socket programming in linux: send()

Question

I'm a beginner in socket programming in linux env.

the code is :

void proccess_server(int s)
{

    ssize_t size =0 ;
    char buffer[1024];
    for(;;)
    {
        printf("proccess:%d proccessing socket :%d\n",getpid(),s);
        size = recv(s,buffer,sizeof(buffer),0);

        if(0 == size)
          return ;
        sprintf(buffer,"Response from server:%d,%d bytes altogether\n",getpid(),size);

        send(s,buffer,strlen(buffer)+1,0);
    }

}

and I don't understand why len param in send() add another byte (strlen(buffer)+1) in send(s,buffer,strlen(buffer)+1,0);

please help me !

This code is all wrong, where did you find it ? It does not check for error conditions (size == -1), and it expect "buffer" to be null terminated without checking for it, so it has a security flaw. — BatchyX
– BatchyX, Commented Jan 1, 2012 at 11:12
er..It's a simple tutorial show the network programming api form my text book. and if I want to fix it ,should I fill buffer with \0 before receive data? — Crystal Cat
– Crystal Cat, Commented Jan 1, 2012 at 14:39

tcak · Accepted Answer · 2012-01-01 09:16:25Z

2

In C language, compiler puts a \0 (NULL) character at the end of every string. So while you are using a string, computer can understand where that string ends. In Pascal language example, compiler puts a byte front of string to store length of string.

This must be the reason why there is +1 there.

answered Jan 1, 2012 at 9:16

tcak

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2 Comments

Crystal Cat Over a year ago

I see,I understand there will be a \0 at the end of the string,as the question is ,+1 means the function strlen() return the length of a string not include \0?

tcak Over a year ago

Yes, that is right. If a string is "abc", strlen("abc") is 3, but in memory length is 4.

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