#include <iostream>
#include <thread>
using namespace std;
int main()
{
for (int i = 0; i < 10000; i++) {
cout << "Hello\n";
}
this_thread::sleep_for(chrono::milliseconds(2000));
cout << "2 seconds have passed" << endl;
return 0;
}
In my code, I didn't call any std::flush or std::endl, but the hello's are printed before the 2 seconds delay. I am expecting to print all the hello's after the 2 seconds delay, but it didn't. My code runs like this:
Hello
Hello
.
.
.
Hello
Hello
(after 2 seconds)
2 seconds have passed
[terminated]
Why is this happening?
First of all, you're writing more output than a typical file buffer will hold, so you'd almost always expect at least some of the output to show up before the sleep.
Second, you're doing a lot of separate output calls, so if cout is unit-buffered, each one is going to be flushed immediately.
Third, you're writing a new-line at the end of each item, so if cout is line-buffered, (yup) each one is going to be flushed immediately.
So, if you want a better chance of seeing at least some of the output showing up after the sleep ends, turn off unit buffering and get rid of the new-lines:
#include <iostream>
#include <thread>
#include <chrono>
using namespace std;
int main()
{
cout << nounitbuf;
for (int i = 0; i < 1000; i++) {
cout << "Hello";
}
// display something different to make it easier for user to see
// whether all output showed up before sleep or not.
cout << "...";
this_thread::sleep_for(chrono::milliseconds(2000));
cout << "2 seconds have passed" << endl;
return 0;
}
But even with this, there's no guarantee the behavior will change. Rather the contrary, most implementations go to some pain to assure that output written to the console shows up as promptly as possible, so even when you take steps toward delaying it, it'll still probably show up before the sleep. But this might improve your chances a little bit anyway.
Related
Following program prints 1 2 3 4 5 at once. That means there is no time delay in printing the output.
#include<iostream>
#include<stdio.h>
#include <thread>
using namespace std;
int main()
{
for(int i = 1; i <= 5; ++i)
{
cout << i << " ";
// Function to sleep the thread
this_thread::sleep_for(500ms);
}
return 0;
}
But this program prints
1
2
3
4
5
one by one that means I’m getting the output with 0.5 sec time delay.
#include<iostream>
#include<stdio.h>
#include <thread>
using namespace std;
int main()
{
for(int i = 1; i <= 5; ++i)
{
cout << i << "\n";
// Function to sleep thread
// for 0.5 sec
this_thread::sleep_for(500ms);
}
return 0;
}
What is literally happening in both of above programs?
Note:You can't see the difference of both the outputs in online compiler because they show the result after termination of program.
The output may be line buffered, in which case only complete lines are sent to the underlying output device.
I'm able to get the desired output, one element at a time appears with 0.5 sec of delay. Used visual studio 2015.
I suggest you to avoid online compilers.
They generally give the complete output once in a text box.
"\n" makes the output go to next line. It is newline character
I would like to make a very basic progress indicator by printing an 'X' char to cout every time a loop progresses another 10%. I am trying to do this as shown in the code pasted below, but it doesn't seem to be working - when really it seems it should.
It's supposed to display steady progress throughout the loop, but instead I get all the X's coming at once after the loop has finished. This is not because the loops are completed too quickly for me to perceive. To verify this you can add an extra 'F' to "TOTAL" to increase the duration of the looping substantially, and you'll see it's not just a question of perception.
Any ideas what might be causing this?
#include <iostream>
#define TOTAL 0xFFFFFFF
using namespace std;
int main(void) {
//Make a counter for counting loops
double counter = 0;
// Set it to trigger after each 10% of progress
double counterMax = TOTAL / 10;
cout << "Starting now..." << endl;
for (double i = 0; i < TOTAL; i++) {
// Do something... anything
i++;
i--;
// Increment the counter
counter++;
// Print an X when the counter exceeds the 10%
// trigger point, and then reset the counter.
if (counter > counterMax) {
cout << 'X';
counter = 0;
}
}
cout << endl << "Done!";
return 0;
}
System input/output calls are usually slow operations. To increase the efficiency of programs, input and output streams are often buffered, to reduce the number of calls to the operating system.
When a program needs "non-buffered" output, one solution is to use the buffered output functions, and simple "flush" the output to ensure the operating system processes any output which has been queued in any buffers.
