Using variables in CURL for urls.

Question

Hello, I am fairly new here and new to programming. I am trying to make a program that will ask for a url and store it in a character array and use that variable as the URL for the url argument, in quotations, using libcurl, of course.
curl_easy_setopt(curl, CURLOPT_URL, "<here>");
However, when I do it the way I would with printf()
curl_easy_setopt(curl, CURLOPT_URL, "%c", &...);
it says prg.c:12:53: error: macro "curl_easy_setopt" passed 4 arguments, but takes just 3
Is there a way to do what I am trying to do?
Last edited by fawx (2008-08-20 13:06:24)

Answer 1

I don't know what u really want to do. And what I found is that there is not an overloaded function like the one you are trying to use, in fact there is no overloaded functions for that one.
http://curl.haxx.se/libcurl/c/curl_easy_setopt.html
CURLcode curl_easy_setopt(CURL *handle, CURLoption option, parameter);
If you want to pass a URL that dynamically fills the parameter variable you can first create a string and copy the info from another one then concatenate another.
Here is an example that maybe could help you.
char url[128];
char *page = "hello.html";
strcpy( url, "www.google.com" );
strcat( url, page );
for this you must include string.h, then you can use url to send as parameter in your function.