output.....

This is a discussion on output..... within the C Programming forums, part of the General Programming Boards category; I am new to c and i have been trying to understand the fundas. i tested a program on arrays ...

  1. #1
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    output.....

    I am new to c and i have been trying to understand the fundas. i tested a program on arrays which goes like this:

    insert
    Code:
    void main(){
    
    	int n[3][3]= {2,4,3,6,8,5,3,5,1};
    	printf("\n%d \n%d \n%d \n%d \n%d",n, *n, &n, *(*(n+1)+0),n[2][2]);
    }
    
    
    The output that i get is this:
    1244692
    1244692
    1244692
    6
    1

    I had a question that n contains the base address of the array and *n should contain the value at address contained in n which should be the first element in the array.

    Also &n is also the same value is this value not supposed to be different?

  2. #2
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    When you refer to an array by name and not to a specific element, it "decomposes" into a pointer to the first element. So n, *n, &n would all return &n[0].

  3. #3
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    Right, so the address of the first element is (apparently) 1244692. The address of n is the same as the address of the first element. And *n is the same as n[0] - which is the first 3 element array in the second dimension. Again it has the same address as n[0][0].

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  4. #4
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    Quote Originally Posted by roaan View Post
    I am new to c and i have been trying to understand the fundas. i tested a program on arrays which goes like this:

    insert
    Code:
    void main(){
    
    	int n[3][3]= {2,4,3,6,8,5,3,5,1};
    	printf("\n%d \n%d \n%d \n%d \n%d",n, *n, &n, *(*(n+1)+0),n[2][2]);
    }
    
    
    The output that i get is this:
    1244692
    1244692
    1244692
    6
    1

    I had a question that n contains the base address of the array and *n should contain the value at address contained in n which should be the first element in the array.

    Also &n is also the same value is this value not supposed to be different?
    n - the name of an array gives the base address of the array.
    *n - Look it like this (*(n+0)+0) which is nothing but the address of n[0][0] if it would have been **n, it would have printed the very first element.
    &n - The array 'n' is itself stored at some place which is the same as from where the first element starts so &n and n print the same thing but it'll be better for you to check it by printing "n+1" and "&n+1", you will know what I mean and what's the differnce between "n" and "&n".
    By the way dont use void main.
    HOPE YOU UNDERSTAND.......

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