Confusing pointer

In my question paper there was a question to find output of the following program

Code:

---

#include<iostream.h>
#include<conio.h>

int main()
{
  int ia[3]={2,5,9};
  int *ptr=ia;
 
  for(int i=0;i<3;i++)
  {
    cout<<*(ia++);//Error Lvalue reqd.
    cout<<" "<<*ptr++;
  }
getch();
return(0);
}

---

On compiling it on my pc i am getting strange error l value reqd.Can anyone tell me why i am getting it. I also request you to compile it on your compiler as i am using old turbo c++.

If you're trying to say *ia + 1, then that isn't what *(ia++) means. Try changing it to *ia + 1

When you say *(ia++) you're trying to use a non-lvalue as an lvalue.

Array names are constant pointers...you can not alter them...read further from here

http://www.fredosaurus.com/notes-cpp...spointers.html

thanks sunny,
It did not strike me during paper.
I have one more doubt.

Code:

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int *p=new int[10];
int p[i]=3;

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how can a pointer change to array here?

Code:

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int *p=new int[10];
p[i]=3;

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This should work fine.

Quote:

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Originally Posted by vaibhav

thanks sunny,
It did not strike me during paper.
I have one more doubt.

Code:

---

int *p=new int[10];
int p[i]=3;

---

how can a pointer change to array here?

---

Your declaring a new variable p in the second line, which should give you a compiler error saying something like "redefinition of p". A variable named p already exists in the first line.

Quote:

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Originally Posted by vaibhav

how can a pointer change to array here?

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Now you are dynamically allocating 10 bytes of memory which is pointed by pointer p.....and it is one of many ways to access the memory.

p[i] is equivalent to *(p+i) and i[p]

Note: Pointers are not arrays and arrays are not pointers

Assuming the the int size is 4 then allocating 10 ints allocates 40 bytes.

sorry for that.....and thanx for correction

Quote:

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Originally Posted by Quantum1024

Code:

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int *p=new int[10];
p[i]=3;

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This should work fine.

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You should probably also free the memory:

Code:

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delete [] p;

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