不熟悉使用“ new”进行内存分配（C ++）

Question

I am working on some legacy code. I have the following data types:

typedef struct {

    char        *name ;

    ColumnType  type ;

    unsigned    pos ;    //column position in table

    CellData    **data ; //ptr to list of cells in column

}Column ;

struct _table {

    char name[TABLE_NAME_LEN+1] ;

    unsigned int num_rows ;

    unsigned int num_cols ;

    Column  **cols ; //ptr to list of columns

};



struct _table m_

In the source code, there is the following statement:

m_table.cols = new Column*[m_table.num_cols];

I am familiar with new[], but I'm no sure what the multiplication operator is doing there - can any explain?

Answer 1

It's not multiplication. The symbol * has many, many completely different meanings in C++, all depending on context.

In your case, you're creating a dynamic array of Column* , ie of pointers to Column .

In other words, you're saying new T[N]; , where T = Column* .

Answer 2

It's not a multiplication operator. It's instead allocating an array of Column* (pointer to Column type). The resulting array stores pointer values

m_table.cols = new Column*[m_table.num_cols];
m_table.cols[0] = Column();  // Error: Expected Column* got Column
m_table.cols[0] = new Column();  // Ok

Answer 3

在您的情况下， Column*是一种类型（用于指向Column类实例的指针），然后创建它们的数组。

Answer 4

It's not a multiplication operator. It's allocating an array of pointer-to- Column , rather than an array of Column .

Answer 5

它不是乘法运算符，它是一个指针规范，即“指向列的指针”。

Answer 6

The * denotes a pointer. So you're creating an array of Column pointers.