[2 days left] What’s wrong with your cloud strategy? Learn why multicloud solutions matter with Nimble Storage.Register Now

x
?
Solved

Compilation error

Posted on 2003-12-04
11
Medium Priority
?
530 Views
Last Modified: 2010-04-15
1. The below code is OK if compiled in ANSI C Compiler but not on normal C Compiler. What's the solution for this problem?

char buf[100];

if ((numbytes=recv(new_fd, buf, 100-1, 0)) == -1) {
   perror("recv");
  exit(1);
}

buf[numbytes] = '\0';
printf("Received: %s\n",buf);

char *search_str = buf; ==> error at this line
--------------------------------------------------
error 1000: Unexpected symbol: "char".
error 1588: "search_str" undefined.
--------------------------------------------------

printf("search_str == %s\n", search_str); ==> error at this line
---------------------------------------------------
error 1549: Modifiable lvalue required for assignment operator.
---------------------------------------------------

2. Can anybody explain to me or suggest to me for any good interactive tutorial on pointer?
0
Comment
Question by:showbix
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 5
  • 5
11 Comments
 
LVL 45

Expert Comment

by:sunnycoder
ID: 9873531
Hi showbix,

I presume you meant not compiling in ANSI C but compiling in "normal" C

1. in C all declarations must precede the first statement...
To be more specific declarations are allowed only at the beginning of a "block"

2. you cannot assign strings like a=b ... nor can you compare them like a==b ... you have to use strcpy and strcmp respectively

http://www.cs.cf.ac.uk/Dave/C/CE.html
is a good tutorial

Cheers!
Sunny:o)
0
 
LVL 45

Expert Comment

by:sunnycoder
ID: 9873562
just noticed that you were not assigning strings there ... just  pointer manipulation ... that will work fine once search_str is declared properly ...

move char * search_str to the beginning of the function and everything should work well.
0
 
LVL 30

Expert Comment

by:Mayank S
ID: 9874047
>> 1. in C all declarations must precede the first statement...

But Sunny, that was old.... the latest accepted standard C99 overrides it. It is now legal in C to declare variables anywhere in a program, in any block. However, all compilers don't support it as yet.
0
Concerto Cloud for Software Providers & ISVs

Can Concerto Cloud Services help you focus on evolving your application offerings, while delivering the best cloud experience to your customers? From DevOps to revenue models and customer support, the answer is yes!

Learn how Concerto can help you.

 
LVL 45

Expert Comment

by:sunnycoder
ID: 9874075
> It is now legal in C to declare variables anywhere in a program, in any block.
Nope mayank ... Variables can be declared only at the *beginning* of a block
C++ permits you to declare variables anywhere in the program but C still requires declarations to be at the beginning of a block
0
 
LVL 30

Expert Comment

by:Mayank S
ID: 9879830
Ok, sorry I got confused with program and block :-), so I thought that you were wrong.

Anyways, Sunny, pls comment on: http://www.experts-exchange.com/Programming/Q_20816080.html
0
 
LVL 45

Expert Comment

by:sunnycoder
ID: 9880190
sorry mayank ... I do not have much idea about java or .net
I am still limited to C and linux
0
 
LVL 30

Expert Comment

by:Mayank S
ID: 9880300
ok, no worries :-)
0
 

Author Comment

by:showbix
ID: 9880797
Hey guys,
Let me explain what I need to achieve here.

I read the data using recv function and insert it into buf variable
---------------------------------------------------------------------
char buf[100];


if ((numbytes=recv(new_fd, buf, 100-1, 0)) == -1) {
   perror("recv");
   exit(1);
}

buf[numbytes] = '\0';
printf("Received: %s\n",buf);

char *search_str = buf;

if (strstr(search_str,"SUCCESS") != NULL)
{
  printf("SUCCESS found\n");
}
  else
{
  printf("SUCCESS not found\n");
}

---------------------------------------------------------------------
I tried to copy the buf value into search_str so that I can perform the sub-string search.
From search_str variable I would like to scan a sub-string (e.g SUCCESS).

The code above works fine in ANSI C compiler but it throw me erros when using normal C Compiler.
The error is at this line --- char *search_str = buf;

Putting the declaration "char *search_str = buf;" at the beginning of function will need me to put another statement(which I dont know) to copy the buf value into the search_str variable.
 

How do I rectify it so that the program works?

FYI, I'm in the UNIX environment.




0
 
LVL 30

Assisted Solution

by:Mayank S
Mayank S earned 75 total points
ID: 9880914
char buf[100] ;
char * search_str = NULL ;

if ((numbytes=recv(new_fd, buf, 100-1, 0)) == -1) {
   perror("recv");
   exit(1);
}

buf[numbytes] = '\0';
printf("Received: %s\n",buf);

search_str = buff ;
0
 
LVL 45

Accepted Solution

by:
sunnycoder earned 75 total points
ID: 9880916
>Putting the declaration "char *search_str = buf;" at the beginning of function will need me to put another statement(which I
>dont know) to copy the buf value into the search_str variable.
just move the declaration to the beginning and let the assignment remain as it is


char buf[100];
char * search_str; //declaration here

if ((numbytes=recv(new_fd, buf, 100-1, 0)) == -1) {
  perror("recv");
  exit(1);
}

buf[numbytes] = '\0';
printf("Received: %s\n",buf);

search_str = buf;     //assignment remains

if (strstr(search_str,"SUCCESS") != NULL)
{
 printf("SUCCESS found\n");
}
 else
{
 printf("SUCCESS not found\n");
}
0
 
LVL 30

Expert Comment

by:Mayank S
ID: 9880917
On second thought:

char buf[100] ;
char * search_str = buf ;

if ((numbytes=recv(new_fd, buf, 100-1, 0)) == -1) {
   perror("recv");
   exit(1);
}

buf[numbytes] = '\0';
printf("Received: %s\n",buf);

// because 'buf' will always have the same base address,so it can be assigned to search_str right at the start, after declaration.
0

Featured Post

VIDEO: THE CONCERTO CLOUD FOR HEALTHCARE

Modern healthcare requires a modern cloud. View this brief video to understand how the Concerto Cloud for Healthcare can help your organization.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Have you thought about creating an iPhone application (app), but didn't even know where to get started? Here's how: ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Important pre-programming comments: I’ve never tri…
Examines three attack vectors, specifically, the different types of malware used in malicious attacks, web application attacks, and finally, network based attacks.  Concludes by examining the means of securing and protecting critical systems and inf…
The goal of this video is to provide viewers with basic examples to understand and use structures in the C programming language.
The goal of this video is to provide viewers with basic examples to understand recursion in the C programming language.

656 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question