Why does setupterm terminate the program? - c

This is sample program from "Beginning Linux Programming" book:
#include <stdio.h>
#include <term.h>
#include <curses.h>
#include <stdlib.h>
int main()
{
setupterm("unlisted", fileno(stdout), (int *)0);
printf("Done.\n");
exit(0);
}
Running it, I have this result:
./badterm
'unlisted': unknown terminal type.
According to setupterm function definition, it must return 0: "No matching entry in terminfo database". Instead of this, program terminates. Why?

It looks like you asked it to do so. From man setupterm on my machine:
If errret is null, setupterm prints an error message upon finding an
error and exits. Thus, the simplest call is:
setupterm((char *)0, 1, (int *)0);
which uses all the defaults and sends the output to stdout.
Presumably, if you want to handle any error return yourself, you must supply a non-NULL pointer value for the errret (third) parameter.

Related

glibc: unable to get unwind information for certain sections

I am currently facing issue with the glibc v2.22 where I am not able to get the proper unwind information.
When there is SIGABRT application, it is calling abort function from glibc. It should be using unwind information which is enabled in the build. However, it is scanning the stack (as indicated by the red line below the address in the screenshot) and providing the misleading information as shown in the screenshot attached (using sentry for analyzing the dump).
Here, do_crash is called which does assert(0) which then aborts the main application. While analyzing the dump, the do_crash function calls the _fini which is never in the main application's stack.
I have enabled unwind for the glibc by using CFLAGS += "-funwind-tables". I also tried with the flags such as -rdynamic and -fno-omit-frame-pointer but it was also of no use.
Am I missing something here? How can I get the complete backtrace of the signals, particularly SIGABRT?
Thanks in advance
When there is SIGABRT application, it is calling abort function from glibc
That is not true, this is not happening, unless you explicitly registered it.
I have enabled unwind for the glibc by using CFLAGS += "-funwind-tables"
It tells the compiler to add the information, it doesn't "enable unwind". What exactly happens when compiling with -funwind-tables?
Here, do_crash is called which does assert(0) which then aborts the main application.
This is not related to receiving SIGABRT signal.
Am I missing something here?
I believe you are making wrong assumptions - that something is called on SIGABRT, that SIGABRT is sent on assert, that abort() is called on SIGABRT. Nothing is called on SIGABRT and the program is terminated when receiving SIGABRT by default (see man 7 signal), assert just terminates the program and doesn't raise SIGABRT, and abort() raises the SIGABRT signal, not receives it.
How can I get the complete backtrace of the signals, particularly SIGABRT?
Register a handler that will do that. See How to automatically generate a stacktrace when my program crashes .
#include <execinfo.h>
#include <signal.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
void handler(int sig) {
void *array[10];
size_t size;
size = backtrace(array, 10);
backtrace_symbols_fd(array, size, STDERR_FILENO);
_Exit(1);
}
int main(int argc, char **argv) {
signal(SIGABRT, handler); // install our handler
raise(SIGABRT);
}
If you want to print stacktrace on assert() that's completely different and you would overwrite the glibc handler for assert to do that:
#include <assert.h>
#include <execinfo.h>
#include <unistd.h>
#include <stdlib.h>
#include <stdio.h>
void print_trace(void) {
void *array[10];
size_t size;
size = backtrace(array, 10);
backtrace_symbols_fd(array, size, STDERR_FILENO);
}
// based on https://code.woboq.org/userspace/glibc/assert/assert.c.html
void __assert_fail(const char *assertion, const char *file, unsigned int line, const char *function) {
extern const char *__progname;
fprintf(stderr, "%s%s%s:%u: %s%sAssertion `%s' failed.\n",
__progname, __progname[0] ? ": " : "",
file, line,
function ? function : "", function ? ": " : "",
assertion);
print_trace();
abort();
}
int main() {
assert(0);
}

How do calls to the execvp system call work?

