Question

I would want to know how is it possible to escape “%*s” in a sprintf call.
For example I have the following piece of code:

sprintf(log_buffer,"1234 567 89");

strcpy(format,"%*s"); 

sprintf(format1, " %%%ds %%%ds ",5,5); 

printf("\n Format is : %s ",format); 

printf("\n Format1 is : %s ",format1); 

strcat(format,format1); 

printf("\n new format is : %s ",format); 

sscanf(log_buffer,format,name,name1);

printf(" Name is 1: %s \n",name);

printf(" Name is 2: %s ",name1);

This works fine.

It gives me:

Format is : %*s 

Format1 is : %5s %5s

new format is : %*s %5s %5s

Name is 1: 567 

Name is 2: 89

Can I possibly not use strcpy and ignore the %*s within the sprintf itself? I tried escaping it inside sprintf but I guess it’s not proper as I get a Segmentation fault.

sprintf(format1, " \%*s %%%ds %%%ds ",5,5);

Answer 1

It looks like you already know how to escape % signs in sprintf (by prefixing with another %) – you’re doing it correctly for %%%ds.

What you want is:

sprintf(format1, " %%*s %%%ds %%%ds ",5,5);

Edit: The reason it’s segfaulting is because \%*s is interpreted as a literal backslash, followed by a NULL-terminated string; so sprintf is trying to interpret 5 as a NULL-terminated string, which it patently isn’t. It’ll be running off into forbidden memory looking for a NULL-byte, and segfaulting.