Hi,
Why the address of variable array is equal the variable array?Code:#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; }
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().
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.
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".Originally Posted by claudiu
To the OP, read: http://c-faq.com/aryptr/index.html
Quzah.
Last edited by quzah; 05-05-2012 at 01:43 PM.
Hope is the first step on the road to disappointment.
Might also read
The C Book — Arrays, the & operator and function
Tim S.
"...a computer is a stupid machine with the ability to do incredibly smart things, while computer programmers are smart people with the ability to do incredibly stupid things. They are,in short, a perfect match.." Bill Bryson
If you have an array as follows:
This is a basic standard of the c- language.Code: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"
