c++ - 在 C++ 中将大型二维数组转换为一维数组的内存分配

Question

A是二维数组，data是一维数组

double **A, *data;

全尺寸为 9222， 内存分配:

A = (double **)malloc((fullsize)*sizeof(double *));         //allocate memory dynamically for matrix A
for (i = 0; i < fullsize; i++)
    A[i] = (double *)malloc((2 * fullsize)*sizeof(double));

data = (double *)malloc((fullsize*fullsize)*sizeof(double *));

转换逻辑:

{
    for (int n = 0; n<fullsize; n++)
    for (int m = 0; m<fullsize; m++)
    {
        data[n*fullsize + m] = A[n][m];
    }
}

错误日志:

The thread 0xad80 has exited with code 0 (0x0).
Unhandled exception at 0x00F5E6E9 in ConsoleApplication1.exe: 0xC0000005: Access violation writing location 0x69E65000.

The program '[50964] ConsoleApplication1.exe' has exited with code 0 (0x0).

Answer 1

您正在使用 double 指针的大小分配 data:

data = (double *)malloc((fullsize*fullsize)*sizeof(double *));
//                                                        ^

它应该使用 sizeof(double) 来分配。

您可以通过使用 vector 在 C++ 中避免使用 malloc。如果您想要分配的确切大小，请改用 new double[...]。无论哪种情况，您最终都会得到更具可读性的代码:

double **A = new double*[fullsize];
for (int i = 0 ; i != fullsize ; i++) {
    A[i] = new double[fullsize];
}
double *data = new double[fullsize*fullsize];