Ok, I am a bit embarrassed to ask such a simple thing but still.
I have command line utility application and need to show progress to the user.
I could write progress into cout, like this:
std::cout << "10%\n";
...
std::cout << "20%\n";
...
std::cout << "30%\n";
... but as a result user will see:
some line printed before
10%
20%
30%
...
... but what i really need is that percentage got updated, like this at the beginning:
some line printed before
10%
...
... and after update:
some line printed before
20%
...
... and after second update:
some line printed before
30%
...
How should I achieve that?
Instead of using '\n', use '\r':
std::cout << "\r10%" << std::flush;
Print newline ('\n') when done.
It's important to use std::flush so the stream contents really is output.
Use a carriage return.
std::cout << "\r10%";
std::cout << "\r20%";
...
Goes to the beginning of the line.
Related
I use two different txt files at work for a task. One of them gets updated any time I need to call someone back (callback.txt), and one that gets updated with all of the callbacks that didn't get finished by the end of the day (recap.txt). I have written a basic python program, as well as a PowerShell script, that updates recap.txt at the end of the day by adding the newest entries to the top of the recap.txt from the day before.
Now, I am trying to write it in C++ just to get some experience with the language, but I can't seem to find why I am running into this issue. Everything works as intended, except that "recapNEW.txt" has 0 and 1 written and the end of the file on separate lines each. For example:
callback.txt:
----------
here's something
here is something else
yadda yadda
recapOLD.txt
----------
old news
still need it though
so we keep it around
recapNEW.txt (blank)
----------
And, after running the program:
recapNEW:
----------
here's something
here is something else
yadda yadda
old news
still need it though
so we keep it around
0
1
This is what I have written:
#include<fstream>
#include<iostream>
#include<string>
void update () {
std::ifstream callback("callback.txt");
std::ifstream recapBAK("recapBAK.txt");
std::ofstream recapNEW("recapNEW.txt");
std::string callLine;
std::string bakLine;
if (!callback || !recapNEW || !recapBAK) {
std::cerr << "Could not open needed files. Exiting.";
std::exit(1);
}
while (std::getline ( callback,callLine )) {
recapNEW << callLine << std::endl ;
}
while (std::getline ( recapBAK,bakLine )) {
recapNEW << bakLine << std::endl ;
}
recapNEW << callback << std::endl << recapNEW << std::endl ;
callback.close();
recapBAK.close();
recapNEW.close();
}
I split the recap.txt file into two separate files ("recapBAK.txt" & "recapNEW.txt") to keep a copy of the previous day's recap and am updating it in a separate function (posted below), but I have narrowed it down that it is the above function that is writing the extra numbers to the end of the file.
void backup () {
std::ofstream backup("recapBAK.txt");
std::ifstream today("recapNEW.txt");
std::string todayLine;
while (std::getline (today,todayLine)) {
backup << todayLine << std::endl;
}
backup.close();
today.close();
}
What is confusing me the most is that I don't see what I am doing differently in the first function that is causing the integers to be tagged on to the end of the file, because no additional integers are added to the updated version of recapUSED.
Apologies if this is not clear enough, first time I've hit a wall hard enough that I posted to a forum lol. I appreciate the help, I'm self-teaching and love having input to work with to see where I can/should improve.
I think the problem is that in:
recapNEW << callback << std::endl << recapNEW << std::endl ;
... you are attempting to output the contents of the streams callback and recapNEW to recapNEW, but it doesn't work that way. Rather, the Boolean "good" status of the file streams is output instead. Prepending to a stream isn't that simple.
Actually, it looks like you have what you need without that line. Have you tried just removing it?
This question already has answers here:
endl and flushing the buffer
(5 answers)
Closed 6 years ago.
std::cout << "Enter two numbers:";
std::cout << std:endl;
This code snippet is followed by two paragraphs and a warning note, among which I understood the first para, but neither the second one nor the note. The text is as follows -
"The first output operator prints a message to the user. That message
is a string literal, which is a sequence of characters enclosed in
double quotation marks. The text between the quotation marks is
printed to the standard output.
The second operator prints endl,
which is a special value called a manipulator. Writing endl has
the effect of ending the current line and flushing the buffer
associated with that device. Flushing the buffer ensures that all the
output the program has generated so far is actually written to the
output stream, rather than sitting in memory waiting to be written.
Warning Programmers often add print statements during debugging. Such statement should always flush the stream. Otherwise, if the
program crashes, output may be left in the buffer, leading to
incorrect inferences about where the program crashed."
So I didn't understand of the part of endl, nor the following warning. Can anyone please explain this to me as explicitly as possible and please try to keep it simple.
