output value not as expected @beginner

Code:

---

#include<stdio.h>
void main()
{
int n[3][3]={{1,2,3},{4,5,6},{7,8,9}};


printf("\n%d %d %d", *n , n[0][0], n[2][2]);
}

---

why am i getting garbage values in the output

What did you expect to get from *n?

isn't *n same as n[0][0]??

*n is NOT the same as anything....

and neither is

Code:

---

void main()

---

your programs should always look like the following unless your main is taking in command line arguments

Code:

---

int main(void)  //void in the () is optional
{
.
.
.
return 0;
}

---

*n is going to be the same as n[0] (which is the same as &n[0][0]), which is a pointer to the first element of your array.

thats his prob, *n is not pointing to the array. hence the garbage output from the printf for the *n position

this code:

Code:

---

#include<stdio.h>
int main()
{
int n[3][3]={{1,2,3},{4,5,6},{7,8,9}};


printf("\n %d %d %d", *n , n[0][0], n[2][2]);
return 0;
}

---

and this code:

Code:

---

#include<stdio.h>
int main()
{
int n[3][3]={{1,2,3},{4,5,6},{7,8,9}};


printf("\n %d %d %d", n[0][0] , n[0][0], n[2][2]);
return 0;
}

---

are giving different outputs!! any explanations??

guess you want that

Code:

---

#include<stdio.h>
int main()
{
int n[3][3]={{1,2,3},{4,5,6},{7,8,9}};


printf("\n %d %d %d", **n , n[0][0], n[2][2]);
return 0;
}

---

Kurt

thanks a lot kurt !! :)
though can you or someone provide a little explanation since n is a double-dim array and *n should denote the value at its base address

*n is a pointer to the first element in the array and that is an array of 3 integers and if you dereference that (**n) you get n[0][0]
Kurt

ok...got it!!
thanks...