[c++]返回数组

一维数组

其实如果支持vector的话，只要用引用就可以了。

void testFun(vector<int> &var)

1.返回函数指针

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在c++中是不允许数组作为函数的返回值的

int [] someFunction( ); //ILLEGAL

要想实现函数返回一个数组，那返回对应数组里面类型的指针

you must return a pointer to the array base type and have the pointer point to the array. So, the function declaration would be as follows:

int* someFunction( ); //Legal

int* doubler( int a[], int size)
{
int* temp = new int[size];
for ( int i =0; i < size; i++)
temp[i] = 2*a[i];
return temp;
}
int main(){
int a[] = {1, 2, 3, 4, 5};
int* b;
b = doubler(a, 5);

delete[] b

return 0;
}

2.

void doubler( int* a, int size)
{
for ( int i =0; i < size; i++)
a[i]=0;
}
int main(){
int a[5];
doubler(a, 5);

return 0;
}

二维数组

#include <iostream>
using namespace std;
void property(int shape[4][4])
{
shape[0][0]=1;
}

int main()
{
int shape[4][4];
property(shape);
cout<<shape[0][0];
return 0;
}

#include <iostream>
using namespace std;
void property(int (*shape)[4])
{
shape[0][0]=1;
}

int main()
{
int shape[4][4];
property(1,shape);
cout<<shape[0][0];
return 0;
}

void fun2(int **ga,int m,int n)
{
ga=new int*[m];
for(int i = 0; i < m; i++)
ga[i] = new int
;
}

int** newArray(int m,int n)
{
int** array=new int*[m];
for(int i = 0; i < m; i++)
array[i] = new int
;
return array;
}

int main()
{
int **shape;
int m,n;
m=2;n=4;
shape=newArray(m,n);
for(int i = 0; i < m; i++)
delete []shape[i];
delete []shape;
}