I was doing a research into the contents of another StackOverflow question and I thought it was a good time to brush up my knowledge of unix system calls.
While experimenting with execvp (WITHOUT fork on purpose) I ran into something that confuses me
I wrote 4 test programs
Program 1
#include <stdio.h>
int main() {
//printf("Doge\n");
execvp("ls");
printf("Foo\n");
return 0;
}
The program works as expected, the contents of the directory are printed and the Foo print statement is not
Program 2
However when I uncomment the first print statement and have the program be this
#include <stdio.h>
int main() {
printf("Doge\n");
execvp("ls");
printf("Foo\n");
return 0;
}
execvp returns a -1 and both print statements are issued. why?
Program 3
I vaguely remember having to use unistd.h when experimenting with unix system calls from college.
So I included it, but not execvp has a different signature and it needed some more args than just the name of the program. So I did this
#include <stdio.h>
#include <unistd.h>
int main() {
printf("Doge\n");
char *const parmList[] = {"ls", NULL};
execvp("ls", parmList);
printf("Foo\n");
return 0;
}
And this works. This has confused me. Why did exec work in the first program?
I also used This as a reference to the system calls.
Finally I wrote
Program 4
#include <stdio.h>
//#include <unistd.h>
int main() {
printf("Doge\n");
char *const parmList[] = {"ls", NULL};
execvp("ls", parmList);
printf("Foo\n");
return 0;
}
Which also works as expected.
Can someone explain what's going on?
With this snippet
#include <stdio.h>
int main() {
execvp("ls");
printf("Foo\n");
return 0;
}
you're invoking undefined behaviour. You're not providing the prototype for execvp which requires an argument list (null terminated) as a second parameter.
Using gcc without any warning option silently uses execvp as implicitly declared, and doesn't check parameters. It just calls the function. The function then looks for a second parameter and encounters... whatever is left of the call stack (or registers, depending on call conventions), that's why a previous printf call can change the behaviour.
Using gcc -Wall gives the following warning:
test.c:5:9: warning: implicit declaration of function 'execvp' [-Wimplicit-function-declaration]
execvp("ls");
Including the proper include (#include <unistd.h>) leads to:
test.c:6:9: error: too few arguments to function 'execvp'
execvp("ls");
^~~~~~
That's why you've got strange behaviour. Don't look further. Use execvp with 2 arguments, period. In your case "Program 3" is the way to go, and always set warning level to the maximum, if possible (gcc and clang: -Wall -Wextra -pedantic -Werror)

Seeing the output of my program

I am having a problem with a program of mine, as I cannot see the output display. Using a Dev C++ compiler to compile my C program, I debug it to see the output. However my program immediately terminates, so I can't see the output properly.
I ended my program with return 0, and Aldo tried getch(), but even with both endings my program terminates quick.
I want to know if my program endings are wrong, and if so what is the correct way to end a program?
you need the window stop to view the output, is it right?
if yes, include this library
#include <stdlib.h>
then add this line at the end of code:
system("PAUSE");
e.g
#include <stdlib.h>
#include <stdio.h>
int main()
{
/* do/print some thing*/
system("PAUSE");
}

How to use System(const char*) in TC++

Today , When i coding, met a question..my Code as follow:
#include<stdlib.h>
void main()
{
system("dir");
getch();
}
The question : The user Screen is nothing..Why ? where is my result?
If you want the output when using system, at least into something you can read in your application, you need to pipe the output:
system("dir > /tmp/output.txt");
FILE *f = fopen("/tmp/output.txt", "r");
char text[1024]; // max sizeof of 1 kb, any more and I'd consider using `malloc()` instead.
fread(text, 1, 1024, f);
printf("%s\n", text);
fclose(f);
There are some problems in your program, at least one of which has already been mentioned.
void main() should be int main(void).
As I recall, the Windows/DOS getch function is declared in <conio.h>; you should have a #include directive for it. Be aware that both <conio.h> and getch are non-standard.
Since main returns int, you should return an int result.
But none of these problems explain the problem you're seeing.
With these changes:
#include <stdlib.h>
#include <conio.h>
int main(void)
{
system("dir");
getch();
return 0;
}
This should work; it should show a directory listing of whatever directory your program runs in (which is determined by TC; I don't know the details).
It's possible that the program is running in an empty directory, which means the dir command wouldn't show any files, but it should still produce some output.
Try commenting out the system() call and adding a printf call (note the added #include <stdio.h>):
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
int main(void)
{
printf("Hello, world\n");
getch();
return 0;
}
This should open a console window, print "Hello, world" in it, and wait for you to type Enter.
If you still don't see any output (either no console window, or a console window with nothing in it), then you have a problem that's not related to the system() call. Most likely the problem has to do with the way you're using Turbo C (I presume that's what "TC" stands for).
The main function in every C program is supposed to return an int you are returning void
Change void to int:
#include<stdlib.h>
int main()
{
system("dir");
getch();
}
When I tested, the dir command ran in my console and printed to standard out.
May be he is the running the program directly in the Turbo C IDE and hence his output is not visible. If he runs the program directly from cmd line it works. I remember you need to run Alt - F5 or some other combination to see the output window in Turbo C++

