Cout gives no output after declaring an std::string variable

Question

I've written a simple program returning the hostname of the IP address passed as an argument. The program uses two functions: getaddrinfo() and getnameinfo(). I'm using Linux Mint, Netbeans IDE and the G++ compiler. The output is alright, there are no errors, but when I declare an

std::string str;

then cout gives no output, nothing is printed on the screen. However when I comment out the std::string declaration or remove it, the statement

std::cout << "hostname: " << hostname << std::endl;

prints the returned hostnames successfully.

What may be the cause of such a strange error?

#include <netdb.h>
#include <netinet/in.h>
#include <sys/socket.h>
#include <iostream>
#include <string>

int main()
{
    struct addrinfo* result;
    struct addrinfo* res;
    int error;
    const char* host;
    // When I comment out this line, cout prints the hostnames succesfully.
    std::string str;

    error = getaddrinfo("127.0.0.1", NULL, NULL, &result);
    res = result;

    while (res != NULL)
    {
        char* hostname;
        error = getnameinfo(res->ai_addr, res->ai_addrlen, hostname, 1025, NULL, 0, 0);
        std::cout << "hostname: " << hostname << std::endl;
        res = res->ai_next;
    }

    freeaddrinfo(result);
    // When I declare an std::string str variable, this cout doesn't either print anything
    std::cout << "TEST" << std::endl;

    return 0;
}

C++ doesn't work that way. It has nothing to do with the string. You can't just wildly declare a char * and hope it points somewhere sensible. Look at this example for some inspiration. — Kerrek SB
– Kerrek SB, Commented Apr 5, 2013 at 21:25

Daniel Williams · Accepted Answer · 2013-04-05 21:26:42Z

3

   The arguments host and serv are pointers to caller-
   allocated buffers (of size hostlen and servlen respectively) into which
   getnameinfo() places null-terminated strings containing the host and
   service names respectively.

http://man7.org/linux/man-pages/man3/getnameinfo.3.html

Your pointers must be actually allocated. The fact that commenting out that line changes anything is probably a coincidence or a strange side effect of optimization.

answered Apr 5, 2013 at 21:26

Daniel Williams

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Comments

Woy · Accepted Answer · 2013-04-05 21:57:27Z

0

Thanks, it works now :). I'd like to find out when to use different ways to allocate memory. As far as I know the main difference between creating an object in the following way:

// Creating objects:
Test t1;
Test* t2 = new Test();

The first object will be created in a heap and it will be automatically deleted when the function is finished running.
The second object will be created in a stack and memory deallocation has to be done manually using the delete/delete[] operators?

So what else should I keep in mind when dealing with pointers? I guess I need to read a good book about computer architecture, as knowledge about memory and microprocessors will give profits :)

answered Apr 5, 2013 at 21:57

Woy

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