Client doesn't recognize '+'!

Hello!
I'm creating a simple pop3 client, but the problem is that my code doesn't recognize the '+' character! I don't get it! It should work!!!

Code:

#include <sys/socket.h>
#include <sys/types.h>
#include <netinet/in.h>
#include <arpa/inet.h>
#include <iostream>
#include <sys/io.h>

using namespace std;

int main(){

	int sock_fd, recv_data_size;
	char buffer[100];
	
	struct sockaddr_in server;
	
	if((sock_fd = socket(AF_INET, SOCK_STREAM, 0)) == -1){
		perror("socket();");
	}

	server.sin_family = AF_INET;
	server.sin_port = htons(324);
	server.sin_addr.s_addr = inet_addr("127.0.0.1");

	if(connect(sock_fd, (struct sockaddr *)&server, sizeof(struct sockaddr)) == -1){
		perror("connect();");
	}
	
	recv_data_size = recv(sock_fd, buffer, sizeof(buffer), 0);
	
	for(int i=0; i<recv_data_size; i++){
		if(strcmp(buffer, "+") == 0){
			cout << "+\n";
		}	
	}

return 0;
}

It doesn't print out '+'! Thank you!

Using string compare to compare 1 character is pretty pointless.

And you're only ever comparing with the start of the buffer.

You could do this

Code:

		if(strncmp( &buffer[i], "+", 1) == 0){
			cout << "+\n";
		}

But this is better

Code:

		if ( buffer[i] == '+' ) {
			cout << "+\n";
		}

Also, use std::endl in place of "\n" inside string constants.
Unlike in C, a "\n" does NOT guarantee that the output buffer will be flushed.

Well you are asking it if the string you just got is the same as "+".
Try this:

Code:

for (int i=0; i < recv_data_size; i++)
if ( buffer[i] == '+' )
  cout<<"+\n";
else
  cout<<buffer[i]

Thank you!!!
This works:

if(strncmp( &buffer[i], "l", 1) == 0){
cout << "+\n";
}

But what does &buffer mean? Why do I need to use &?

and this works allso!
if(buffer[i] == '+'){
cout << "+\n";
}
Thank you!!!

&buffer[i] is a pointer to the i'th character of the buffer.

strcmp() needs pointers to characters.

strcmp(buffer,"foo") and strcmp(&buffer[0],"foo) are the same thing.

Instead of &buffer[i] wouldn't (buffer + i) be better then you don't have to do any dereferencing?

> Instead of &buffer[i] wouldn't (buffer + i) be better then you don't have to do any dereferencing?
Since neither is a dereference, what's your question?
Both statements are the same, and compute the address of the i'th element of buffer.

Well buffer[i] would get the value, since array names are oftened stored as pointers we are derefencing the name to get the value. Then the & would be applied. I guess any good compiler would optimize it so that it really becomes (buffer+i)

> since array names are oftened stored as pointers
You so need an "arrays are not pointers" refresher course.
http://www.eskimo.com/~scs/C-faq/s6.html

> we are derefencing the name to get the value.
Nothing in &foo[bar] implies the result of foo[bar] ever being evaluated as a dereference.

> I guess any good compiler would optimize it so that it really becomes (buffer+i)
Every compiler understands them as being equivalent - quality and optimisation has nothing to do with it.

I'm shocked at this nonsense!

NM: I know what I mean but don't know how to explain it. I already know those points you mentoined.