I referred this article to setup compiler and execution tool.
http://codingfox.com/1-9-how-to-configure-gcc-in-notepad-for-c11/
The problem I am facing is that when I try to compile the program I have made, compiler shows up for milliseconds and disappears.Thus, I am not able to see what's wrong in that program. I want compiler to stop by and wait for my response.
Code is as follows
header files used are stdio.h and conio.h
int main()
{
printf("Hello World");
return 0;
getch();
}
In case, there are more ways of what I am trying to do, please let me know.
Help will be appreciated.
Post your code. This is need to understand if it closes because of the progam is ended, or if compiler stops on errors. In any case the article gives you instruiction to run your code, but you can also compile it, as #Sourav suggested you. If the program simply ends, as #Igor wrote, you can add a getch() before the end of main and the execution stops and wait for an input.
Another way is to launch the compiled program into cmd shell: it will not close the shell at the end of the program.
The issue is that your output window disappears before you ever get a chance to look at it.
You took it a bit too literally to use getch at the end of your program.
We mean, "at the end of the program, before the return statement"
What you wrote
int main()
{
printf("Hello World"); // This does exactly what you expect
return 0; // Your program ends here.
getch(); // This line is NEVER run.
}
What you should write
int main()
{
printf("Hello World"); // This does exactly what you expect
getch(); // Pause the program for input.
return 0; // Your program ends here.
}
Related
I've come across this weird situation i can't wrap my head around. In the following code, the Program is apparantly enetering the infinite loop, but doesn't execute the code that comes before it.
#include <stdio.h>
#include <stdlib.h>
int main() {
char buf[100];
if (scanf("%s", buf)==EOF) return 0;
printf("This does not get printed");
while(1) {} //infinite loop
return 1;
}
Somehow, the printf command does not get executed, even after pressing enter or using Ctrl-d.
However, it seems as if the code would end up in the infinite loop.
Does anyone explain whats going on here? I'm using gcc.
Standard output is line buffered by default. If you write less than a line to the output, some or all of it can be delayed until more output occurs.
I'm a novice programmer getting introduced to C and I'm missing something fundamental about the way my scanf() works. I want to read a single int from the keyboard with code like this:
#include <stdio.h>
#include <stdlib.h>
int main(void) {
int userBookSelection;
scanf("%i", &userBookSelection);
printf("Printing userBookSelection: %i", userBookSelection);
return EXIT_SUCCESS;
}
When I run the code, the console stays black until I stop debugging. There is never a cursor waiting for keyboard input. When I stop debug I can see this output in the console, same every time:
Printing userBookSelection: 2130567168
I'm debugging in Eclipse with MinGW GCC compiler on Windows. The code syntax seems to be correct -- is it possible there's something wrong in my build path to make this happen? I need to know why scanf() isn't reading for keyboard input.
So I've gotten a line of code from my professor which takes care of this bug -- whether it's a necessary solution particular to Eclipse and/or MinGW I'm not sure. In any case, here's the code with the additional line:
int main(void) {
int userBookSelection;
setvbuf (stdout, NULL, _IONBF, 0);//<---The magic line
scanf("%i", &userBookSelection);
printf("Printing userBookSelection: %i", userBookSelection);
return EXIT_SUCCESS;
}
I'd appreciate any additional wisdom on what's going on, what setvbuf() is doing and how scanf() works more fundamentally.
so I have using lex tool in linux and fell into an embarrassing position position , I couldn't break from a while loop.
e.g : I wrote ;
while(1)
{
int x = yylex();
switch(x):
case(ID):printf("ID");
case(NUM):printf("NUM");
}
Now I am reading from a file using yyin ; the problem is that the while loop does not fail after reading the whole file and keeps asking for more inputs at the terminal . Therefore I could not invoke other functions after calling the while loop . I know I am missing something basic , it will be great if somebody can provide an insight into this problem .
P.S# Thanks all for your answers ; tension and lack of sleep before homework submission date ; btw - I figured my answer already .
