I'm learning c++ again after having not touched it for the last few years, and I've run into a rather peculiar bug that I can't seem to figure out.
When I run the code below, it will accept 10 inputs as I expect it to, but immediately after the first for loop, the program exits. I have run it ingdbto try and figure out the issue, but it reported that the process 'exited normally'.
I compiled using g++ -std=c++11
#include <iostream>
#include <string>
using namespace std;
int main() {
//User inputs
string input[10];
//Get the inputs
for(int i = 0; i < 10; i++) {
//Get this input
printf("%i> ", i);
getline(cin, input[i]);
}
//The application does not make it to this point
//Collected
printf("Thank you for submitting your data.\n");
//Print inputs
for(int a = 0; a < 10; a++) {
//Show the input
printf("%i> %s\n", a, input[a].c_str());
}
}
Based on what you're describing, it sounds like stdout is not being flushed before the program ends. That's unusual; normally, stdout is automatically set up for line-buffered operation in which case it will be flushed as soon as a newline is encountered.
Your best bet is to follow #NathanOliver's advice and use cout << ... rather than printf. The printf command is a throwback to C, and you're using a C++ compiler and C++ features. In fact, you're not even including the header that's usually required for printf, so I'm a little surprised it even compiles.
FWIW, if you choose to continue using printf maybe try manually flushing stdout at the end like so:
fflush(stdout);
Your application does what is supposed to do:
Application will pause at getline() (10 times) (because getline is blocking execution), then it will do some for loops, print something and end (probably closing console window). If you add something at the end to block execution (such as cin.get(), which waits for enter key press) you will see results (because application won't terminate). If you run your code somewhere where output is not removed after program has ended you will see what was printed.
Related
The code is as follows, when I enter "101010^Z000", my output becomes "000". Obviously, my input is invalid after it becomes "^Z". However, why can I continue typing after typing "^Z"? According to the code, shouldn't it have jumped out of the loop and ended the program at this time? I'm curious.
int main()
{
const int num = 20;
int a[num];
int i = 0;
while((a[i]=cin.get())!=EOF)
{
a[i] = cin.get();
cout.put(a[i]);
i++;
}
cout << a;
}
like this:
And, after this I keep typing "ssss" and the program still outputs "ss" as if the loop is continuing.
Input is usually buffered. There is nothing in C++ that says it must be buffered but usually it is. What this means is that when your program is waiting for input it waits for a whole line of input. That whole line of input goes into a buffer and subsequent reads take characters from the buffer until it is empty. Then the next read will cause the program to wait again, and again it will wait for a whole line of input to be entered.
If you want unbuffered input then I've afraid there is no way to get that in standard C++. You have to use platform specific functions for that.
I've just stepped into the world of competitive programming and therefore, was reading few articles to grasp the best strategies a programmer should follow while writing code in c++.
I read somewhere that one should use '\n' in lieu of std::endl at the end of the line because std::endl forces the buffer to flush.
I didn't know what 'flushing of buffer' meant so I started searching for it and came across this answer What does flushing the buffer mean? After reading the answer the idea that I got 'flushing the buffer' was to print & wipe out everything in the output buffer to the console.
So, to check the validity of the answer I made this program on my machine.
#include <iostream>
#include <time.h>
using namespace std;
int main()
{
struct timespec tim, tim2;
tim.tv_sec = 3;
tim.tv_nsec = 0;
for(int i=0; i<5; i++)
{
cout << i << endl;
nanosleep(&tim , &tim2);
}
return 0;
}
The results I got on executing the program came as expected. The loop printed 0 to 4 with a delay of 3 seconds after each iteration.
However, when I changed endl to '\n' I was expecting the results to show on the console all at once after 15 seconds. But, the digits got outputted after the same delay it got outputted in the previous case. Now, if '\n' doesn't flush the output buffer, shouldn't I've got the results as I was expecting?
std::endl forces a flush of the stream. This does not mean that no flush will happen without it.
