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Asked: 2026-05-10T15:05:09+00:00

1167 ptr = (void*)getcwd(cwd, MAX_PATH_LENGTH-1); (gdb) n 1168 if (!ptr) { (gdb) print ptr

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1167      ptr = (void*)getcwd(cwd, MAX_PATH_LENGTH-1); (gdb) n 1168      if (!ptr) { (gdb) print ptr $1 = 0xbff2d96c '/media/MMC-SD/partition1/aaaaaaaaaaa' (gdb) print &cwd $2 = (char (*)[3500]) 0xbff2d96c (gdb) print strlen(cwd) $3 = 36 (gdb) print '%s',cwd $4 = '/media/MMC-SD/partition1/aaaaaaaaaaa', '\0' <repeats 912 times>, '��O�001\000\000\000\000��027\000\000\000�3����EL鷠3�000��027\000\000\000\000\000\000\000\027\000\000\000\000��/�027\000\000\000�3����N����\230���鷠3�000��027\000\000\000\000\000\000\000��000\000\000\000\001\000\000\000��M鷠3����\000\000\000\000.\231�027��w\005\b\001\000'... (gdb) print '%s', ptr $5 = 0xbff2d96c '/media/MMC-SD/partition1/aaaaaaaaaaa' (gdb) Quit

Why is ptr printing the string correctly but cwd not; this also affects the program and it crashes if I try to use the cwd…

[edit: turns out that crash was caused by a stupid buffer overflow on this var… grr…not gdb, but the print question was still valid]

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1 Answer

  1. The reason that cwd is printed differently in gdb is because gdb knows that ptr is a char * (I guess) and that cwd is an array of length 3500 (as shown in your output). So when printing ptr it prints the pointer value (and as a service also the string it points to) and when printing cwd it prints the whole array.

    I don’t see any reason why using cwd instead of ptr would lead to problems, but I would need to see some code to be sure.

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