Imagine you have some code that crashes somewhere, and you don't know where. So you insert some print statements to narrow the problem down:
std::cout << "Before everything\n";
f1();
std::cout << "f1 done, now running f2\n";
f2();
std::cout << "all done\n";
Assuming that the program crashes during the evaluation of either f1() or f2(), you may not see any output, or you may see partial output that is misleading -- e.g. you could see only "Before everything", even though the crash happened in f2(). That's because the output data may be waiting in a buffer and hasn't actually been written to the output device.
The Primer's recommendation is therefore to flush each output, which you can conveniently achieve with endl:
std::cout << "Before everything" << std::endl;
f1();
std::cout << "f1 done, now running f2" << std::endl;
f2();
std::cout << "all done" << std::endl;
An alternative is to write debug output to std::cerr instead, which is not buffered by default (though you can always change the buffering of any ostream object later).
A more realistic use case is when you want to print a progress bar in a loop. Usually, a newline (\n) causes line-based output to be printed anyway, but if you want to print a single character for progress, you may not see it printed at all until after all the work is done unless you flush:
for (int i = 0; i != N; ++i)
{
if (i % 1000 == 0)
{
std::cout << '#'; // progress marger
std::cout.flush();
}
do_work();
}
std::cout << '\n';
Well, simply:
std::cout << "Hello world!";
will print "Hello world!" and will remain in the same line. Now if you want to go to a new line, you should use:
std::cout << "\n";
or
std::cout << std::endl;
Now before I explain the difference, you have to know 1 more simple thing: When you issue a print command with the std::cout stream, things are not printed immediately. They are stored in a buffer, and at some point this buffer is flushed, either when the buffer is full, or when you force it to flush.
The first kind, \n, will not flush, but the second kind std::endl, will go to a new line + flush.
Operating systems do buffered IO. That is, when your program outputs something, they dont necessarily put it immediately where it should go (i.e. disk, or the terminal), they might decide to keep the data in an internal memory buffer for some while before performing the actual IO operation on the device.
They do this to optmize performance, because doing the IO in chunks is better than doing it immediately as soon as there are a few bytes to write.
Flushing a buffer means asking the OS to perform immediately the IO operation without any more waiting. A programmer would do this this when (s)he knows that waiting for more data doesn't make sense.
The second note says that endl not only prints a newline, but also hints the cout to flush its buffer.
The 3rd note warns that debugging errors, if buffered and not flushed immediately, might not be seen if the program crashes while the error messages are still in the buffer (not flushed yet).
I was wondering what is the most efficient performant way to output a new line to console. Please explain why one technique is more efficient. Efficient in terms of performance.
For example:
cout << endl;
cout << "\n";
puts("");
printf("\n");
The motivation for this question is that I find my self writing loops with outputs and I need to output a new line after all iterations of the loop. I'm trying to find out what's the most efficient way to do this assuming nothing else matters. This assumption that nothing else matters is probably wrong.
putchar('\n') is the most simple and probably fastest. cout and printf with string "\n" work with null terminated string and this is slower because you process 2 bytes (0A 00). By the way, carriage return is \r = 13 (0x0D). \n code is Line Feed (LF).
You don't specify whether you are demanding that the update to the screen is immediate or deferred until the next flush. Therefore:
if you're using iostream io:
cout.put('\n');
if you're using stdio io:
std::putchar('\n');
The answer to this question is really "it depends".
In isolation - if all you're measuring is the performance of writing a '\n' character to the standard output device, not tweaking the device, not changing what buffering occurs - then it will be hard to beat options like
putchar('\n');
fputchar('\n', stdout);
std::cout.put('\n');
The problem is that this doesn't achieve much - all it does (assuming the output is to a screen or visible application window) is move the cursor down the screen, and move previous output up. Not exactly a entertaining or otherwise valuable experience for a user of your program. So you won't do this in isolation.
But what comes into play to affect performance (however you measure that) if we don't output newlines in isolation? Let's see;
Output of stdout (or std::cout) is buffered by default. For the output to be visible, options include turning off buffering or for the code to periodically flush the buffer. It is also possible to use stderr (or std::cerr) since that is not buffered by default - assuming stderr is also directed to the console, and output to it has the same performance characteristics as stdout.
stdout and std::cout are formally synchronised by default (e.g. look up std::ios_base::sync_with_stdio) to allow mixing of output to stdout and std::cout (same goes for stderr and std::cerr)
If your code outputs more than a set of newline characters, there is the processing (accessing or reading data that the output is based on, by whatever means) to produce those other outputs, the handling of those by output functions, etc.
There are different measures of performance, and therefore different means of improving efficiency based on each one. For example, there might be CPU cycles, total time for output to appear on the console, memory usage, etc etc
The console might be a physical screen, it might be a window created by the application (e.g. hosted in X, windows). Performance will be affected by choice of hardware, implementation of windowing/GUI subsystems, the operating system, etc etc.