In general, the standard library will flush std::cout (and, yes, other streams are treated differently) even if the output buffers are not full if std::cout points to a tty and the buffers ends with \n. In your second case, that is precisely what's happening.
If you want to see your code working as you expect it, try redirecting the output to a file (and then watch the file with tail -f). That will cause the standard output not to be a tty, and thus not cause a flush on \n.
Also, IIRC, std::cerr is flushed on \n whether it is a TTY or not. If you write to a file, it should never flush unless you explicitly flush (e.g., with std::endl).
Edited to add
I couldn't get it to work with either stderr or stdout, but the following does print all numbers at the end:
#include <iostream>
#include <fstream>
#include <time.h>
using namespace std;
int main()
{
struct timespec tim, tim2;
tim.tv_sec = 3;
tim.tv_nsec = 0;
ofstream out("output");
for(int i=0; i<5; i++)
{
out << i << "\n";
nanosleep(&tim , &tim2);
}
return 0;
}
You dont have full control over when the buffer is flushed. Some systems flush it when you insert a \n into the stream, others dont.
Just because flush does flush the stream it does not mean that not calling flush will cause the stream to not flush.
I have this strange problem with stdout/stderr.
I want to apologize for not being able to put here the original code, it's too long / too many libraries dependent etc...
So please let me know if you ever encountered anything like it, or what may cause this issue without getting the original code, only the idea and examples of the simple things I tried to do:
I'm using g++ (GCC) 4.4.6 20120305 (Red Hat 4.4.6-4) on RHEL 6.3
I couldn't isolate the problem for putting it here, I'll give code examples of what I did.
fprintf() / printf() / std::cout stops working after a while.
I'm using boost::asio::io_service with deadline_timer in order to call a my_print() function.
This my_print() function prints to screen every 1 second some information.
In order to print, I use alignments, like the following:
fprintf(stdout, "%*s\n", -printWidth, someEnumToStr[i]);
fprintf(stdout, "%s\n", aString);
fprintf(stdout, "%u\n", num);
While aString is a std::string. Sometimes I construct aString from std::ostringstream.
Sometimes I construct it with snprintf().
I have an std::map with information, exactly 16 elements inside the map. I iterate over it, and for each element I try to print data with the example of fprintf() above.
For an unknown reason, the line of element 16 isn't printed.
If I call the executable, and redirect stdout to a file (./a.out > aaa.txt) the line of element 16 is getting printed.
If I open a new FILE* and fprintf() to this file, again, everything is getting printed (all lines, including line of element 16)
Before using fprintf() I tried to use std::cout (and alignments with std::cout.width(printWidth) << std::left...), The same behavior happened, but when line 16 wasn't drawn, stdout got stuck (I mean, the program still worked, but nothing was printed to stdout never again. I had to call std::cout.clear() for it to work again). Since a point in the code, which I couldn't lay my hands on, std::cout.failbit and badbit were 1.
If I run the code with valgrind this behavior doesn't happen. valgrind doesn't say anything wrong.
If I run it with gdb it happens, but gdb doesn't say anything wrong.
If I run it in an IDE (clion) in debug mode, it doesn't happen.
If I run it in IDE, without debug, it happens.
I figure it depends on the printWidth I give for the alignment in fprintf() - When printWidth is bigger, it happens sooner (when it's smaller, line 16 is randomly getting printed).
Another important thing: it happens more frequently when there is more to print.
I tried to give std::cout a bigger buffer (not his default) and it didn't work.
I tried to buffer all of the output into a buffer (instead of printing each line), then to only fprintf() once. Same behavior happens.
I didn't find anywhere in the code I try to print a NULL pointer.
I print with \n every couple of fprintf()s, and do fflush() in the end of my_print()
Please let me know if you know anything.
Illustration:
deadline_timer..... every 1 sec... my_print()
boost::asio::io_service.run
my_print() {
for(std::map<>::iterator... begin, end, ++it....) {
fprintf()s....
}
}
Non printable characters may be breaking terminal.
fprintf(stdout,"%s", astdstring.cstr() );
Is how to print std::string
I use boost::asio, I have a callback to read from stdin. this read is nonblocking - happens with async_read_some().
