Fix path resolution for symlinks #23072
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If `/a/link` is a symlink to directory `/b/c` then `/a/link/..` should be resolved to `/b/c/..` which is `/b`. Currently, we cancel out `link/..` to get `/a`. The problem is that we apply `PATH_FS.resolve` and `PATH_FS.normalize` to paths. These functions cancel `..` incorrectly. This at least partially handles the situation. `lookupPath` is modified to avoid calls that call `PATH_FS.normalize()`. We check that `mkdir`, `open`, and `stat` now work correctly.
sbc100
reviewed
Dec 5, 2024
| if (!dir.endsWith("/")) { | ||
| dir += "/" | ||
| } | ||
| return dir + path; |
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How is this different to the old code?
| if nodefs: | ||
| self.skipTest('NODEFS in WasmFS') | ||
| self.set_setting('FORCE_FILESYSTEM') | ||
| self.do_runf('fs/test_symlink_resolution.c', 'success', emcc_args=args) |
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Can you name this file that same as the test (i.e. include the fs_ part)
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| assert(truncate("a/link/../x.txt", 0) == 0); | ||
| assert(chmod("a/link/../x.txt", 0777) == 0); | ||
| printf("success\n"); |
| @@ -826,7 +829,6 @@ var SyscallsLibrary = { | |||
| path = SYSCALLS.calculateAt(dirfd, path); | |||
| // remove a trailing slash, if one - /a/b/ has basename of '', but | |||
| // we want to create b in the context of this function | |||
| path = PATH.normalize(path); | |||
| if (path[path.length-1] === '/') path = path.substr(0, path.length-1); | |||
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Should this now be while instead of if to handle multiple tailing slashes?
sbc100
reviewed
Dec 5, 2024
| @@ -192,6 +194,12 @@ FS.staticInit(); | |||
| break; | |||
| } | |||
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| if (parts[i] === "..") { | |||
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Single quotes (here and elsewhere)
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If
/a/linkis a symlink to directory/b/cthen/a/link/..should be resolved to/b/c/..which is/b. Currently, we cancel outlink/..to get/a. The problem is that we applyPATH_FS.resolveandPATH_FS.normalizeto paths. These functions cancel..incorrectly.This at least partially handles the situation.
lookupPathis modified to avoid calls that callPATH_FS.normalize(). We check thatmkdir,open,stat,truncate, andchmodnow work correctly.