`bash: ./a.out: No such file or directory` on running executable produced by `ld`

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温柔的废话
温柔的废话 2020-12-15 23:57

Here is a Hello World code in C:

// a.c
#include 

int main() {
    printf(\"Hello world\\n\");
    return 0;
}

I compile it

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  • 2020-12-16 00:39

    Use that:

      ld -o a.out -dynamic-linker /lib/ld-linux.so.2 /usr/lib/crt1.o /usr/lib/crti.o -lc c.o /usr/lib/crtn.o
    
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  • 2020-12-16 00:55

    The other answers only address how to avoid this, not the actual question of what happened.

    The gcc -c a.c; ld -lc a.o commands you gave produce a pretty obvious warning:

    ld: warning: cannot find entry symbol _start; defaulting to 0000000000400260
    

    So even if this file could be executed, it will probably crash right away. See @EmployedRussian's answer for an explanation of what you should have done.


    The question of why it can't even be executed is still interesting:

    $ strace ./a.out 
    execve("./a.out", ["./a.out"], [/* 72 vars */]) = -1 ENOENT (No such file or directory)
    

    execve(2) returns ENOENT because it can't find the interpreter (which I figured out from file and so on, see below). You'd get the same error from trying to run a file that started with

    #!/usr/non-existant-path/bin/bash
    

    As you discovered, the usual reason for this error message is when running an ELF binary on a system without the right dynamic linker and dynamic libraries installed (e.g. a 64bit system without 32bit support installed). In your case, it's because you used a bad link command and made a dynamic executable with a bad interpreter path.


    I'm on Ubuntu 15.10, where GNU file version 5.22 reports:

    a.out: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib/ld64.so.1, not stripped
    

    There is no /lib/ld64.so.1 on my system. ldd output is confusing, because ldd uses its default ELF interpreter, not the one specified by the binary.

    $ ldd a.out
            linux-vdso.so.1 =>  (0x00007ffc18d2b000)
            libc.so.6 => /lib/x86_64-linux-gnu/libc.so.6 (0x00007f0e0a79f000)
            /lib/ld64.so.1 => /lib64/ld-linux-x86-64.so.2 (0x0000559dbc9d2000)
    

    So it assumes that the runtime interpreter in the binary resolved to the one ldd used itself, I guess.

    Your ldd output is probably from an old version too, since it just shows /lib64/ld-linux-x86-64.so.2 for that line. Not taking a bad guess is probably better behaviour, for a weird case like this, but doesn't help you see that your binary has a weird interpreter path.

    readelf -l a.out
    

    will decode the ELF headers for you, including the interpreter path. (Thanks to @EmployedRussian's comment for pointing this out.)

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  • 2020-12-16 01:01

    ld -lc a.o

    There are several things wrong with this command line:

    1. In general, user-level code should never use ld directly, and always use appropriate compiler front end (gcc here) to perform the link.

      As you have discovered, the link command line that gcc constructs is quite complicated, and the command line that you've accepted in Joan Esteban's answer is wrong.

      If you want to see the actual link command, examine output from gcc -v a.o.

      Also note that link command changes significantly when you change gcc command only slightly (e.g. some OSes require different crt1.o depending on whether you are linking multi-threaded executable or not), and the command line is always OS-specific (which is one more reason to never use ld directly).

    2. Libraries should follow object files on command line. So ld -lc a.o is never correct, and should always be (a variant of) ld a.o -lc. Explanation.

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