The problem was stdin was turned to be nonblocking, and it also caused stdout to be nonblocking as well because they point to the same file description (explanation).
It caused the fprintf() calls to fail (returned -1 with errno 11) and not all of the output got printed out on the screen.
It has no relation to boost.
I succeeded isolating the problem, the following code creates this problem:
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <fcntl.h>
#include <errno.h>
#include <stdlib.h>
using namespace std;
int main(int argc, char *argv[]) {
const int problem = 8000;
const int myBuffSize = 32000;
char *myBuff = new char[myBuffSize];
int myoffset = 0;
memset(myBuff, '-', myBuffSize);
int flags;
bool toogle = true;
bool running = true;
// Comment from here
if ((flags = fcntl(STDIN_FILENO, F_GETFL, 0)) < 0) {
printf("error fcntl()\n");
return 0;
}
if (fcntl(STDIN_FILENO, F_SETFL, flags | O_NONBLOCK) < 0) {
printf("error fcntl()\n");
return 0;
}
// Comment until here
while(running) {
toogle = toogle ? false : true;
if (toogle) {
snprintf(myBuff + problem, myBuffSize - problem, "fin\n\n");
} else {
snprintf(myBuff + problem, myBuffSize - problem, "end\n\n");
}
fprintf(stdout, "%s", myBuff);
sleep(1);
}
delete[] myBuff;
return 0;
}
If you'll comment the // Comment from here to // Comment untill here, it will print all of the output (fin and end will be printed).
One solution to this problem is to open another fd to the current tty using fopen(ttyname(STDOUT_FILENO), "w") and to print into it.
I believe another solution is to async_write() into screen.
The output might be stuck in a buffer, and not flushed before program termination.
Try adding exit(0) at the end of the program, and see if it helps.
http://www.cplusplus.com/reference/cstdlib/exit/
All C streams (open with functions in <cstdio>) are closed (and flushed, if buffered), and all files created with tmpfile are removed.
In case the title is not precisely revealing what I want to do, the following is my problem.
I want to write a c++ program in a Linux or Mac terminal to print numbers that keeps counting from 1, 2, 3 ... at the same position under a command line mode. For example, it is like displaying the number of percentage when your work is progressing(e.g. downloading something, installing software...).
I wrote a simple for-loop to print numbers and use usleep(1000); for a delay of 1 second before printing next number. Then I use cout << "\b"; trying to move cursor back to display coming number at the same position. However I fail to create the effect that I want, the numbers are printed in a line.
I am not a skillful c++ programmer and know very limited on programming in a terminal environment. Can anyone help to give me hint or sample code for this function? Thanks!!
If you are in Linux Terminal you can also use the following code,
system("clear");
cout<<"\b";
cout<<Your_Number;
// Repeat this in a loop and call the delay function
This works in terminal for me (am using linux)
#include
#include
using namespace std;
int main(int argc, char *argv[]) {
int i;
for(i=1;i<100;i++)
{
cout<<"\b\b\b"<<i;
cout.flush();
sleep(1);
}
return 0;
}
One of the first things I learned in C++ was that
#include <iostream>
int main()
{
std::cout<<"Hello, World!\n";
return 0;
}
would simply appear and disappear extremely quickly without pause. To prevent this, I had to go to notepad, and save
helloworld.exe
pause
ase
helloworld.bat
This got tedious when I needed to create a bunch of small test programs, and eventually I simply put while(true); at the end on most of my test programs, just so I could see the results. Is there a better wait function I can use?
you can require the user to hit enter before closing the program... something like this works.
#include <iostream>
int main()
{
std::cout << "Hello, World\n";
std::cin.ignore();
return 0;
}
The cin reads in user input, and the .ignore() function of cin tells the program to just ignore the input. The program will continue once the user hits enter.
Link
Please note that the code above was tested on Code::Blocks 12.11 and Visual Studio 2012
on Windows 7.
