Incrementing array pointer

Question

Is it possible to increment the array passed to a function?

I have a single dimensional array a[4] = {10,20,30,40}, I want to call a function and increment the array to point to 3rd element and print it from main().

For ex:

int main()
{
  int a[4] = {10,20,30,40};
  cout << *a; // 10
  increment(); // function call
  cout << *a; // 40
}

How can we accomplish this?If array is passed as a pointer, only the modification to the value stored in array get reflected. What is the prototype of the function if we have to increment the array in function?

There are a lot of information about arrays. What did you try? — Denis Ermolin
– Denis Ermolin, Commented Dec 3, 2012 at 10:53
Look up references. Or return a pointer from the function. The last is probably best as then you won't loose the original "pointer". — Some programmer dude
– Some programmer dude, Commented Dec 3, 2012 at 10:55
What do you mean "increment the array"? Arrays are not pointers. Arrays are arrays. Obtain a pointer to an element in the array and increment that. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Dec 3, 2012 at 11:20
What I tried was passing array as a pointer, what I wanted was passing to the function that takes a reference to a pointer. — user1177586
– user1177586, Commented Dec 3, 2012 at 14:22

Nim · Accepted Answer · 2012-12-03 10:57:16Z

3

Alternative to as suggested is:

#include <iostream>

void foo(int *&p) // <-- take pointer by reference...
{
  ++p;
  ++p;
}

int main(void)
{
  int a[] = {1,2,3,4};

  int*p = a;

  std::cout << *p << std::endl;

  foo(p);

  std::cout << *p << std::endl;

}

Use a second pointer to access the original array and modify that in the function...

Further alternative is to use either std::vector<int>, std::array<int, 4> and then pass an iterator to the function (and get it to return an iterator)...

answered Dec 3, 2012 at 10:57

Nim

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Comments

NPE · Accepted Answer · 2012-12-03 10:54:51Z

0

Not with your current syntax. You could, however, return the new pointer from increment(), and store it in another variable.

int main()
{
  int a[4] = {10,20,30,40};
  cout << *a;
  int* new_a = increment(a)
  cout << *new_a;
}

answered Dec 3, 2012 at 10:54

NPE

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Comments

Luca Davanzo · Accepted Answer · 2012-12-03 10:58:56Z

0

You can access at your pointer position in this way:

int main() {
  int a[4] = {10,20,30,40};
  cout << *a;        print ---> 10 
  cout << *(a + 3);  print ---> 40
}

You can't do what you want without lose information

answered Dec 3, 2012 at 10:58

Luca Davanzo

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Comments

Ken Cheung · Accepted Answer · 2012-12-03 11:05:40Z

0

The pointer a in your program points to the data space of your program in compile time (or link time depends on how you think).

If you have something like

void function(int* &x) {
  x+=2;
}

int main() {
  int *a = new int(4);
  int *b = a;
  a[0] = 10; a[1] = 20; a[2] = 30; a[3] = 40;
  cout << *a;
  function(b);
  cout << *b;
  delete a;
}

GNU G++ will check and not allowing you to move the pointer variable a because no more pointer will be referencing to the head of this memory block. Haha

answered Dec 3, 2012 at 11:05

Ken Cheung

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