So...here is my code that refer to C++ Primer 1.4.4 section.
#include <iostream>
int main()
{
int currVal = 0, val = 0;
if (std::cin >> currVal) {
int cnt = 1;
while (std::cin >> val) {
if (val == currVal) {
++cnt;
} else {
std::cout << currVal << " occurs "
<< cnt << " times." << std::endl;
currVal = val;
cnt = 1;
}
}
std::cout << currVal << " occurs "
<< cnt << " times." << std::endl;
}
return 0;
}
I have to admit that the simple program basically work normally. But for the further discussion, I'd like refer to a asciinema recording in the first place.
In the recording I run ./a.out three times. In the first time, I typed numbers and press enter the first two recording printed and only after ctrl + D was pressed the third (last) result was printed. From previous sections in the book, I know ctrl + D is used to feed some symbols to stop the while loop. But I really wonder if it (ctrl + D here cause a strange stop) works just like as same as authors' expectation for there is no mention on how keys was pressed exactly about that example in the book.
In the last two runs, my first approach is press ctrl + D (the first result printed) and enter (the second result printed), then following several enters (no respond except more lines). Lastly, I pressed ctrl + D (the third result printed).
My last approach is press ctrl + D for two times. The first result printed on the first press and the last two came out on the second press.
With these two runs comes my confusing that after press ctrl + D (I think) there should be a stopping signal in the end of cin stream. So why did the while loop not stop? I'd appreciate a lot if anyone can explain the process behind it.
My working environment is urxvt on Manjaro Linux with zsh and oh-my-zsh installed.
Related
The 《c++ primer 5th》1.4.4 code example is like this
#include <iostream>
int main()
{
// currVal is the number we're counting; we'll read new values into val
int currVal = 0, val = 0;
// read first number and ensure that we have data to process
if (std::cin >> currVal) {
int cnt = 1; // store the count for the current value we're processing
while (std::cin >> val) { // read the remaining numbers
if (val == currVal) // if the values are the same
++cnt; // add 1 to cnt
else { // otherwise, print the count for the previous value
std::cout << currVal << " occurs "
<< cnt << " times" << std::endl;
currVal = val; // remember the new value
cnt = 1; // reset the counter
}
} // while loop ends here
// remember to print the count for the last value in the file
std::cout << currVal << " occurs "
<< cnt << " times" << std::endl;
} // outermost if statement ends here
return 0;
}
if i input :
11
11
13
13
13
14
I think it should execute like this :
when i input
11
11
the console should show "11 occurs 2 times".
Then i can continue to input
13
13
13
then the console should show " 13 occurs 3 times".
But the result is only when i have finished typing all the number , the console output the result once . why ?
Thanks for your help .
Input from terminal is line buffered.
The first std::cin >> currVal blocks until input is available at standard input.
That doesn't happen until you press <Enter>. (Up to that <Enter>, the characters you entered are still residing in the line buffer of your terminal / CMD box. You can backspace, edit etc.; only when you press <Enter> does the terminal / CMD box actually send those characters to the program's standard input.)
For your expected behaviour to happen, try pressing <Enter> after each number.
Both std::cin >> currVal in your sample will block the standard input, std::cin means a value that comes from standard input, in that case your keyboard is your standard input.
To confirm the input from keyboard you need to push Enter key, then values are removed from the buffer of the keyboard and processed by your code.
If you want to get the values for each press of key you need to use something like std::getchar, that function reads the buffer immediately like you expect.
I am going through C++ Primer (5th ed). In section 1.4.4, there is the following example:
#include <iostream>
int main()
{
// currVal is the number we're counting; we'll read new values into val
int currVal = 0, val = 0;
// read first number and ensure that we have data to process
if (std::cin >> currVal) {
int cnt = 1; // store the count for the current value we're processing
while (std::cin >> val) { // read the remaining numbers
if (val == currVal) // if the values are the same
++cnt; // add 1 to cnt
else { // otherwise, print the count for the previous value
std::cout << currVal << " occurs " << cnt << " times" << std::endl;
currVal = val; // remember the new value
cnt = 1; // reset the counter
}
} // while loop ends here
// remember to print the count for the last value in the file
std::cout << currVal << " occurs " << cnt << " times" << std::endl;
} // outermost if statement ends here
return 0;
}
When you run it with the given input
42 42 42 42 42 55 55 62 100 100 100
It prints
42 occurs 5 times
55 occurs 2 times
62 occurs 1 times
However, in order to get the final output line
100 occurs 3 times
you must press CTRL+D. Then that is printed and the program exits.
Why is this? To me, it looks like this last line should be printed and the program exited with the others. It seems I am misunderstanding how the control flow is executed so can someone please clarify?
ps I am aware of this Incorrect output. C++ primer 1.4.4 and C++ Primer fifth edtion book (if statement) is this not correct? However, neither of these explain WHY you must ctrl+d to print the final statement.
That's because of this part:
while (std::cin >> val)
In order to terminate reading the input stream, you have to terminate it with an EOF, which is supplied by Ctrl-D.
Think about it: cin skips over whitespace by default and every time you enter a number you separate it with whitespace (a space, a tab or a newline).
How will the program ever terminate the input? The answer is when it reads an EOF character - which, as stated before, is supplied by Ctrl-D.
