named input
that holds pointers to single words at a time followed by a pointer NULL
at the end of the array. 在下面的代码片段中,我正在创建一个名为input
的char*
数组,该数组一次保存指向单个单词的指针,然后在该数组的末尾保留一个指针NULL
。 This is the only time (I believe) I allocate memory in my code. (我相信)这是我唯一一次在代码中分配内存的时间。
char* input[999];
//exec commands
for(unsigned int i = 0; i < commands.size(); i++)
{
string current = "";
string word = "";
int k = 0;
for(unsigned int j = 0; j < commands.at(i).size(); j++) //iterate through letters
{
current = commands.at(i);
//cout << "current: " << current << endl;
if(current[j] == ' ')
{
input[k] = new char[word.size() + 1];
strcpy(input[k], word.c_str());
k++;
word = "";
}
else
word += current[j]; //add letter
//cout << "word: " << word << endl;
}
input[k] = new char[word.size() + 1];
strcpy(input[k], word.c_str());
k++;
input[k] = new char[1]; //add the NULL char *
input[k] = NULL;
...
Later on, I attempt to de-allocate this memory with this snippet of code: 稍后,我尝试使用以下代码片段取消分配此内存:
for(int z = 0; z < k; z++)
{
delete[] input[z];
}
I am iterating through my char*
array and deleting the memory allocated at each index. 我正在遍历char*
数组并删除在每个索引处分配的内存。 Without using this snippet, valgrind informs me of memory leaks. 如果不使用此代码段,则valgrind会通知我内存泄漏。 Using this snippet, valgrind informs me of less memory leaks than before. 使用此代码段,valgrind通知我的内存泄漏少于以前。 I am still stuck with the issue of memory still being definitely lost . 我仍然对内存的问题仍然毫无疑问感到困惑 。
I am unsure what I am missing to remove the rest of the allocated memory (or if the cause of the rest of the memory leaks is in fact somewhere else in my code). 我不确定要删除剩余的分配内存(或者如果导致内存泄漏的原因实际上是我代码中的其他原因),我不确定。 I appreciate any and all help, suggestions, and tips. 我感谢所有帮助,建议和提示。
I think, your problem is in below case, 我认为,您的问题出在以下情况中,
input[k] = new char[1]; //add the NULL char *
input[k] = NULL;
here, without free
-ing input[k]
, you're assigning NULL
to it. 在这里,没有free
input[k]
,您正在为其分配NULL
。 Thus, the previous input[k]
is lost. 因此,先前的input[k]
丢失了。
If you really want input[k]
to hold NULL
, ( maybe as a sentinel value ), you can simply do 如果您真的想让input[k]
保持NULL
,( 也许是一个哨兵值 ),您可以简单地做
input[k] = NULL;
No need to allocate memory separately. 无需单独分配内存。
You don't need to create a new char on where it says input[k] = new char[1]; //add the NULL char *
您不需要在input[k] = new char[1]; //add the NULL char *
地方创建新字符input[k] = new char[1]; //add the NULL char *
input[k] = new char[1]; //add the NULL char *
Just leave the NULL assignment in and it should work fine. input[k] = new char[1]; //add the NULL char *
只需保留NULL赋值,它就可以正常工作。
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