Why the address of variable array is equal the variable array?

Hi,

Code:

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#include <stdio.h>


int main()
{
    int vetor[5], *pointer;


    printf("vetor = %p\n",vetor);
    printf("&vetor = %p\n",&vetor);
    // vetor == &vetor!

    pointer = vetor;
    printf("pointer = %p\n",pointer);
    printf("&pointer = %p\n",&pointer);
    // pointer != &pointer

    return 0;
}

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Why the address of variable array is equal the variable array?

The name of variable of array is a pointer, but with pointers is different!

Thanks

First of all you need to cast pointers to void* when using variadic functions like printf().

Quote:

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Originally Posted by claudiu

First of all you need to cast pointers to void* when using variadic functions like printf().

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First of all, when you use "first of all", there needs to be a second or next thing. Next, this has nothing to do with it being a varadic function, and everything to do with the format specifier he's using. Finally, it's good to finish up with a "finally", or maybe a "last but not least". ;)

To the OP, read: http://c-faq.com/aryptr/index.html


Quzah.

Might also read
The C Book — Arrays, the & operator and function

Tim S.

If you have an array as follows:

Code:

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p[5],*pointer;

// The following are the same

p;
&p[0];

// That is

p == &p[0]

// Therefore,

pointer = p;

// Puts the address of p[0] in the pointer "pointer"

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This is a basic standard of the c- language.