从 C++ 中的函数返回一个二维数组
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returning a two dimensional array from a function in c++
提问by MrHs
I want to use a two dimensional int array which is returned from a function how should I define the function return value ? I used int** but the compiler gave error:
我想使用从函数返回的二维 int 数组，我应该如何定义函数返回值？我使用了 int** 但编译器给出了错误：
int** tableCreator(){
int** table=new int[10][10];
for(int xxx=1;xxx<10;xxx++){
for(int yyy=1;yyy<10;yyy++){
table[xxx][yyy]=xxx*yyy;
}
}
return(table); //Here:cannot convert from 'int (*)[10]' to 'int **'
}
回答by mohaps
Try this:
尝试这个：
#include <cstdio>
#include <cstdlib>
int** createTable(int rows, int columns){
int** table = new int*[rows];
for(int i = 0; i < rows; i++) {
table[i] = new int[columns];
for(int j = 0; j < columns; j++){ table[i][j] = (i+j); }// sample set value;
}
return table;
}
void freeTable(int** table, int rows){
if(table){
for(int i = 0; i < rows; i++){ if(table[i]){ delete[] table[i]; } }
delete[] table;
}
}
void printTable(int** table, int rows, int columns){
for(int i = 0; i < rows; i++){
for(int j = 0; j < columns; j++){
printf("(%d,%d) -> %d\n", i, j, table[i][j]);
}
}
}
int main(int argc, char** argv){
int** table = createTable(10, 10);
printTable(table, 10, 10);
freeTable(table, 10);
return 0;
}
You need the second loop to allocate a 2-d array in C and similar operation to free it. a two-D array is in essence an array of arrays so can be expressed as a pointer array. the loop initializes the arrays pointed to the pointers.
您需要第二个循环在 C 中分配一个二维数组，并需要类似的操作来释放它。二维数组本质上是数组的数组，因此可以表示为指针数组。循环初始化指向指针的数组。
Clarifying as per conversation with @Eric Postpischil below: changed createTable to take row/column count for truly dynamic allocation.
根据下面与@Eric Postpischil 的对话进行澄清：更改 createTable 以获取行/列计数以进行真正的动态分配。
回答by taocp
int** table=new int[10][10];
this is wrong. you cannot allocate space for 2D dynamic array in this way in C/C++.
这是错误的。您不能在 C/C++ 中以这种方式为二维动态数组分配空间。
Meanwhile, you declared array size as 10, so indices are from 0-9, but you are trying to assign values to index 10in your nested for loops, which is not right too.
同时，您将数组大小声明为10，因此索引来自0-9，但是您试图10在嵌套的 for 循环中为索引分配值，这也不正确。
You may do the following for allocation:
您可以执行以下操作进行分配：
int** table = new int*[10];
for (int i = 0; i < 10; ++i)
{
table[i] = new int[10];
}
回答by Eric Postpischil
Usually, the type used to point to an array is a pointer to an element of the array. Since a two-dimensional array of intis an array of array of int, you want a pointer to array of int. The C++ syntax for this type is int (*)[N], for some dimension N. This code demonstrates:
通常，用于指向数组的类型是指向数组元素的指针。由于二维数组 ofint是数组 of 的数组int，所以您需要一个指向数组的指针int。int (*)[N]对于某些维度，此类型的 C++ 语法是N。此代码演示：
#define N 10
int (*tableCreator())[N]
{
int (*table)[N] = new int[N][N];
for (int i = 0; i < N; ++i)
for (int j = 0; j < N; ++j)
table[i][j] = i*j;
return table;
}
#include <iostream>
int main()
{
int (*t)[N] = tableCreator();
for (int i = 0; i < N; ++i)
{
for (int j = 0; j < N; ++j)
std::cout << t[i][j] << ' ';
std::cout << '\n';
}
delete [] t;
return 0;
}
回答by MrHs
I. Arrays are not pointers.
I.数组不是指针。
II. Why not vector<vector<int> >?
二、为什么不vector<vector<int> >呢？
III. If not, then:
三、如果没有，那么：
typedef int Int10Array[10];
Int10Array *arr = new Int10Array[10];
IV. Why write past the bounds? Do you want explicit nasal demons?
四、为什么要写越界？你想要明确的鼻妖吗？
for(int xxx = 0; xxx < 10; xxx++)
^^^ ^^^^
