Thread: Calling struct member error

Code:

typedef struct call_info_t
{
    char id[20];

} *call_info;


void cs_logging(char *logstr)
{
    char logline[50];
    snprintf(logline, sizeof logline, "%s%s%s, ":id:", call_info.id, logstr);

    FILE *file;
    file = fopen("output_trace.log","a+");
    fprintf(file, "%s\n", logline);
    fclose(file);
}

Why am I getting this error
error: expected expression before 'call_info'

Look at the syntax highlighting on the code you posted:

Code:

snprintf(logline, sizeof logline, "%s%s%s, ":id:", call_info.id, logstr);

Why does the blue string highlighting stop before :id:? Because you have an extra " in the middle of your string. If you want it there, you have to escape it with \". You should be using a editor that has decent syntax highlighting. It's immensely helpful.

Also, you declared call_info to be a pointer to struct call_info_t, so you need to use -> to access the members, not the dot operator.

Code:

snprintf(logline, sizeof logline, "%s%s%s", ":id:", call_info->id, logstr);

sorry, I missed up in pasting the code it should be "%s%s%s".

I have now tried with -> operator but it is still giving the same error message. Would it have anything to do with string formatting?

Doh! I misread the struct definition. You are actually defining a type, there (see the keyword 'typedef'?), not a variable. You created a new type, called call_info, that is a pointer to a struct call_info_t type. The -> operator expects a variable on the left side. You need to declare a variable of the appropriate type, and use it's name instead of call_info, e.g. variable_name->id or variable_name.id.

Note, most people frown on typedef'ing a pointer unless you're making a totally opaque data type, but I don't think that's the case here. Typedef'ing pointers just makes it harder to keep track of what's going on, how much dereferencing you should do, etc. It makes pointer bugs much more likely.