C getting a portion of string within a string

Question

I'm trying to write code that parses a HTTP GET request and checks if the "Host" is www.bbc.co.uk.

This is my working code:

char data[] = "GET /news/ HTTP/1.1\nHost: www.bbc.co.uk\nConnection: keep-alive";
    unsigned int size = strlen(data);

    if (size>3 && data[0] == 'G' && data[1] == 'E' && data[2] == 'T'){ //If GET Request
        int host_index = -1;

        for (int i=4; i<size-4; i++){
            if (data[i] == 'H' && data[i+1] == 'o' && data[i+2] == 's' && data[i+3] == 't'
                    && data[i+4] == ':' && data[i+5] == ' '){
                host_index = i+6;
            }
        }

        if ( host_index != -1 && size > host_index+11 &&
                data[host_index] == 'w' && data[host_index+1] == 'w' && data[host_index+2] == 'w' &&
                data[host_index+3] == '.' && data[host_index+4] == 'b' && data[host_index+5] == 'b' &&
                data[host_index+6] == 'c' && data[host_index+7] == '.' && data[host_index+8] == 'c' &&
                data[host_index+9] == 'o' && data[host_index+10] == '.' && data[host_index+11] == 'u' &&
                data[host_index+12] == 'k')
        {
            printf("BBC WEBSITE!\n");
        }

    }

I think this is a lot of code for not a lot. How can I make this code more compact?

[Please keep it to plain C. No 3rd party libs]

MANY THANKS!

It depends on how pedantic you want to be and this detail is entirely missing in the post. How much of the string needs to be in the expected format? If you do not care, just do if strstr(data,"www.bbc.co.uk")) printf("BBC WEBSITE!\n"); — chux
– chux, Commented Dec 4, 2014 at 21:02

Severin Pappadeux · Accepted Answer · 2014-12-04 19:09:26Z

4

Why don't you use strstr() ?

Split big string into chunks using strstr(), and then parse smaller chunks by separate routines

answered Dec 4, 2014 at 19:09

Severin Pappadeux

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Comments

Joseph Quinsey · Accepted Answer · 2014-12-04 21:53:35Z

Your code can be written more compactly as:

   if (!strncmp(data, "GET ", 4) && strstr(data, "\nHost: www.bbc.co.uk\n"))
       printf("BBC WEBSITE!\n");

However, while this may work 99.9% of the time, it doesn't handle arbitrary white space after the colon. Regular expressions would be helpful, but this would need a third-party library, which you can't have.

One solution is:

  if (!strncmp(data, "GET ", 4)) {
      const char *p = data;
      char buf[99 + 1];
      buf[0] = 0;
      while ((p = strchr(p, '\n')) && sscanf(++p, "Host: %99s", buf) != 1)
          ;
      if (!strcmp(buf, "www.bbc.co.uk"))
          printf("BBC WEBSITE!\n");
  }

Edit: The above solution permits any number of CRs and/or LFs after "Host:". I am unclear whether the HTTP/1.1 LWS (Linear White Space) permits this. If not, and to allow only zero or more spaces and tabs, change the sscanf to read:

   (sscanf(++p, "Host:%*[ \t]%99[^ \t]", buf) == 1 || 
    sscanf(p,   "Host:%99[^ \t]",        buf) == 1)

As you can see, it starts to get messy.

Blagovest Buyukliev · Accepted Answer · 2014-12-04 19:16:53Z

2

Using only standard library functions, you could do:

char data[] = 
    "GET /news/ HTTP/1.1\n"
    "Host: www.bbc.co.uk\n"
    "Connection: keep-alive";

char *found_host = strstr(data, "Host: ");

if (found_host != NULL) {
    found_host += sizeof("Host: ") - 1;

    char *end_of_host = strpbrk(found_host, "\r\n");

    if (end_of_host != NULL) {
        int equal = strncmp(found_host, "www.bbc.co.uk", end_of_host - found_host);
    }
}

Note that this does not account for an arbitrary amount of whitespace between the colon and the value.

answered Dec 4, 2014 at 19:16

Blagovest Buyukliev

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Comments

BLUEPIXY · Accepted Answer · 2014-12-04 19:15:19Z

1

char data[] = "GET /news/ HTTP/1.1\nHost: www.bbc.co.uk\nConnection: keep-alive";
unsigned int size = strlen(data);
char buff[size];
sscanf(data, "%*[^:]:%s", buff);
if(strcmp(buff, "www.bbc.co.uk")==0)
    puts("BBC");

answered Dec 4, 2014 at 19:15

BLUEPIXY

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3 Comments

chux Over a year ago

Maybe if (sscanf(data, "%*[^\n] Host:%s", buff) == 1) ...`?

BLUEPIXY Over a year ago

@chux It by the needs.

Yahya Uddin Over a year ago

can you explain the sscanf part or comment just that line

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