Thread: 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

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;
}

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

*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...