I'm new to C++ and I have some problem with memory allocation.
vector<int *> V;
for(int i=0;i<3;i++){
int A[3];
cin>>A[0]>>A[1]>>A[2];
V.push_back(A);
}
for(int i=0;i<3;i++){
cout<<V[i][0]<<V[i][1]<<V[i][2]<<endl;
}
I want to create arrays and push them into vector as below. But if I input
1 1 1
2 2 2
3 3 3
The output will be
3 3 3
3 3 3
3 3 3
If I replace int A[3] with int *A=new int() then I get the right answer. I want to know why?
When you use int A[3]; you are invoking undefined behaviour because the array you are storing in the vector goes out of scope after the loop.
If you use int *A = new int();, you are allocating one integer, and then using three, which is also invoking undefined behaviour.
You should be using int *A = new int[3]; or thereabouts to allocate an array of three int.
In the cold light of morning, the int *A = new int[3]; solution leaks horribly. The vector destructor doesn't know it needs to free the sub-arrays.
The diagnosis of the trouble remains accurate; the appropriate remedy needs rethinking.
I need to work on what the correct (leak-free) solution is — and I've got other things to do right now. One plausible option is vector< vector<int> >, but I've not coded that up so it works yet.
int A[3] and int *A=new int[3]. I think they both allocate memory for 3 integer to the pointer A??
int A[3] allocate memory, but because of the life time of var A[3], the memory will be deallocated when goes out of the for loop.Although their both generate a array with 3 element, their memory space are maintained in different behave.
int A[3] will generate a local array which will be destroyed after loop. Hence, the array in the vector is destroyed and you will get a strange result.
For int *A=new int[3], the array will not be destroyed until the delete operator is used. Hence all data in the vector will be hold corrected.
