Chmod calculator

Both, and it works out which from what you type. Three or four octal digits convert to symbolic, so 755 gives rwxr-xr-x. A nine-character symbolic string converts to octal, so rwxr-xr-x gives 755. A leading file-type character is allowed and ignored, which means you can paste -rwxr-xr-x or drwxr-xr-x straight out of ls -l.

Yes, in both directions. A fourth leading octal digit carries them: 4 for setuid, 2 for setgid, 1 for sticky. In symbolic form they appear in the execute position as s or t when execute is also set, and as capital S or T when it is not, which is the convention ls uses. Reading the other way, an s, S, t or T is decoded back into that leading digit.

No. This converts between the two notations for a complete mode, not the relative form chmod accepts on the command line. u+x, go-w and a=r are all rejected with a message saying the input is neither a valid octal mode (e.g. 755) nor a symbolic mode (e.g. rwxr-xr-x). Write out the full mode you want instead.

Usually length or a stray character. After any leading d, l or dash, the body must be exactly nine characters, three each for user, group and other. Each triple is r or dash, then w or dash, then one of x, s, t, S, T or dash. A short string gives "expected a 9-character permission string (e.g. rwxr-xr-x)" with your input quoted back at you.

Yes. Three and four digits are both accepted, and a leading zero simply means no special bits are set. Each digit must be 0 to 7, so 855 and 79 do not parse, and five digits are not accepted either. When converting the other way, a mode with no special bits is printed without its leading zero.