You must press CTRL+D because otherwise the program does not know when your stdin stream has finished. It will otherwise just sit there at while (std::cin >> val forever without terminating.
friends. I have a problem.
Problem: the computer must pick randomly one string out of an array of 36 strings. If by any chance it picks strings #34 or #35 (the two last ones), it has to draw two more random strings from the same array. I tried a do-while solution, and it "almost" works (see code below).
The randomization works fine - called srand inside main(). There is a forced "x2" draw (for testing reasons), so the computer draws two more strings. These two new random picks are NOT "x2", but still the loop kicks again - but just one more time! This time the computer picks two more "chits", which aren't "x2" either, so, as expected, it returns the "The chits have been drawn" sentence and the function is terminated. Why is the same code running twice with the same results but different if/else behavior? Thank you very much in advance.
string mortalityCalc ()
{
string mortalityChits[36] = {"1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30","-","-","-","-","x2","x2"};
int mortalityResult;
// mortalityResult = rand() %36;
mortalityResult = 35; // for testing only. Delete afterwards.
string drawnChit = mortalityChits[mortalityResult];
string drawnChit1;
string drawnChit2;
if (drawnChit != "-" && drawnChit != "x2")
{
string returnText = string("The computer has drawn the chit '") + drawnChit + "'.";
return returnText;
}
else if (drawnChit == "-")
{
string returnText = string("The computer has drawn the chit '") + drawnChit + "'. No senators died this year.";
return returnText;
}
do
{
cout << "The computer has drawn the 'x2' chit." << endl;
cout << "Two more chits will be drawn.\n" << endl;
mortalityResult = rand() %36;
drawnChit1 = mortalityChits[mortalityResult];
cout << "The first draw is the chit '" << drawnChit1 << "'. ";
mortalityResult = rand() %36;
drawnChit2 = mortalityChits[mortalityResult];
cout << "The second draw is the chit '" << drawnChit2 << "'." << endl;
} while (drawnChit1 == "x2" || drawnChit2 == "x2");
return "The mortality chits have been drawn. The corresponding senators are dead.";
}
UPDATE: Tried running this code isolated from the rest of the program and it behave as expected. So I guess it's important to post what comes before it:
cout << "If you are a lazy bastard, the computer can pick one senator randomly for you.\nAre you a lazy bastard? [y/n]" << endl;
string lazyBastard;
cin >> lazyBastard;
cout << endl;
if (lazyBastard == "y" || lazyBastard == "Y" || lazyBastard == "yes" || lazyBastard == "YES" || lazyBastard == "Yes")
{
mortalityCalc ();
cout << mortalityCalc () << endl;
cout << "Very well. Now, imminent wars become active (Only one of each match)." << endl;
cout << "Get ready for the next phase." << endl;
My guess, from reading some other questions here, is that somehow the cin is messing with the loop behavior, even though they are not related and there's no user input whatsoever in the loop's statements or conditions. Is that possible? If so, why and how to remedy it?
Thank you again.
In the first loop you are forcing an 'x2' so your are entering the do-while loop. The result of the two calls for 'rand())%36' is always 19 and a number between 30 and 34. The point is that the random number generator generates always the same sequence of numbers, if you don't give him a seed 'srand(...)'.
do {
// ...
cout << rand()%36;
// ...
} while( /*...*/ )
See http://ideone.com/zl8ggH
You have to create random numbers and your code does what you expect.
Finally! I thought it would be a stupid thing! I just realized that I called the mortalityCalc() function twice! That's why it was looping twice!
Thanks to all who tried to help!
I have a for loop, a very simple one, in my program, and I want it to loop through and do something for some minimum number of times. However, the loop simply...stops. But does not move on to the next thing in the program. For instance, when min is 9, it runs for i=0 to i=8, then freezes. It should exit the for loop, but it does not execute the next print instruction, nor does it execute the loop again. It just stops. The program hangs, doing absolutely nothing as far as I can tell. I don't understand why this is.
The merged.put() function I want to execute just puts x or y in merged, depending on the condition. That part works. This is just a small part of a much larger program. sp1, sp2, and merged are all defined elsewhere.
int i;
int x;
int y;
for(i=0; i < min; i++)
{
cout << " here " + convert(i);
x = sp1.get_num(i);
y = sp2.get_num(i);
if(x >= y) {
merged.put(x);
}
else {
merged.put(y);
}
cout << " end" << endl;
}
cout << "out";
EDIT: I'm not posting the entire code, it's several hundred lines long. Type of min is int. The reply down there was helpful, when << endl was added to the last print statement, it printed. My problem now appears to be here, getting stuck on the second while, because I was not incrementing i. Shame on me...thanks for the help. (This comes directly after the above code)
if (sp_large == 2) {
cout << "1" << endl;;
while (i < sp2.get_size()) {
merged.put(sp2.get_num(i));
}
}
else {
while (i < sp1.get_size()) {
merged.put(sp1.get_num(i));
}
cout << "2" << endl;
}
EDIT: Problem solved, thanks for the help.
I'm betting that it's actually a later part of the program that is hanging.
This line:
cout << "out";
just puts "out" on the output-buffer, and won't actually print "out" until the output-buffer gets flushed. (Which could happen immediately, but is not likely to.) Change that line to this:
cout << "out" << endl;
and "out" will be printed as soon as that line is run. This will help you figure out if the program is hanging before it gets to that line, or somewhere later on.
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.