const int ADJ_MATRIX[VERTEX_NUM][VERTEX_NUM]={
{0,1,1,0,0,0,0,0},
{1,0,0,1,1,0,0,0},
{1,0,0,0,0,1,1,0},
{0,1,0,0,0,0,0,1},
{0,1,0,0,0,0,0,1},
{0,0,1,0,0,0,1,0},
{0,0,1,0,0,1,0,0},
{0,0,0,1,1,0,0,0}
};
typedef struct {
int vertex;
int matrix[VERTEX_NUM][VERTEX_NUM];
int vNum;
int eNum;
}Graph;
void buildGraph(Graph *graph){
graph->vNum = VERTEX_NUM;
graph->eNum = EDGE_NUM;
graph->matrix = ADJ_MATRIX;
}
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错误发生在这句话中:
graph->matrix = ADJ_MATRIX;
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我是c ++的新手.请告诉我为什么会出现这个问题以及如何解决?
我想将ADJ_MATRIX分配给矩阵struct.
如前所述,您无法在C++中分配数组.这是因为编译器是一个意思,因为编译器可以.它只是不会让你这样做......
......除非你欺骗它;)
template <typename T, int N>
struct square_matrix {
T data[N][N];
};
square_matrix<int, 10> a;
square_matrix<int, 10> b;
a = b; // fine, and actually assigns the .data arrays
a.data = b.data; // not allowed, compiler won't let you assign arrays
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抓到了吗?现在代码需要一些小东西:
const square_matrix<int, VERTEX_NUM> ADJ_MATRIX={{
// blah blah
}}; // extra set of braces
typedef struct {
int vertex;
square_matrix<int, VERTEX_NUM> matrix;
int vNum;
int eNum;
}Graph;
void buildGraph(Graph *graph){
graph->vNum = VERTEX_NUM;
graph->eNum = EDGE_NUM;
graph->matrix = ADJ_MATRIX; // no change
}
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而要访问细胞,现在我们需要使用graph->matrix.data[1][2].这可以通过重载operator[]或operator()for 来减轻square_matrix.但是,现在这已经非常接近新std::array类或Boost等价物boost::array,因此考虑这些可能是明智之举.
不幸的是(或者幸运的是,谁知道......)你不能只用C++将一个数组分配给另一个数组.
如果要复制数组,则需要将每个元素逐个复制到新数组中,或者使用以下memcpy()函数:
for( int i = 0; i < VERTEX_NUM; i++ )
for( int j = 0; j < VERTEX_NUM; j++ )
graph->matrix[i][j] = ADJ_MATRIX[i][j];
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要么
memcpy( graph->matrix, ADJ_MATRIX, VERTEX_NUM * VERTEX_NUM * sizeof(int) );
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