I am learning a new language called "D" but i have a problem when trying to write a simple program
import std.stdio;
void main()
{
double gradeOne;
writeln("Please enter the First Test Grade: ");
readf(" s", &gradeOne);
}
Why does my program ask me for the input first before the output message?
I think its just the DDT problem, when i run the program in command prompt its working fine
Output to Eclipse buffers output by larger data blocks rather than lines. To force output to appear, insert calls to stdout.flush(); before asking for input to ensure it shows up when you want it.
See also: Eclipse console writes output only after the program has finished
Related
After last week's updated to Xcode 8.3, in a C/C++ program the output from a printf statement no longer appears on the screen without a newline. Thus I can't prompt for the user to enter a number, and have them type in that number on the same line following the input prompt.
Neither flushing the output buffer [fflush(stdout) or cout << endl] nor setting the output buffer to NULL [setbuf(stdout, NULL)] addresses this problem, but rather is a question specifically about Xcode 8.3 seemingly being broken.
With the scanf commented out, the output of the program below is:
Enter a value for x: Value of x is: 0
With the scanf in place, the output from the first printf never shows up. If you go ahead and type in a value and press enter, only then does it show up. Output is:
3
Enter a value for x: Value of x is: 3
Full test program is here:
#include <iostream>
using namespace std;
int main() {
int x=0;
printf("Enter a value for x: ");
//scanf("%d", &x);
printf("Value of x is: %d\n", x);
return 0;
}
My work-around has been to revert back to Xcode 8.2.1, downloaded from developer.apple.com/xcode/downloads/
8.3.2 was announced last night and addresses this supposedly:
It is standard behavior in C for stdout to be flushed when an input function is called, such as scanf(), regardless of whether a newline was output prior to the call or not. This ensures that all appropriate output is displayed before the input operation takes place. Therefore, the update might have broken something in Xcode. Although I'm not currently sure what the exact nature of the problem is, a (temporary) workaround is to run your application on the command line. This has worked for my projects. It also reveals that this problem is with the Xcode output window, and not the compiler or something else.
In response to tell's comment: no, flushing stdout does not correct the issue within Xcode. This implies even more strongly to me that the issue is definitely in the Xcode interface itself. When running the application from the command line, calls to fflush() work as expected.
Also, printing to stderr makes no difference from within Xcode. Basically, stdout should be flushed in this case without appealing to stderr or any other gimmicks because the OP is calling scanf(). It works perfectly from the command line... just not in the Xcode output window.
And please note this this question is not a duplicate: it has nothing to do with anyone's misunderstanding of how C input and output work, and everything to do with the fact that a recent update to Xcode broke something.
EDIT:
Thanks, joe_04_04. The update certainly seems to have fixed the problem.
I am very new to C++. I have my code and it displays my desired output in a win32 console.
However, my Instructor wants the output to be run through a .txt file. We have done this before with a program that had input already written within the coding.
ex.
cout << "example1....example2";
We achieved this with his exact instructions:
*1) Probably the easiest way to obtain a hard copy of the generated
program output on a Microsoft Windows platform is to run your
program from a command prompt, redirecting the output to a file.
The command line syntax would like like this:
lab1prog >lab1.txt*
My problem, however, is that I did this again for lab2 and redirected the output to lab2.txt but I need user input for this time around. When I run the lab2.exe file, my lab2.txt file outputs my "cout" statement and waits for input but I cannot enter input through a .txt file.
Please help if you can.
When you redirect the output of the program with
lab1prog >lab1.txt
then the user is still able to input data. The only problem is that she doesn't know when to enter what data. This is usually done by prompt outputs that are hidden whith the >lab1.txt.
To circumvent the problem you could abuse the error output what is not redirected with the command above.
cerr << "prompt";
To avoid this abuse you should use a "tee" program like wintee instead of redirecting with a simple >lab1.txt.
If it's acceptable to get the input from a file instead from the user you can use the input redirection.
labprog1 <input.txt >lab1.txt
If you specifically want to use a text file as input into your program, the easiest way (as Kamil Mikolajczyk mentioned) is to run it this way:
> lab1prog >lab1.txt <input.txt
On the other hand, if you need to run the program as it is, but have complete control over what to output into the file, I'd suggest using a file handle. Check out this question on how to use file handles. You could even duplicate the output on the command prompt and the file when you want by outputting it as:
fout << "My output";
cout << "My output";
You can as well output the user input into your output file for your convenience.
Shouldn't this work? I mean, the code is merely a test and is meant so that the dialogue goes this way : What is your name? name here, Hello name here, and yet it does not show the last line of Hello after I type in my name and click enter it just dissapears. Here is the code.
