what is zed doing here?

[Alex Sturza]

in this code?from where i need to understand that is deleting something? void Database_delete(struct Connection *conn, int id) { struct Address addr = {.id = id, .set = 0}; conn->db->rows[id] = addr;

[CPallini]

As far as I know, this is not valid C++ code.

THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?! -- C++ FQA Lite

[Alex Sturza]

this is C code!!!

[jeron1]

This[^] thread might help. It looks as though there are members of the Address struct one called set and it's getting initialized to 0, and another called id and it's getting initialized to the given integer id.

"the debugger doesn't tell me anything because this code compiles just fine" - random QA comment "Facebook is where you tell lies to your friends. Twitter is where you tell the truth to strangers." - chriselst

[Alex Sturza]

yes,but by the name of function says(database_delete),this function must delete something,and i don't know how it is deleting something

[jeron1]

Just a guess, perhaps set is some kind of flag marking a record as inactive or something, as opposed to actually deleting a record. What a functions name is and what it actually does may be two different things. What's needed here I believe is some useful comments.

"the debugger doesn't tell me anything because this code compiles just fine" - random QA comment "Facebook is where you tell lies to your friends. Twitter is where you tell the truth to strangers." - chriselst

[Alex Sturza]

can i give you the entire code?maybe you will find my answer:)

[Pete OHanlon]

This ties into the code you posted below. Specifically, it's related to this method:

void Database_set(struct Connection *conn, int id, const char *name, const char *email)
{
struct Address *addr = &conn->db->rows[id];
if(addr->set) die("Already set, delete it first");

addr->set = 1;

Basically, it looks like it's controlling whether or not you've already set the value in this row - so, to set a value here, you would call Database_delete first, which changes the value in id from 1 to 0, which means that the rest of Database_set can continue.

[CPallini]

Yes, you are right. :-O

THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?! -- C++ FQA Lite

[Alex Sturza]

i understand now,thank you very much!

[Pete OHanlon]

You're welcome. :-D