I see some issues with the C code, independent of what's going on inside the yylex() function. You quote the code as:
while(1)
{
int x = yylex();
switch(x):
case(ID):printf("ID");
case(NUM):printf("NUM");
}
This clearly isn't the source you compiled because it is not acceptable as C. You need to replace the first colon (after switch(x)) with an open brace {, and you need another close brace } at the end, to yield:
while(1)
{
int x = yylex();
switch(x)
{
case(ID):printf("ID");
case(NUM):printf("NUM");
}
}
This is syntactically valid C code, but there are still problems:
There should probably be a break; after each case.
There should probably be a default: clause, which might be used to terminate the loop (but you can't use break to do that inside the scope of a switch).
The printf() statements should print a newline so that the data appears.
Since yylex() returns 0 when it reaches the end, you should probably revise the code so it is more like:
void function(void)
{
int x;
while ((x = yylex()) != 0)
{
switch (x)
{
case ID:
printf("ID\n");
break;
case NUM:
printf("NUM\n");
break;
default:
printf("Other: %d\n", x);
break;
}
}
}
This will at least show you what's going on, and does not have an infinite loop unless you've written your lex analyzer such that it does not terminate properly. I opted not to have the default case terminate the loop since the loop is bounded by yylex() returning 0 instead.
When you write while (1), you are writing an infinite loop. It is incumbent on you to consider whether the loop is really infinite or not. If it is not, then you should aim to be able to make a test at the top of the loop that controls whether there is another cycle to the loop or not.
yylex() returns 0 at the logical end of input. Try this:
int x = -1;
while(!(x == 0))
{
x = yylex();
switch(x):
{
case(0): printf("that's all, folks!"); break;
case(ID): printf("ID"); break;
case(NUM): printf("NUM"); break;
}
}
Well, while I'll admit I'm not familiar with what lex is, I do know quite a bit about the linux environment and the terminal, so...
Try typing 'Ctrl' and 'c' at the same time while in the terminal. If that doesn't work, use 'Ctrl' + 'z'. If you don't mind having the process stopped in the background until the computer turns off and it goes away entirely, after Ctrl + Z you don't need to do anything else. If Ctrl + C worked, then the program's ended and you're fine.
If, for whatever reason, neither work, if you know the name of the program you wrote, you can type into the terminal:
pkill -9 [insert name of program here]
...which will send an unignorable kill signal to the process, and return all its memory (RAM- it won't undo changes made to disk) to the parent process.
If you DON'T know the name of the program, but you think you'll probably be able to recognize it, you can try:
top
which will bring up all the currently running processes in the terminal. Once you locate the name of your still-running program, press q to exit top, and then type in the kill command above with the name you recognized in top. If you don't know the program's name, and Ctrl + Z worked, but you don't want the stopped job to keep hogging its memory until you've turned off the computer, you can type:
ps -a
while in the terminal session from which you ran (and stopped) the program, and the job which shows up labeled 'stopped' is, in all likelihood, yours. Again, using the pkill command above will kill it. If, for whatever reason, you know the process ID but not the name (I don't know why you would, but there you go...), you could use the command 'kill' instead of 'pkill' (instructions for how to use these commands are all over the internet so I won't mention them here. They also have helpful help manual pages which can be accessed via:
man [the name of the program you want help with; 'pkill' or 'kill', in this case.]
To add to David's answer, here's a different version:
do
{
int x = yylex();
if (!x)
break;
switch(x):
{
case(ID): printf("ID"); break;
case(NUM): printf("NUM"); break;
}
} while(1);
This was only to show that you can break out of a while(1). You keep the x in local scope, and you don't evaluate the first pass or switch unnecessarily. Of course someone will most likely pull out the trump card of "it isn't as readable"...
I have written a program in c. However once the program is finished it stops (duh). Is there a simple script that allows me to let the user start the program over again?
Why not use loop (for, while) in the main itself: ( if the program is simple!)
main()
{
while( Exit condition)
{
//logic
}
}
char cont_prog = 'n';
do {
/* main program in here */
printf("Do you want to start again? (y/n): ");
cont_prog = getchar();
} while (cont_prog == 'y' || cont_prog == 'Y');
Essentially, you want to put you main prog in a loop, asking the user if they want to continue. You have to deal with the user entering in too much data (they type, 'yes', for example) and your buffer being full next time through the loop.
