2维数组释放 [英] 2-Dimensional array deallocation

问题描述

作为一个前奏，我用C ++在Visual Studio 2010中，编制了64位。我有一个使用2维数组来存储数据，通过一个C风格的函数，我没有控制权运行的程序：

 浮**的结果;
结果=新的浮动* [行]
的for（int i = 0; I＆LT;行++我）{
    结果[I] =新的浮动[列];
}INT **数据;
数据= INT新* [行]
的for（int i = 0; I＆LT;行++我）{
    数据由[i] =新的INT [列];
}//发送数据的功能和填充值的结果
ExternalFunction（*数据，*结果）;//删除一切
的for（int i = 0; I＆LT;行-1 ++ I）{
    删除[]和放大器;结果[I]。
    删除[]和放大器;数据[I]
}
删除[]的结果;
删除[]的数据;
 

这导致VS10通过一个调试断言失败与_BLOCK_TYPE_IS_VALID（PHEAD - > nBlockUse）。发生这种情况的节目的结束而不管实际发生在包含删除的最后几行。这是什么意思是什么呢？我究竟做错了什么？我觉得这是很简单，但我一直在寻找这个code太久。

- 编辑 -
问题解决了由于dasblinkenlight乐于助人的轻推到我的大脑！

 浮动*结果=新的浮动[*行列];
浮动*数据=新的浮动[*行列];ExternalFunction（安培;数据[0]，＆安培;结果[0]）;删除[]的结果;
删除[]的数据;
 

解决方案

您code崩溃，因为你是传递一个地址的地址为删除[] ，这是不是你分配什么。更改code这样：

 的for（int i = 0; I＆LT;行++我）{
    删除[]结果[I]。
    删除[]数据[I]
}
 

将不再崩溃。

在这个规则很简单：因为你分配新的的结果[..] 到结果[I] ，你应该通过结果[I] ，而不是＆安培;结果[I] ，到删除[] 。同样适用于数据。

另外请注意，这code删除您分配的所有行，包括最后一个（现在的循环条件为 I＆LT; N ，而不是 I＆LT; N-1 ）。感谢bjhend！

As an intro, I'm using C++ in Visual Studio 2010, compiling for x64. I have a program that's using 2-Dimensional arrays to store data for running through a C style function that I have no control over:

float **results;
results = new float*[rows];
for (int i = 0; i < rows; ++i){
    results[i] = new float[columns];
}

int **data;
data = new int*[rows];
for (int i = 0; i < rows; ++i){
    data[i] = new int[columns];
}

//send data to the function and populate results with values
ExternalFunction(*data, *results);

//delete everything
for (int i = 0; i < rows-1; ++i){
    delete [] &results[i];
    delete [] &data[i];
}
delete [] results;
delete [] data;

This causes VS10 to through a Debug Assertion Failure with _BLOCK_TYPE_IS_VALID(pHead -> nBlockUse). This happens by the end of the program regardless of what really happens in the last few lines containing the deletes. What does this mean exactly? What am I doing wrong? I feel like it's really simple, but I've been looking at this code for too long.

--EDIT--- Problem solved thanks to dasblinkenlight's helpful nudge to my brain!

float *results = new float[rows * columns];
float *data = new float[rows * columns];

ExternalFunction(&data[0], &results[0]);

delete [] results;
delete [] data;

解决方案

Your code crashes because you are passing an address of an address to delete[], which is not what you allocated. Change your code to this:

for (int i = 0; i < rows ; ++i){
    delete [] results[i];
    delete [] data[i];
}

It will no longer crash.

The rule on this is simple: since you assigned the results of new[..] to results[i], you should be passing results[i], not &results[i], to delete []. Same goes for data.

Also note that this code deletes all rows that you allocated, including the last one (the loop condition is now i < n, not i < n-1). Thanks bjhend!

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