For forcing your programme stop or wait, you have several options :
sleep(unsigned int)
The value has to be a positive integer in millisecond.
That means that if you want your programme wait for 2 seconds, enter 2000.
Here's an example :
#include <iostream> //for using cout
#include <stdlib.h> //for using the function sleep
using namespace std; //for using cout
int main(void)
{
cout << "test" << endl;
sleep(5000); //make the programme waiting for 5 seconds
cout << "test" << endl;
sleep(2000); // wait for 2 seconds before closing
return 0;
}
If you wait too long, that probably means the parameter is in seconds. So change it to this:
sleep(5);
For those who get error message or problem using sleep try to replace it by _sleep or Sleep especially on Code::Bloks.
And if you still getting problems, try to add of one this library on the beginning of the code.
#include <stdio.h>
#include <time.h>
#include <unistd.h>
#include <dos.h>
#include <windows.h>
system("PAUSE")
A simple "Hello world" programme on windows console application would probably close before you can see anything. That the case where you can use system("Pause").
#include <iostream>
using namespace std;
int main(void)
{
cout << "Hello world!" << endl;
system("PAUSE");
return 0;
}
If you get the message "error: 'system' was not declared in this scope" just add
the following line at the biggining of the code :
#include <cstdlib>
cin.ignore()
The same result can be reached by using cin.ignore() :
#include <iostream>
using namespace std;
int main(void)
{
cout << "Hello world!" << endl;
cin.ignore();
return 0;
}
cin.get()
example :
#include <iostream>
using namespace std;
int main(void)
{
cout << "Hello world!" << endl;
cin.get();
return 0;
}
getch()
Just don't forget to add the library conio.h :
#include <iostream>
#include <conio.h> //for using the function getch()
using namespace std;
int main(void)
{
cout << "Hello world!" << endl;
getch();
return 0;
}
You can have message telling you to use _getch() insted of getch
Lots of people have suggested POSIX sleep, Windows Sleep, Windows system("pause"), C++ cin.get()… there's even a DOS getch() in there, from roughly the late 1920s.
Please don't do any of these.
None of these solutions would pass code review in my team. That means, if you submitted this code for inclusion in our products, your commit would be blocked and you would be told to go and find another solution. (One might argue that things aren't so serious when you're just a hobbyist playing around, but I propose that developing good habits in your pet projects is what will make you a valued professional in a business organisation, and keep you hired.)
Keeping the console window open so you can read the output of your program is not the responsibility of your program! When you add a wait/sleep/block to the end of your program, you are violating the single responsibility principle, creating a massive abstraction leak, and obliterating the re-usability/chainability of your program. It no longer takes input and gives output — it blocks for transient usage reasons. This is very non-good.
Instead, you should configure your environment to keep the prompt open after your program has finished its work. Your Batch script wrapper is a good approach! I can see how it would be annoying to have to keep manually updating, and you can't invoke it from your IDE. You could make the script take the path to the program to execute as a parameter, and configure your IDE to invoke it instead of your program directly.
An interim, quick-start approach would be to change your IDE's run command from cmd.exe <myprogram> or <myprogram>, to cmd.exe /K <myprogram>. The /K switch to cmd.exe makes the prompt stay open after the program at the given path has terminated. This is going to be slightly more annoying than your Batch script solution, because now you have to type exit or click on the red 'X' when you're done reading your program's output, rather than just smacking the space bar.
I assume usage of an IDE, because otherwise you're already invoking from a command prompt, and this would not be a problem in the first place. Furthermore, I assume the use of Windows (based on detail given in the question), but this answer applies to any platform… which is, incidentally, half the point.
The appearance and disappearance of a window for displaying text is a feature of how you are running the program, not of C++.