The above is just a selection, but there are numerous factors that determine what might be considered more or less performance.
On Ubuntu 15.10, g++ v5.2.1 (and an older vxWorks, and OSE)
It is easy to demonstrate that
std::cout << std::endl;
puts a new line char into the output buffer, and then flushes the buffer to the device.
But
std::cout << "\n";
puts a new line char into the output buffer, and does not output to the device. Some future action will be needed to trigger the output of the newline char in the buffer to the device.
Two such actions are:
std::cout << std::flush; // will output the buffer'd new line char
std::cout << std::endl; // will output 2 new line chars
There are also several other actions that can trigger the flush of the std::cout buffering.
#include <unistd.h> // for Linux
void msDelay (int ms) { usleep(ms * 1000); }
int main(int, char**)
{
std::cout << "with endl and no delay " << std::endl;
std::cout << "with newline and 3 sec delay " << std::flush << "\n";
msDelay(3000);
std::cout << std::endl << " 2 newlines";
return(0);
}
And, per comment by someone who knows (sorry, I don't know how to copy his name here), there are exceptions for some environments.
It's actually OS/Compiler implementation dependent.
The most efficient, least side effect guaranteed way to output a '\n' newline character is to use std::ostream::write() (and for some systems requires std::ostream was opened in std::ios_base::binary mode):
static const char newline = '\n';
std::cout.write(&newline,sizeof(newline));
I would suggest to use:
std::cout << '\n'; /* Use std::ios_base::sync_with_stdio(false) if applicable */
or
fputc('\n', stdout);
And turn the optimization on and let the compiler decide what is best way to do this trivial job.
Well if you want to change the line I'd like to add the simplest and the most common way which is using (endl), which has the added perk of flushing the stream, unlike cout << '\n'; on its own.
Example:
cout << "So i want a new line" << endl;
cout << "Here is your new line";
Output:
So i want a new line
Here is your new line
This can be done for as much new lines you want. Allow me to show an example using 2 new lines, it'll definitely clear all of your doubts,
Example:
cout << "This is the first line" << endl;
cout << "This is the second line" << endl;
cout << "This is the third line";
Output:
This is the first line
This is the second line
This is the third line
The last line will just have a semicolon to close since no newline is needed. (endl) is also chain-able if needed, as an example, cout << endl << endl; would be a valid sequence.
I have code such as the following:
std::cout << "Beginning computations..."; // output 1
computations();
std::cout << " done!\n"; // output 2
The problem, however, is that often output #1 and output #2 appear (virtually) simultaneously. That is, often output #1 does not get printed to the screen until after computations() returns. Since the entire purpose of output #1 is to indicate that something is going on in the background (and thus to encourage patience from the user), this problem is not good.
Is there any way to force the std::cout buffer to get printed before the computations() call? Alternatively, is there some other way (using something other than std::cout) to print to standard out that would fix this problem?
Just insert std::flush:
std::cout << "Beginning computations..." << std::flush;
Also note that inserting std::endl will also flush after writing a newline.
In addition to Joseph Mansfield answer, std::endl does the flush too (besides a new line).
Inserts a endline character into the output sequence os and flushes it as if by calling os.put(os.widen('\n')) followed by os.flush().
I'm working on a project where I call a function which triggers a segfault. I fixed this, but during the process I noticed the following.
When my code is of the format;
main(){
...
std::cout << "Looking for segfault\n"; // this does not print
buggyFunction(); // crashes in here
...
}
buggyFunction(){
...
thing_that_causes_segfault;
...
}
The line "Looking for segfault" doesn't print to STD, and the program crashes in buggyFunction. Fine, but when I add a cout line inside buggyFunction();
main(){
...
std::cout << "Looking for segfault\n"; // this now *does* print
buggyFunction();
...
}
buggyFunction(){
...
std::cout << "Now we're INSIDE buggy function\n"; // this prints too
thing_that_causes_segfault;
...
}
Inside buggy function, both lines print (and then it crashes).
Why do we see this difference in ouput, depending on the addition of this extra output call? Is it related to the handling of streams, or something else? I'm using g++ (Ubuntu 4.4.3-4ubuntu5) 4.4.3.
The reason for this is that cout has a buffer and it will only pass to the system function and write to the console when the buffer is full. Your second use of cout happens to overflow the buffer and so it calls the operating system and empties the buffer. If you want to guarantee that the output has left the buffer, you must use std::flush. You can both end a line and flush the buffer with std::endl.
Because your line might not immediately printed out because it's cached and the cache is not "flushed". std::endl is an newline + a flush thus forces immediate printout:
std::cout << "Looking for segfault" << std::endl;
It has to do with buffering. Things that you write to cout are appended to an internal buffer that only gets flushed periodically. You can explicitly flush the buffer by writing std::flush to your stream, or replacing the "\n" with << std::endl.