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.

"manual enumeration"? Can you define that?, you're probably not going to escape loops. — Ryan
– Ryan, Commented Jun 2, 2016 at 21:10
@self printf(string-mask, param1,param2,param3, etc...) is a manual enumeration. — 0x1337
– 0x1337, Commented Jun 2, 2016 at 21:12
Using a loop is not mutually exclusive with printing it once. — user3386109
– user3386109, Commented Jun 2, 2016 at 21:20
Would you like the compiler to guess the parameters for you, so you don't have to enumerate them? ;-) — Martin Törnwall
– Martin Törnwall, Commented Jun 2, 2016 at 21:21

2501 · Accepted Answer · 2016-06-02 21:50:38Z

5

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

edited Jun 2, 2016 at 21:50

answered Jun 2, 2016 at 21:44

2501

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