Linking a dynamically linked executable with ld

Question

i'm trying to create a dynamically linked executable (elf_i386) without gcc. The program is very simple (only a printf)...here the commands:

$ gcc -c simple.c
$ ld -o simple -dynamic-linker /lib/ld-linux.so.2 --entry main /usr/lib/crt1.o /usr/lib/crti.o simple.o -lc /usr/lib/crtn.o

The executable is created and also file command and ldd command show the right output... However when i launch the program after the call to printf i get a segmentation fault...i've examined with objdump the executable and i think the problem is about the dtors...seems that compiling with:

$gcc -o simple simple.c

a section .dtors is present while it is not present inside the executable created directly with ld :(

Any ideas?

You'll need to show your code if you want anyone to help.

johnsyweb
– johnsyweb

2011-05-24 11:02:59 +00:00
Commented May 24, 2011 at 11:02 — johnsyweb
– johnsyweb, Commented May 24, 2011 at 11:02

Robᵩ · Accepted Answer · 2011-07-11 21:38:50Z

7

Lose the --entry main. main isn't your entry point, _start is. Try this:

$ gcc -c hello.c
$ ld -o hello -dynamic-linker /lib/ld-linux.so.2 /usr/lib/crt1.o /usr/lib/crti.o hello.o -lc /usr/lib/crtn.o
$ ./hello
hello, world
$

answered Jul 11, 2011 at 21:38

Robᵩ

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2 Comments

ACyclic Over a year ago

And for x86_64 + C++, use this : ld -o hello hello.o --dynamic-linker /lib64/ld-linux-x86-64.so.2 /usr/lib64/crt1.o /usr/lib64/crti.o -lstdc++ -lc /usr/lib64/crtn.o -L/usr/lib/gcc/x86_64-redhat-linux/4.1.2/ -lgcc_s -L /usr/lib64 Replace the .../x86_64-redhat-linux/ dir with the right path to your libstd++.so, use "g++ -print-search-dirs"

Eduen Sarceño Over a year ago

thank you, the magic is -dynamic-linker /lib/ld-linux* /lib/crt*

Tarunn · Accepted Answer · 2016-09-24 08:48:03Z

2

It is not necessary to include the C run time environment i guess unless you are using return from your main().

We can strip the CRT and just link using :

ld -o hello -lc -dynamic-linker /lib/ld-linux.so.2 hello.o -e main

Will work.

edited Sep 24, 2016 at 8:48

Tarunn

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answered Sep 24, 2016 at 3:03

mahesh waldiya

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2 Comments

Peter Cordes Over a year ago

This works on Linux with glibc because glibc uses dynamic linking hooks to get its startup functions called. If you statically link glibc and call functions like printf or malloc without having manually called its init functions, you will segfault. stackoverflow.com/questions/36861903/…

Peter Cordes Over a year ago

Other platforms, like MinGW, don't have this, and you need to manually call libc init functions from _start before you can use some functions.

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