very basic pointer questions--pls help

Hello, I have a very basic pointer questions.

int *i;
int i_len=0;
i[i_len++]=1;
i[i_len++]=2;

In the above scenerio, does it mean 'i' is being set as a pointer to 1 dimensional array, which sets 1,2 one after another sequentially into the array? The coder later pass the 'i_len' value as return from function. I understand 'i_len' is just an array index but could it mean something else?

If someone can kindly clarify me it would be very helpful. Many thanks.

your code extends to:

Code:

---

int *i;
int i_len=0;

i[i_len]=1;
i_len += 1;

i[i_len]=2;
i_len += 1;

---

But be certain that that code as it is will surely crash!

Quote:

---

Originally Posted by Sipher

your code extends to:

Code:

---

int *i;
int i_len=0;

i[i_len]=1;
i_len += 1;

i[i_len]=2;
i_len += 1;

---

But be certain that that code as it is will surely crash!

---

Thanks, Sipher. I do realize the index increasing part, just wanted to confirm with someone on the array part. Thanks a lot for that. Since you have mentioned, why it would crash? A little more tutoring pls.....

Also, on my second query, if the length of the array is returned, does it point to any address of the array? Or just
simply array length?

Many thanks.

Code:

---

/* i is a pointer to an int, and has not been given a valid address yet */
int *i;

/* i_len is an int, and has been assigned the value of zero */
int i_len=0;

/* i_len is an int still, and has been incremented to 1. i_len is being used as an
  index in an array, but i has not yet been given memory space. This is an error,
  and the results are undefined. If it's kept up. It will crash the program.

  The place in memory that is one int past i in memory, has been assigned the
  value of one, although the pointer i doesn't own this memory location.
*/
i[i_len++]=1;

/* Same as above, except one more int's size up from the space i is pointing to.
  This is undefined, and may cause anything to happen, from a program crash, to 
  a wrong answer, to nothing noticeable.

  The more it's continued, the more likely the "anything" will be more severe.
*/

i[i_len++]=2;

---

Quote:

---

In the above scenerio, does it mean 'i' is being set as a pointer to 1 dimensional array, which sets 1,2 one after another sequentially into the array? The coder later pass the 'i_len' value as return from function. I understand 'i_len' is just an array index but could it mean something else?

---

There is a close relationship between pointers and arrays, in C. i is a pointer, but not an array, and has been given no memory ("real estate") of it's own. It is a "squatter", if you will.

If i were allocated (given) some memory, ( with malloc() or calloc() )it could be used as a "dynamic" array - which is *VERY* close (but never quite the same thing), as a "real array".

Code:

---

int *i;  //a pointer, which could be allocated memory, and become a "dynamic" array.
int i[50]; //a real array

---

Explained completely. :)

Quote:

---

Originally Posted by Adak

Code:

---

/* i is a pointer to an int, and has not been given a valid address yet */
int *i;

/* i_len is an int, and has been assigned the value of zero */
int i_len=0;

/* i_len is an int still, and has been incremented to 1. i_len is being used as an
  index in an array, but i has not yet been given memory space. This is an error,
  and the results are undefined. If it's kept up. It will crash the program.

  The place in memory that is one int past i in memory, has been assigned the
  value of one, although the pointer i doesn't own this memory location.
*/
i[i_len++]=1;

/* Same as above, except one more int's size up from the space i is pointing to.
  This is undefined, and may cause anything to happen, from a program crash, to 
  a wrong answer, to nothing noticeable.

  The more it's continued, the more likely the "anything" will be more severe.
*/

i[i_len++]=2;

---

There is a close relationship between pointers and arrays, in C. i is a pointer, but not an array, and has been given no memory ("real estate") of it's own. It is a "squatter", if you will.

If i were allocated (given) some memory, ( with malloc() or calloc() )it could be used as a "dynamic" array - which is *VERY* close (but never quite the same thing), as a "real array".

Code:

---

int *i;  //a pointer, which could be allocated memory, and become a "dynamic" array.
int i[50]; //a real array

---

---

Thanks Adak, very kind of you. You have spent a good amount of time explaining me. God bless.

You're welcome, and welcome to the forum, TKM.

sizeof(i) would be << drum roll please >>






wait for it...








have you thought of it yet?








wait one...






the size of a pointer to int. :D

Quote:

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Originally Posted by tkm

Also, on my second query, if the length of the array is returned, does it point to any address of the array? Or just
simply array length?

Many thanks.

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You mean i_len ?
i_len is 2 after the last assignment which is the length of your array. No it does not point to any legitimate address if you did i[i_len] because it is past the last value you assigned.