When the output buffer is filled, or when the stream is closed, it is automatically flushed. The standard output is also automatically flushed by certain sequences, like endl. But you can trigger a flush of the standard output at any point by called cout.flush() or by using the flush manipulator, like:
cout << 'X' << flush;
I want some random characters to be printed to console and then deleted by "\b". After this I want to put my own variable so it will be looking like "randomizing". The problem is that it is happening too fast. I wanted to delay the output by using usleep or sleep function but when I'm using it, nothing is printed into console.
Short example:
#include <iostream>
#include <unistd.h>
using namespace std;
int main()
{
char chars[]={'a','b','c','g','h','u','p','e','y'};
for(int i=0; i<8; i++)
{
cout << chars[i];
usleep(200000);
cout << "\b";
}
}
Problem is, std::cout is line-buffered. It stores all input in a buffer until a newline is encountered (or the program terminates). Use std::flush to flush std::cout explicitly:
cout << chars[i] << flush;
Notes:
since C++11, multithreading and time are standardized. That brings the std::this_thread:sleep_for function with it, which you should use for a portable >= C++11 program:
std::this_thread::sleep_for(std::chrono::milliseconds(200));
Try my little program slowtty from github.
It allows you to simulate in a pty the behaviour of an old rs232c line, by delaying the output per character as stty(1) command allows to set the baudrate.
You call it with
$ slowtty
$ stty 1200
$
and the terminal begins to write characters at a slow pace (like a 1200baud line)
On many systems output is buffered.
To ensure that what you sent out to cout has really been flushed out of buffers you need to call
cout.flush();
before the sleep
My code look like below
int i=0;
while(i<10){
cout<<"Hello";
sleep(1);
i++
}
In Windows the code prints on each loop but in Linux it prints everything after exiting while loop . And also if I put an endl at the last of cout then it prints on each loop. Why this happening ?. Can anyone explain this behavior?.
Try to use cout.flush(); maybe the two OS has different policy in term of buffering the stdout.
For efficiency reasons, sometimes the standard streams will be implemented with a buffer. Making lots of tiny writes can be slow, so it will store up your writes until it gets a certain amount of data before writing it all out at once.
Endl forces it to write out the current buffer, so you'll see the output immediately.
#include <iostream>
using namespace std;
int main()
{
int i = 0;
while(i < 10){
cout << "Hello" << endl;
sleep(1);
++i;
}
}
I'm writing a code to output fibonacci series in C++, which is simple enough, but since I'm new to programming, I'm considering ways to control
I was wondering if there's a way to control time for when outputs come out without processing time being included (e.g. If it takes .0005 seconds to process my input and I want it to repeat the output in 1 second rather than 1.0005 seconds).
I also am wondering if there's a way to just have it output (say 1) and then have it wait for me to press enter, which will make it output the second part (2).
Also, I'd appreciate any other suggestions on how to control output.
If you're on Windows, a quick way to pause is to call system("pause"); This isn't a very professional way of doing things, however.
A way to pause with the std namespace would be something like this:
cout << "Press Enter to continue . . ." << endl;
cin.ignore(numeric_limits<streamsize>::max(), '\n');
As for your main question... If you're just dealing with text output to the user, it's not possible for them to notice a five microsecond delay. They can notice if your interval lengths fluctuate by tens of milliseconds, however. This is why sleep(..) functions sometimes fail.
Let's take your example of wanting to output another number in the Fibonacci sequence once per second. This will work just fine:
#include <ctime>
#include <limits>
#include <iostream>
void pause() {
std::cout << "Press Enter to continue . . ." << std::endl;
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}
int main() {
clock_t next = clock();
int prev1, prev2, cur = 0;
for (int i = 0; i < 47; ++i) {
if (i < 2) cur = i;
else cur = prev1 + prev2;
prev2 = prev1;
prev1 = cur;
while (next > clock());
std::cout << (i+1) << ": " << cur << std::endl;
next += CLOCKS_PER_SEC;
}
pause();
return 0;
}
The next number in the sequence is computed and ready to print prior to the wait, thus the computation time will not add any delay to the timed output.
If you want your program to continue outputting at a fixed rate while your program works in the background, you'll need to look into threads. You can have your results added to a queue in one thread, while another thread checks for results to print once per second.