See this page as a slide show
IOCCC One Liner
Step 1
The original code.
- OK, really, it’s not the original code.
- The original code no longer compiles. ☹
Step 2
Add some formatting, and we get:
Note the old-style function header.
Step 3
Add #include
Step 4
- convert
main arguments to new style
- Explicitly declare
main’s return type
Step 5
Give variables traditional names
Step 6
argv is an array, so it’s always non-zero,
so !argv==0, and !!argv==1.
Make this substitution:
Step 7
a[b] is the same as *(a+b) is the same as *(b+a)
is the same as b[a].
Step 8
- Replace
*argv with argv[0]
- Replace
**argv with argv[0][0]
Step 9
Replace !!<true quantity> with 1
Step 10
Convert to a while loop
Step 11
Pull argv[0][1] out of the while loop
Step 12
Assume that if execlp is ever called, it won’t return,
so pull it out of the loop.
Step 13
Since --argv is always true, make it a separate statement.
Step 14
The write call gets a length of zero after the first time,
so the loop is a red herring. Nobody ever looks at argc.
Step 15
Remove ++ and --, adjust the values of argv.
Step 16
Don’t have to zero out argv[1][0] any more.
Step 17
The trailing paren is just to confuse % in vi, so remove it
and adjust the check.
Step 18
The argument string is only needed if no arguments are given.
Step 19
Make a proper sentinel value
Step 20
main returns an int.
Step 21
Add comments
Why not the original code?
Original code:
for (v[c++]="Hello, world!\n)";
New code:
char msg[]="Hello, world!\n)";
for (v[c++]=msg;
- The original code had
argv pointing to a string constant.
- The new code has
argv pointing to a string variable.
- What’s the difference?