#include <iostream>
#include <string>
int main (void)
{
using std::cin;
using std::cout;
using std::string;
string name = "";
cout << "What is your name, pls?\n";
cin >> name;
cout << "\nHello " << name.c_str() << "\n";
return 0;
}
My guess is that you are running from the debugger, or double clicking the executable. In either of those cases, when the program ends, the console will close. So, the program produced output, but you just could not see it before the console closed.
Run the program from a pre-existing console so that the console remains after your program ends. Or, just whilst debugging, arrange that your program does not terminate immediately after emitting its final output. A simple way to do that is to place a break point at the end of the program.
It probably showed it right before it disappeared. If you're going to write console programs, and if you're going to send output to a console, you should run them from a console so the output has some place to go.
After you are done with your program, press Ctrl + F5 ( Run without debugging). This will prompt before closing the window and this is what you want.
Make sure you put a breakpoint before main goes out of scope. I guess your console disappears under VS?
Also, you don't need to extract the char* in the last cout statement:
cout << "\nHello " << name << endl;
Open a terminal (or a command prompt window).
Navigate to the folder that contains the executable.
Run it.
It's not disappearing. It is just running really fast.
Every IDE has a keyboard shortcut that allows you to run code and pause after the execution has finished.
This keyboard shortcut is Ctrl-F5 in Visual Studio.
I have no idea what IDE you're running, but that is your basic problem.
The other thing you can do is to test your code in ideone : ideone.com/hb4Cel (it's the same code. There is no point pasting it here)
A dirty workaround is to add something like this
cin >> name;
at the end, just before return 0;. It forces the window to wait for input (i.e. hitting return) before returning (which closes the program).
This isn't necessarily good design, but if all you want to do is run some tests then it'll do the trick.
Basically when you enter your name it displays your last line and exits after return 0.
Here are the following things to avoid that
1- use command line to run the application
Start->accessories->command prompt
Go to folder in which your application is using cd command
c:>cd c:\path\foldername
Now run the application by typing the program name e.g
c:\path\foldername>my_application.exe
It will display your last line.
2- Now if your are using microsoft visual c++ press ctrl+F5 to run your program
3- This is not recommended but you an use it as long as your are debugging then remove it from the code afterwards. Include conio.h header file and add getch(); line before return statement. It would hold the screen for you till you press a key.
Hey so I am trying to learn Fortran basics so that I can use it for a basic physics project. I am having trouble with getting input properly. My code is:
program main
write(*, *) "Enter n:"
read(*, *) n
print *, "Number is ", n
end program main
It is quite simple. Except that when I compile and run it, nothing happens until I enter the input in which it looks like this
gfortran num.f90 -o num
./num
(nothing happens until I type) 3
Enter n:
Number is: 3
Is there a problem with my memory allocation? Or could it be my compiler? I honestly have no clue.
Your program is fine, on my machine it prints the Enter n: before reading the variable. If you do not see the message until you enter a value (and hit Enter), it could have to do with the buffering of your command window you use. For checking this, you could just open an xterminal (type xterm in your command window), and run the program there.
Good day,
I wrote a Java program that starts multiple C++ written programs using the Process object and Runtime.exec() function calls. The C++ programs use cout and cin for their input and output. The Java program sends information and reads information from the C++ programs input stream and outputstream.
I then have a string buffer that builds what a typical interaction of the program would look like by appending to the string buffer the input and output of the C++ program. The problem is that all the input calls get appended and then all the output calls get posted. For example, and instance of the StringBuffer might be something like this...
2
3
Please enter two numbers to add. Your result is 5
when the program would look like this on a standard console
Please enter two numbers to add. 2
3
Your result is 5
The problem is that I am getting the order of the input and output all out of wack because unless the C++ program calls the cout.flush() function, the output does not get written before the input is given.
Is there a way to automatically flush the buffer so the C++ program does not have to worry about calling cout.flush()? Similiar to as if the C++ program was a standalone program interacting with the command console, the programmer doesn't always need the cout.flush(), the command console still outputs the data before the input.
Thank you,
In case someone comes looking for a way to set cout to always flush. Which can be totally fair when doing some coredump investigation or the like.
Have a look to std::unitbuf.
std::cout << std::unitbuf;
At the beginning of the program.
It will flush at every insertion by default.
I can't guarantee that it will fix all of your problems, but to automatically flush the stream when you're couting you can use endl
e.g.:
cout << "Please enter two numbers to add: " << endl;
using "\n" doesn't flush the stream, like if you were doing:
cout << "Please enter two numbers to add:\n";
Keep in mind that using endl can be (relatively) slow if you're doing a lot of outputting
See this question for more info