Run in a persistent command line environment, or include windowing support in your program, or use sleep or wait on input as shown in other answers.
the equivalent to the batch program would be
#include<iostream>
int main()
{
std::cout<<"Hello, World!\n";
std::cin.get();
return 0;
}
The additional line does exactly what PAUSE does, waits for a single character input
There is a C++11 way of doing it. It is quite simple, and I believe it is portable. Of course, as Lightness Races in Orbit pointed out, you should not do this in order to be able to see an Hello World in your terminal, but there exist some good reason to use a wait function. Without further ado,
#include <chrono> // std::chrono::microseconds
#include <thread> // std::this_thread::sleep_for
std::this_thread::sleep_for(std::chrono::microseconds{});
More details are available here. See also sleep_until.
Actually, contrary to the other answers, I believe that OP's solution is the one that is most elegant.
Here's what you gain by using an external .bat wrapper:
The application obviously waits for user input, so it already does what you want.
You don't clutter the code with awkward calls. Who should wait? main()?
You don't need to deal with cross platform issues - see how many people suggested system("pause") here.
Without this, to test your executable in automatic way in black box testing model, you need to simulate the enter keypress (unless you do things mentioned in the footnote).
Perhaps most importantly - should any user want to run your application through terminal (cmd.exe on Windows platform), they don't want to wait, since they'll see the output anyway. With the .bat wrapper technique, they can decide whether to run the .bat (or .sh) wrapper, or run the executable directly.
Focusing on the last two points - with any other technique, I'd expect the program to offer at least --no-wait switch so that I, as the user, can use the application with all sort of operations such as piping the output, chaining it with other programs etc. These are part of normal CLI workflow, and adding waiting at the end when you're already inside a terminal just gets in the way and destroys user experience.
For these reasons, IMO .bat solution is the nicest here.
What you have can be written easier.
Instead of:
#include<iostream>
int main()
{
std::cout<<"Hello, World!\n";
return 0;
}
write
#include<iostream>
int main()
{
std::cout<<"Hello, World!\n";
system("PAUSE");
return 0;
}
The system function executes anything you give it as if it was written in the command prompt. It suspends execution of your program while the command is executing so you can do anything with it, you can even compile programs from your cpp program.
Syntax:
void sleep(unsigned seconds);
sleep() suspends execution for an interval (seconds).
With a call to sleep, the current program is suspended from execution for the number of seconds specified by the argument seconds. The interval is accurate only to the nearest hundredth of a second or to the accuracy of the operating system clock, whichever is less accurate.
This example should make it clear:
#include <dos.h>
#include <stdio.h>
#include <conio.h>
int main()
{
printf("Message 1\n");
sleep(2); //Parameter in sleep is in seconds
printf("Message 2 a two seconds after Message 1");
return 0;
}
Remember you have to #include dos.h
EDIT:
You could also use winAPI.
VOID WINAPI Sleep(
DWORD dwMilliseconds
);
Sleep Function(Windows)
Just a note,the parameter in the function provided by winapi is in milliseconds ,so the sleep line at the code snippet above would look like this "Sleep(2000);"
getchar() provides a simplistic answer (waits for keyboard input).
Call Sleep(milliseconds) to sleep before exit.
Sleep function (MSDN)
You can use sleep() or usleep().
// Wait 5 seconds
sleep( 5 );
Well, this is an old post but I will just contribute to the question -- someone may find it useful later:
adding 'cin.get();' function just before the return of the main() seems to always stop the program from exiting before printing the results: see sample code below:
int main(){
string fname, lname;
//ask user to enter name first and last name
cout << "Please enter your first name: ";
cin >> fname;
cout << "Please enter your last name: ";
cin >> lname;
cout << "\n\n\n\nyour first name is: " << fname << "\nyour last name is: "
<< lname <<endl;
//stop program from exiting before printing results on the screen
cin.get();
return 0;
}
Before the return statement in you main, insert this code:
system("pause");
This will hold the console until you hit a key.
#include<iostream>
#include<string>
using namespace std;
int main()
{
string s;
cout << "Please enter your first name followed by a newline\n";
cin >> s;
cout << "Hello, " << s << '\n';
system("pause");
return 0; // this return statement isn't necessary
}
The second thing to learn (one would argue that this should be the first) is the command line interface of your OS and compiler/linker flags and switches.