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views:

202

answers:

3

I'm confused as to why the following return separate sHA1s

$ perl -MDigest::SHA1 -E'say Digest::SHA1::sha1_hex("http://i.aultec.com/v/8066/Originals/1FTVX12585NA9832010.jpg");'
e1133fa3b7ea0bfb8ffa4d877932ed6c6fa10cef

$ echo "http://i.aultec.com/v/8066/Originals/1FTVX12585NA9832010.jpg" | sha1sum 
5c3731e83ae0184ed93b595b9f5604863dd331e6  -

Which one is right? Am /I/ doing it wrong?

$ perl -MDigest::SHA -E'say Digest::SHA::sha1_hex("http://i.aultec.com/v/8066/Originals/1FTVX12585NA9832010.jpg");'
e1133fa3b7ea0bfb8ffa4d877932ed6c6fa10cef

You can see the digest is right in the successor (Digest::SHA)

+13  A: 

Both are right. Your echo command includes a newline at the end. (and the perl string doesn't) Try with echo -n ...

viraptor
+3  A: 

This is such a frequent mistake and I've made it many times. The echo command is also returning a newline.

PP
echo is not "returning" a newline. It is outputting a newline, and (probably) returning 0.
William Pursell
+9  A: 

Perl is giving you the hash of the literal string you entered, whereas echo is appending a newline. If you tell echo to not add a newline, you'll get the same result:

drewfus:~$ perl -MDigest::SHA1 -E'say Digest::SHA1::sha1_hex("foo");'
0beec7b5ea3f0fdbc95d0dd47f3c5bc275da8a33
drewfus:~$ echo -n "foo" | sha1sum
0beec7b5ea3f0fdbc95d0dd47f3c5bc275da8a33  -
Drew Stephens