If you really want to re-launch the program without exiting (though I can't see why):
Save argv (and I'll assume that argv[0] actually points to your executable, even though that is not guaranteed) if you want the same command line arguments.
Consider saving the environment, if you might change it, and also want it to be repeated.
man execv or execle. Just replace the currently running image with a new one that has the same command line
Frankly, looping would be easier, and can have the same semantics if you avoid global state, or arrange to be able to re-set it.
Sure, but how would you get the user to run that script? Wouldn't it be simpler to have the user simply re-run the program?
#include <stdlib.h>
#ifdef WIN32
#define EXECUTABLE ".exe"
#else
#define EXECUTABLE
#endif
int main(void) {
for (;;) system("executable_in_c" EXECUTABLE);
return 0;
}
Compile this program, rename your old executable to "executable_in_c[.exe]"; rename this one to the name of your old executable ... voila!
can anyone tell me whats wrong in the following program that accepts 1 or 2 digit integers untill it encounters the number 42 after which it prints the previously entered numbers??when i upload this to the sphere online judge site it says compilation successful but runtime error (SIGSEGV).
#include <stdio.h>
int main()
{
int i;
FILE *fp;
fp=fopen("\\db.txt","w+");
if(fp==NULL)
{
printf("file not exitsts and cant be created");
system("exit");
}
while(1)
{
scanf("%d",&i);
if(i==42)
break;
else
{
fprintf(fp,"%d\n",i);
}
}
fclose(fp);
fp=fopen("\\db.txt","r+");
if(fp==NULL)
{
printf("file not exitsts and cant be created");
system("exit");
}
fscanf(fp,"%d",&i);
printf("%d\n",i);
while((!feof(fp)))
{
fscanf(fp,"%d",&i);
if(!feof(fp))
printf("%d\n",i);
}
fclose(fp);
return 0;
}
It seems like you're trying to answer this: http://www.spoj.pl/problems/TEST/ . This problem certainly does not require you to read or write anything from a file, and their server may not allow you to open files on its disk. Even if it does, you're trying to use a windows-style path (with a backslash) on what may be a non-Windows server. And even if it does allow file creation and windows-style path separation, you are trying to create your file in the filesystem root directory, and they almost certainly do not allow file creation there.
Combined with the system("exit") issue that everyone pointed out where it doesn't actually exit the program, this will cause you to receive a NULL file pointer and crash when you try to use it.
Re-read the problem description - you're over-thinking it. It doesn't say anywhere that you have to wait until you get a 42 to print out the other numbers. All you have to do is print back what is entered until you get a 42. That should make the solution much simpler. It's not supposed to be even a mildly challenging problem; it's just supposed to familiarize you with their system.
I don't know what you think:
system("exit");
will do, but the way to exit a program in C is:
exit(1);
You should replace
system("exit");
with
exit(1);
or, because you're already in main:
return 1;
I suspect the SIGSEGV is caused because you cannot write to the file \\db.txt, but the program continues because system("exit") is not causing it to terminate.
On an semi-related note, SIGSEGV is usually a Unix signal, and path separators on Unix are / rather than \.
I don't know precisely the cause of the SEGV, but I guess it is because the input doesn't match what you expect. In any case, this line doesn't do what you think it does:
system("exit");
at which line do you receive the error?
is your empty #include intended? i think it should mean #include
have you got the error for every input or just for 42?
regards
SIGSEGV is an access violation error, which indicates a null pointer. Since system("exit") isn't doing anything, fp is getting set to null, and then when you try to use that pointer (for example with fprintf())... boom, your program crashes.
Replace system("exit") with return 1 (or whatever error code you desire), that should fix it.
$ gcc -Wall test.c -o test
test.c: In function ‘main’:
test.c:8: warning: implicit declaration of function ‘system’
$ ./test
1
2
3
10
42
1
2
3
10
But yes, I do agree that the system("exit") does not what you expect. What you are exiting from, with that call, is a subshell that is spawned by your program, and then goes on. From the system man page
The system() function hands the
argument command to the command
interpreter sh(1). The calling
process waits for the shell to finish
executing the command, ignoring SIGINT
and SIGQUIT, and blocking SIGCHLD.