How to pass list of arguments in function with variable number of parameters?

Question

I have a string-mask that looks something like this:

                  +--\
                  |   \
                  |    \
              +---|     \
              +---|      \
+                 |       \
|\  +---------------------------------+\
| \ | %d|    %d|    %d|    %d|    %d| | \
|  \| %d|    %d|    %d|    %d|    %d| | |\
|   | %d|    %d|    %d|    %d|    %d| | | \
|---|                                 | |  \
|---|                                 | |  /
|   | %d|    %d|    %d|    %d|    %d| | | /
|  /| %d|    %d|    %d|    %d|    %d| | |/
| / | %d|    %d|    %d|    %d|    %d| | /
|/  +---------------------------------+/
+                 |       /
              +---|      /
              +---|     /
                  |    /
                  |   /
                  +--/

I need to printf it - printf(string-mask, param1,param2,param3, etc...), but number of parameters is a huge (in real string it is about 40). Is there way to avoiding manual enumeration of parameters?

P.S. I'm using pure C.

P.S.S. params are storing in array.

Answer 1

Iterate the array (the string), until you hit a print specifier. Then print the string from where you previously left of, to, including, the specifier, while passing a single argument from the array of values.

This is a quick and dirty solution without error checking that assumes every specifier is exactly %d and there are exactly param_count of them. Also the string must be modifiable.

const size_t param_count = 30;
char* first = string;
char* last = string;
for( size_t i = 0 ; i < param_count ; i++ )
{
    last = strchr( last , '%' ); //find the specifier 
    last += 2 ;  //skip the specifier
    const char temp = *last;
    *last = '\0';  //terminate the 'sub-string'
    printf( first , param[i] );
    *last = temp;   //restore the 'string'
    first = last;
}
printf( first ); //print the remaining string

Here is the output: https://ideone.com/zIBsNj