CODEWITHSUNDEEP
c++arrays
Dipendra Pokharel
Dipendra Pokharel

Reputation: 507

c++ array and memory location

I have a piece of code that deals with C++ array.

 using namespace std;
 #include <iostream>

 int main(){
 int *p;
 p = new int[3];
 for(int i = 0; i < 3; i++){
    p[i] = i;
 }
 //delete[] p;
 for(int i = 0;i <3; i++){
    std::cout << *(p+i) << std::endl;
 }
}

How does this code work? How does the memory location *(p+i) work? How is it different from using p[i]. What are the differences on the code if we uncomment the line delete[] p.

Upvotes: 2

Views: 1022

Answers (1)

Kunal Puri
Kunal Puri

Reputation: 3427

1) When you do this:

p = new int[3];

Now, p points to the first element of the dynamically allocated array.

When you do, *(p + i) will lead to simple pointer arithmetic. It will boil down to: value of (<address pointed by p> + <size of type pointed by p> * i) which is equivalent to doing p[i].

That's why it works.

2) In C++, unlike java, you have to explicitly clear the dynamically allocated memory using delete, as there is no GC in C++ (and will never be, as per Bjarne Stroustrup). Otherwise, the memory area will remain acquired for the application lifetime, thereby causing memory leak.

Suggestion:

Place your delete at the end of the program. Otherwise, the loop below it may give SIGSEGV.

Also, Avoid using new and delete as much as you can.

Upvotes: 4

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