二维数组释放 [英] 2-Dimensional array deallocation

问题描述

作为介绍，我在 Visual Studio 2010 中使用 C++，针对 x64 进行编译.我有一个程序使用二维数组来存储数据以运行我无法控制的 C 风格函数:

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;

这会导致 VS10 通过 _BLOCK_TYPE_IS_VALID(pHead -> nBlockUse) 的调试断言失败.无论在包含删除的最后几行中实际发生了什么，这都会在程序结束时发生.这究竟是什么意思?我究竟做错了什么?感觉真的很简单，但是这段代码看的太久了.

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.

--编辑---由于 dasblinkenlight 对我大脑的帮助，问题解决了！

--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;

推荐答案

您的代码崩溃，因为您将地址的地址传递给 delete[]，而这不是您分配的.将您的代码更改为:

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];
}

它不会再崩溃了.

这方面的规则很简单:由于您将 new[..] 的结果分配给 results[i]，您应该传递 results[i]，而不是 &results[i]，要删除 [].data 也是如此.

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.

还要注意，这段代码会删除您分配的所有行，包括最后一行(循环条件现在是 i ，而不是 i >).谢谢 bjhend！

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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