sprintf usage with %.s

Question

While using sprintf like this

sprintf("%.40s",str);

I want to give a value like strlen(str) in place of 40.How do I do that? I tried replacing 40 with strlen(str), doesn't work. Giving a value like

#define len strlen(str)

and using %.len doesnt work either since %.len is used in " ".

Answer 1

Use * character.

sprintf(/* ... */, "%.*s", (int) strlen(str), str);

If you use C99, snprintf can also be suitable.

man printf (3)
The width is not specified in the format string, but as an additional integer value argument preceding the argument that has to be formatted.

Answer 2

As GajananH points out this is a pointless endevor:

For strlen to work, str must be NULL terminated.

If str is NULL terminated then:

sprintf(str2, "%.*s", strlen(str), str);

Is just an over complication of:

sprintf(str2, "%s", str);

Which itself is just an over complication of:

strcpy(str2, str);