Thread: Cant understand the outout

  1. 06-14-2011 #1
    mistu4u
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    Cant understand the outout

    A simple program is:
    Code:
    //Array Access
#include<stdio.h>
void main()
{
   int n[3][3]={
                 2,4,3,
                 6,8,5,
                 3,5,1
                };
 int i,*ptr;
 ptr=&n;
 for(i=0;i<9;i++)
 {
   printf("%d",*(ptr+i));     //should print 2,6 and 3 as n[0],n[1] and n[2]
 }
}
    Now my understanding is a 2 dimensional array is nothing but a collection of one dimensional array within another one dimensional array.So when i am pointing n,the address is held in ptr like n[0][0]..
    and *(ptr+i)=ptr[i]=i[ptr]
    So ptr[i] should=ptr[0],ptr[1],etc which is invalid as it is not pointing to any particular element like ptr[0][1] or ptr[1][2].So how the output is printed?
    Help me in correcting the concept!!
    I also know we can access each element of such array if we have already initialised a pointer to an array.But here nothig is done like that!!
  2. 06-14-2011 #2
    claudiu
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    First of all it's int main(void) and it returns 0, not void main().

    Secondly, ptr is a pointer to an int, which means that it holds the memory address of an integer. n is not an integer. Had you done something like ptr = &n[1][1] that would have been correct.

    I suggest you read the tutorials on this site on pointers and arrays before proceeding further.
    1. Get rid of gets(). Never ever ever use it again. Replace it with fgets() and use that instead.
    2. Get rid of void main and replace it with int main(void) and return 0 at the end of the function.
    3. Get rid of conio.h and other antiquated DOS crap headers.
    4. Don't cast the return value of malloc, even if you always always always make sure that stdlib.h is included.
  3. 06-14-2011 #3
    mistu4u
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    no,I didnot use &n[1][1].....Because just writing n, means i am giving the address n[0][0] in case of array.
  4. 06-14-2011 #4
    claudiu
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    Quote Originally Posted by mistu4u View Post
    no,I didnot use &n[1][1].....Because just writing n, means i am giving the address n[0][0] in case of array.
    And what type is that? (hint: it's not int*)
    1. Get rid of gets(). Never ever ever use it again. Replace it with fgets() and use that instead.
    2. Get rid of void main and replace it with int main(void) and return 0 at the end of the function.
    3. Get rid of conio.h and other antiquated DOS crap headers.
    4. Don't cast the return value of malloc, even if you always always always make sure that stdlib.h is included.
  5. 06-15-2011 #5
    laserlight
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    Quote Originally Posted by mistu4u
    So ptr[i] should=ptr[0],ptr[1],etc which is invalid as it is not pointing to any particular element like ptr[0][1] or ptr[1][2].So how the output is printed?
    As claudiu reminded you, ptr is a pointer to an int, not a pointer to an array. What you did was to make it point to the same address as n[0][0]. The elements of n are contiguous, and the elements of each element of n is also contiguous, thus you can iterate over all the ints in n through ptr just by incrementing ptr.

    Quote Originally Posted by mistu4u
    no,I didnot use &n[1][1].....Because just writing n, means i am giving the address n[0][0] in case of array.
    You understand that the address of an array and the address of its first element are the same in value, but you don't understand type. This lack of understanding is the root of the problem.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way
  6. 06-15-2011 #6
    mistu4u
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    Thank yo guys for making your points clear......
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