How to diff a commit with its parent

Question

Aside from writing an alias or script, is there a shorter command for getting the diff for a particular commit?

git diff 15dc8^..15dc8 

If you only give the single commit id git diff 15dc8, it diffs that commit against HEAD.

Answer 1

Use git show $COMMIT. It'll show you the log message for the commit, and the diff of that particular commit.

Answer 2

Use:

git diff 15dc8^! 

as described in the following fragment of git-rev-parse(1) man page (or in modern Git gitrevisions(7) man page):

Two other shorthands for naming a set that is formed by a commit and its parent commits exist. The r1^@ notation means all parents of r1. r1^! includes commit r1 but excludes all of its parents.

This means that you can use 15dc8^! as a shorthand for 15dc8^..15dc8 anywhere in Git where revisions are needed. For the diff command, the git diff 15dc8^..15dc8 is understood as git diff 15dc8^ 15dc8, which means the difference between parent of commit (15dc8^) and commit (15dc8).

^{Note: the description in git-rev-parse(1) man page talks about revision ranges, where it needs to work also for merge commits, with more than one parent. Then r1^! is "r1 --not r1^@" i.e. "r1 ^r1^1 ^r1^2 ..."}

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Also, you can use git show COMMIT to get the commit description and diff for a commit. If you want only the diff, you can use git diff-tree -p COMMIT.