Here is a Hello World code in C:
// a.c
#include
int main() {
printf(\"Hello world\\n\");
return 0;
}
I compile it
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
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.
$ 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.)
ld -lc a.o
There are several things wrong with this command line:
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).
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.