convert to integer...

Hi friends.
this is my code

Code:

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#include<stdio.h>
#include<string.h>

int main()
{
        char buf[4]={'5','a','4','b'};
        int i;
        i=(int)buf;
        printf("%d\n",sizeof(int));
        printf("%d\n",sizeof(buf));
        printf("%c",buf[0]);
        printf("%c",buf[1]);
        printf("%c",buf[2]);
        printf("%c\n\n",buf[3]);
        printf("%d",i);
        return 0;

}

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what i wanted to know is ...when there is an array of 4 charecters...can i take that to an integer..(where int is 4 bytes). say the array is '5' 'a' '4' 'b' and when i take this array to a single integer of 4 bytes i want the exact "5a4b" in the integer...is this possible anyway...just a thing struck my mind....

Many machines have alignment requirements for various types - for example, allowable int addresses may only be one every 4 bytes (assuming an int is 4 bytes). In this case, if you try to cast a char pointer to an int pointer and dereference, you will often get a segfault, since you only have a 1 in 4 chance of having a valid int address. To avoid this, you can instead define an array of ints, and then cast the int pointers to char pointers. I don't know if just respecting this makes things portable though.

http://cboard.cprogramming.com/showthread.php?t=87461

wow... thanks vart :) i wonder how u guys got these tricks and knowledge in C...
thanks a lot for that link... :)

Quote:

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Originally Posted by robatino

since you only have a 1 in 4 chance of having a valid int address.

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sorry robatino i didnt get what u are telling....

Well, I don't know about the probability, but if the alignment restriction holds, only one out of every 4 possible char addresses is a valid int address (if sizeof(int) == 4). Typically it would be just those addresses that are divisible by 4.
For example, on my machine the output of

Code:

---

#include <iostream>

int main() {
  double d;
  std::cout << "double d address is " << (void *)&d << "\n";
  float f;
  std::cout << "float f address is " << (void *)&f << "\n";
  char c;
  std::cout << "char c address is " << (void *)&c << "\n";
}

---

is

double d address is 0xbff79798
float f address is 0xbff79794
char c address is 0xbff79793

(I remember reading recently that one should cast pointers to void * to print them, and the char pointer printed as garbage until I did this.) Notice that the addresses of d, f, and c are divisible by 8, 4, and 1, corresponding to the sizes.

Edit: Sorry for the code, I keep forgetting when I post to the C board instead of the C++ board. I don't remember offhand how to print a pointer using printf() but you should be able to look it up.

oh.. thanks robatino...i got it... :)