I'm currently working on an implementation of MD5 just because I'm curious and want to get a better understanding of how hashing works. I have 1 file with all the code in it, called easy_md5.c and I'm compiling it with the following Makefile:

easy_md5: easy_md5.c
    gcc -c easy_md5.c -o easy_md5

clean:
    -rm -f easy_md5.o
    -rm -f easy_md5

I ran chmod x easy_md5 on the executable to give it permission, and yet I'm getting the following error when I try to run it:

[gabe@void easy_md5]$ ./easy_md5 hello
bash: ./easy_md5: cannot execute binary file: Exec format error

Running file easy_md5 shows the following:

easy_md5: ELF 64-bit LSB relocatable, x86-64, version 1 (SYSV), not stripped

and running arch confirms my machine has x86-64 architecture. Since its a 64-bit executable, there's no reason it should throw a format error when I try to run it on my 64-bit machine. What is going wrong?

CodePudding user response：

You didn't create an executable at all. The -c option tells gcc to compile only, and output a relocatable object file. You'd normally name that file easy_md5.o and use it when linking, but the -o option forced it to be named easy_md5 which looks like an executable name even though it isn't.

To compile and link in one step, just drop the -c: gcc easy_md5.c -o easy_md5.

There were a couple other hints:

I ran chmod x easy_md5 on the executable

When you actually create an executable, this isn't necessary. gcc sets execute permissions on it for you.

Running file easy_md5 shows the following:

easy_md5: ELF 64-bit LSB relocatable, x86-64, version 1 (SYSV), not stripped

This indicates an object file. An actual executable is reported as something like

a.out: ELF 64-bit LSB shared object, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, BuildID[sha1]=7108d6b0e7767b20e8db974561938d3bfcbbe12a, for GNU/Linux 3.2.0, not stripped