typecasting

z99 · 07-03-2011, 09:44 AM

hello everyone,
i'm totally a newbie,i'm trying to learn network programming with c,i right a very simple program which use libpcap,

Code:

#include <stdio.h>
#include <stdlib.h>
#include <pcap.h>  
#include <errno.h>
#include <sys/socket.h>
#include <netinet/in.h>
#include <arpa/inet.h>

int main()
{
char *dev;
char err[PCAP_ERRBUF_SIZE];
int r;
bpf_u_int32 ip;
bpf_u_int32 net;
struct in_addr addr;
char *rip;
char *rnet;

dev=pcap_lookupdev(err);
if (dev==NULL){
printf("%s",err);
exit(1);
}
else
{
printf("Device Name Is: %s\n",dev);
}

r=pcap_lookupnet(dev,&ip,&net,err);
if(r==-1){
printf("%s,err");
exit(1);
}

addr.s_addr=ip;
rip=inet_ntoa(addr);
printf("%s\n",rip);

when i use

Code:

char rip; 
char rnet;
return 0;
}

[without *] in my code,after compiling,it gives an error about typecasting,

Quote:

pcap.c:36:6: warning: assignment makes integer from pointer without a cast [enabled by default]
pcap.c:37:10: warning: comparison between pointer and integer [enabled by default]

is it because,that the ip and net are pointer so that i must use a pointer too,i mean this:

Code:

char *rip; 
char *rnet;

if i want to type cast it i mean not to use a pointer just:

Code:

char rip; 
char rnet;

what should i do?
how can i typecast it?
sorry for the long post,
thanks in advance

MTK358 · 07-03-2011, 11:01 AM

First of all, indent your code. It's almost impossible to read un-indented code, and it's very easy to to indent it as you write.

Quote:

Originally Posted by z99

when i use

Code:

char rip; 
char rnet;
return 0;
}

[without *]

I don't see that in your code. Could you give some more details?

And what do you want to do, cast a "char*" to a "char"? That's not really valid, since they store very different kinds of data.

z99 · 07-03-2011, 11:34 AM

thanks for helping,
i mean instead of using

Quote:

char *rip;
char *rnet;

use:

Quote:

char rip;
char rnet;

is it possible,no casting will be needed?

MTK358 · 07-03-2011, 11:51 AM

Quote:

Originally Posted by z99

thanks for helping,
i mean instead of using

use:

is it possible,no casting will be needed?

Why do you want to do that? You should really learn about how pointers work.

Basically, a plain "char" holds one character (not a string). A "char*" contains a memory address. Why "char" before the asterisk, then? That means that when you "dereference" the pointer (get the value at the memory address it stores), it will be assumed that it's a "char" (You can use "void*" to just refer to a memory address without specifying any type).

So what does this have to do with strings? The idea is that a string is a sequence of "char"s in memory (with a NULL character at the end to indicate the end of the string), one after another, and the "char*" points to the first one. To get a character at the specified index, you add the character's index to the pointer and dereference it:

Code:

second_char = *(str + 1);

(The "*" operator dereferences a pointer)

And this is so common, that there's special syntax for it:

Code:

second_char = str[1];

(This is identical to the above example)

Hopefully now you understand a bit about how they work, and why "char" and "char*" are completely different things that are not interchangeable.

sundialsvcs · 07-03-2011, 11:54 AM

It seems to me that you need to acquire some deeper understanding of what these various declarations mean.

You need to understand what "char rip; means, and how it differs from "char *rip;" and what your source-code is saying to the compiler. It isn't enough to "change the code to make the syntax errors go away." (Either the symbol rip refers to a character, or it refers to a pointer to a character ... which one is it? You have to know.)

Typecasting, as a concept, is exactly the same thing. You have to understand what the use of this construct is saying to the compiler, and whether it actually matches what you want the computer to do and how the data is arranged in memory. "Nonsense in, garbage out."