Thread: Am I missing something here? (Problem with read())

>but it doesnt seem to do that
What does it seem to do?

>printf("Avklarat %s,",buffer);
I don't see a *buffer = '\0' anywhere, so you're probably overrunning your buffer.

Oups wrong version.
The *buffer = '\0' is in there.
Let me clarify. It doesnt print anything in the printf. It prints "Avklarat ,". And if the buffer that gets returned from this function is past as an argument to another function the sizeof says its only 4 bytes long.

>It doesnt print anything in the printf.
Well, that's probably because you've been incrementing buffer and now it points to an empty string because the first character is '\0'. Maybe something more like this:

Code:

int laes(int socket,char *buffer,int langd) {
  int counter = 0;
  int laes = 1;
  int storlek = 0;
  char c;
  char *pbuffer = buffer;
 
  while(counter < langd) {
    if( (storlek = read(socket,&c,1)) == 1) {
      if(c == '\n') {
	*pbuffer = '\0';
	printf("Avklarat %s,",buffer);

	return counter;
      } else {
	*pbuffer = c;
      }
    } else {
      printf("Avklarat annars"); 
     return counter;
      
    }

    printf("Ett varv\n");
    pbuffer++;
    counter++;
  }
  
  return counter;
}

As for sizeof giving you 4, that's perfectly normal. You aren't working with an array anymore (assuming buffer was declared as such originally). In value context, an array is converted to a pointer to the first element. And on your system, pointers are 4 bytes.

By the way, this:

Code:

int laes = 1;

probably won't work very well. laes is the name of the function.

Thanks. This was driving me insane.