The difference between "printf" and "print sprintf"

Question

The following two simple perl programs have different types of behavior:

#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 thought these two programs are equivalent. I wonder if there is a way to make file2 (using "sprintf") behave like file1.

Answer 1

The built-in sprintf function has a prototype:

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

It treats the first argument as a scalar. Since you provided the @ARGV argument, it was forcibly converted to a scalar, passing the number of @ARGV elements @ARGV .

Since the printf function must support the syntax printf HANDLE TEMPLATE,LIST , as well as printf TEMPLATE,LIST , it cannot support the prototype. Therefore, he always considers his arguments as a flat list and uses the first element in the list as a template.

One way to make its second script work correctly is to call it as

 $tmp = sprintf shift @ARGV, @ARGV 

Another difference between printf and sprintf is that print sprintf adds $\ to the output, and printf doesn't (thanks, ysth).

Answer 2

@ARGV contains the arguments passed in the form of a list script. printf accepts this list and prints it as is.

In the second example, you use sprintf with an array and assign it to a scalar. This basically means that it stores the length of the array in the $tmp variable. Therefore, you get 3 as a result.

Answer 3

From perl docs (jaypal said it already)

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