Seg fault, I do not understand

I am working on a project and one of the elements within the project is not working as expected. So I wrote a short program to test the functionality.

Code:

#include<fcntl.h>
#include <stdio.h>
#include <sys/stat.h>
#include <errno.h>

int main(int argc, char*argv ){
	//Assume that /home/student/m/marescal/Desktop/echo.c is a file.
	chdir("/home/student/m/marescal/Desktop");
	printf("%s, \n",get_current_dir_name());
	int v=open("/echo.c",O_RDONLY,0);//Put a file in your home directory here!
	if (errno==0){
		printf("Success");
	}
	else{ printf("an error has occurred \n");
	}
}

when I comment out the chdir() function it works fine (I am in the directory I am trying to access the file from) so I have 2 questions
1. This never gives the success message, why??
2. When the else is in there, it gives a seg fault.

Did you check the return from chdir() for an error?
(that is; did you change to the directory you wanted, if not you will not find the file you are looking for, so why try to open it?)

When using strings and you must use 2 slashes ('\\') to denote one slash (and the backslashes).

ie "c:\\SomeFolder\\AnotherFolder\\TheFile.ext"

This is because a backslash denotes a special character pair, the second in the pair denoting which special character it is (ie \t is TAB, \n is new line, \\ is \)

This appear to be a linux environment, so the \\ advice may not apply. That said, however, you are sure that the problem isn't "/echo.c"? Seeing as the "/" tells it to look in the root folder.

Yep! it looks like the o/p omitted the the leading dot while specifying the relative path ie "./echo.c"

nope, still does not work!

Then echo.c doesn't exist in your home dir; btw why aren't you checking the return value "v" from open() for a valid file descriptor instead of testing errno.

gcc -g mycode.c
gdb ./a.out

when gdb starts, type "run". You will find out exactly what causes the segfault.

Testing errno to determine if an error happened is incorrect. Instead you must check the return values of the functions themselves. ONLY IF the function returns an error are you allowed to check errno.

In other words, the fact that errno != 0 is totally meaningless since you haven't bothered to check the return status of open(). Of course, there are other problems in your code that were already pointed out.