Use of exit() function

I want to know how and when can I use the exit() function like the program in my book:
#include<stdio.h>
void main()
{
int goals;
printf("enter number of goals scored");
scanf("%d",&goals);
if(goals<=5)
goto sos;
else
{
printf("hehe");
exit( );
}
sos:
printf("to err is human");
}
When I run it, it shows ERROR: call to undefined function exit().
Also, I want to know how I can create an option to close the window in which the program runs? For example, I made a menu-driven program which had several options and one of them was "exit the menu". How can I make this exit the program (i.e. close the window)?
Try using exit(0); instead. The exit function expects an integer parameter. And don't forget to #include <stdlib.h>.
The exit function is declared in the stdlib header, so you need to have
#include <stdlib.h>
at the top of your program to be able to use exit.
Note also that exit takes an integer argument, so you can't call it like exit(), you have to call as exit(0) or exit(42). 0 usually means your program completed successfully, and nonzero values are used as error codes.
There are also predefined macros EXIT_SUCCESS and EXIT_FAILURE, e.g. exit(EXIT_SUCCESS);
exit(int code); is declared in stdlib.h so you need an
#include <stdlib.h>
Also:
- You have no parameter for the exit(), it requires an int so provide one.
- Burn this book, it uses goto which is (for everyone but linux kernel hackers) bad, very, very, VERY bad.
Edit:
Oh, and
void main()
is bad, too, it's:
int main(int argc, char *argv[])
Try man exit.
Oh, and:
#include <stdlib.h>
int main(void) {
/* ... */
if (error_occured) {
return (EXIT_FAILURE);
}
/* ... */
return (EXIT_SUCCESS);
}
The exit() function is a type of function with a return type without an argument. It's defined by the stdlib header file.
You need to use ( exit(0) or exit(EXIT_SUCCESS)) or (exit(non-zero) or exit(EXIT_FAILURE) ).
The following example shows the usage of the exit() function.
#include <stdio.h>
#include <stdlib.h>
int main(void) {
printf("Start of the program....\n");
printf("Exiting the program....\n");
exit(0);
printf("End of the program....\n");
return 0;
}
Output
Start of the program....
Exiting the program....
You must add a line with #include <stdlib.h> to include that header file
and exit must return a value so assign some integer in exit(any_integer).
In addition to return an exit code to parent process -
In UNIX, an important aspect that I think has been left out is, that exit() at first calls (in reverse order) all those functions, which were registered by atexit() call.
Please refer to SUSv4 for details.
on unix like operating systems exit belongs to group of system calls. system calls are special calls which enable user code (your code) to call kernel code. so exit call makes some OS specific clean-up actions before returning control to OS, it terminates the program.
#include <stdlib.h>
// example 1
int main(int argc, char *argv){
exit(EXIT_SUCCESS);
}
// example 2
int main(int argc, char *argv){
return 0;
}
Some compilers will give you the same opcode from both of these examples but some won't. For example opcode from first function will not include any kind of stack positioning opcode which will be included in the second example like for any other function. You could compile both examples and disassemble them and you will see the difference.
You can use exit from any part of your code and be sure that process terminates. Don't forget to include integer parameter.
Write header file #include<process.h> and replace exit(); with exit(0);. This will definitely work in Turbo C; for other compilers I don't know.
Bad programming practice. Using a goto function is a complete no no in C programming.
Also include header file stdlib.h by writing #include <iostream.h>for using exit() function. Also remember that exit() function takes an integer argument . Use exit(0) if the program completed successfully and exit(-1) or exit function with any non zero value as the argument if the program has error.
Include stdlib.h in your header, and then call abort(); in any place you want to exit your program. Like this:
switch(varName)
{
case 1:
blah blah;
case 2:
blah blah;
case 3:
abort();
}
When the user enters the switch accepts this and give it to the case 3 where you call the abort function. It will exit your screen immediately after hitting enter key.
Use process.h instead of stdlib and iostream... It will work 100%.

Resources