Difference between "printf" and "print sprintf"

Question

The following two simple perl programs have different behaviors:

#file1
printf @ARGV;

#file2
$tmp = sprintf @ARGV;
print $tmp;

$> perl file1  "hi %04d %.2f" 5  7.12345
   #output:  hi 0005 7.12

$> perl file2 "hi %04d %.2f" 5  7.12345
   #output:  3

Why is the difference? I had thought the two programs are equivalent. Wonder if there is a way to make file2 (using "sprintf") to behave like file1.

Answer 1

The builtin sprintf function has a prototype:

$ perl -e 'print prototype("CORE::sprintf")'
$@

It treats the first argument as a scalar. Since you provided the argument @ARGV, it was coerced into a scalar by passing the number of elements in @ARGV instead.

Since the printf function has to support the syntax printf HANDLE TEMPLATE,LIST as well as printf TEMPLATE,LIST, it cannot support a prototype. So it always treats its arguments as a flat list, and uses the first element in the list as the template.

One way to make it the second script work correctly would be to call it like

$tmp = sprintf shift @ARGV, @ARGV

Another difference between printf and sprintf is that print sprintf appends $\ to the output, while printf does not (thanks, ysth).

Answer 2

@ARGV contains the arguments passed to the script in list form. printf takes that list and prints it out as is.

In second example you are using sprintf with the array and assigning it to scalar. Which basically means it stores the length of the array in your variable $tmp. Hence you get 3 as output.

Answer 3

From the perl docs (jaypal said it already)

Unlike printf, sprintf does not do what you probably mean when you pass it an array as your first argument. The array is given scalar context, and instead of using the 0th element of the array as the format, Perl will use the count of elements in the array as the format, which is almost never useful.