#include<stdio.h>
#include<conio.h>
void main()
{
static int i=5;
clrscr();
if(--i)
{
printf("%d",i);
main();
}
getch();
}
When I am running this code, it is not giving any output and when I am removing getch() then after running when I am switching to output screen then it is showing ouput . Why?
I am using Turbo C++.
You would have seen output with debuger if you break on each iteration, but since you are obviously just executing your program what you see is the result of the last clrscr();. Since i is 0 you're not getting into the if, where the printing happens and you get right at the getch(). Hope that answers your question.
The printf function buffers output internally until either
the buffer is full
it's asked to print a newline (\n)
stdout is flushed with fflush()
You are not doing either of 2 or 3 and the buffer is surely bigger than 5 integers.
You probably want
printf("%d\n",i);
If you want the ints to be printed without a new line,
printf("%d ",i);
fflush(stdout);
Related
I'm trying to run a simple C program in Eclipse and I am new to Eclipse.
When running the command I see output of the program for a very brief moment in the console, but instead for the program to wait for input, output disappears and and an empty console is shown instead. Is this a problem of the program or something related to the settings of the Eclipse IDE?
Edit: On the console menu there is a terminate button and when I pres it the output of the program shows but it doesn't wait for input and the program is terminated. Any help?
Image:
Here is the code
//The input is a series of numbers and the number X. Find
//how many times the number X is met in the series
#include<stdio.h>
int main(){
int m[20];
int n, X, cnt=0;
printf("Insert maximum length of the series (n<=20)\n");
scanf("n= %d",&n);
printf("Insert X the number you are looking for\n");
scanf("X= %d",&X);
while(n<1||n>20){
printf("Bad input, n should be between 1 and 20\n");
scanf("n= %d",&n);
}
for(int i = 0 ; i < 20 ; i++){
scanf(" %d", &m[i]);
if(m[i]==X)cnt++;
}
printf("X is found %d times\n", cnt);
return 0;
}
and here is the empty console:
Edit: I've tried rewriting the code using cout and cin but then the compiler in Eclipse complains on the line #include<iostream.h> saying that there is an unresolved inclusion. Sounds like a separate question to me.
The answer is here: click
Basically the output of the standard input output is buffered and setvbuf(stdout, NULL, _IONBF, 0); solves the problem.
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.
I am using Microsoft Visual Studio 2012 Professional. I have a simple CPP code:
#include "stdio.h"
int main () {
int a = 0, b = 0, c = 0;
printf("\nEnter two numbers:\n");
scanf("%d %d", &a, &b);
c = a+b;
printf("\nc = %d", c);
return 0;
}
The problem I am facing is that the first printf statement is getting printed but the second one is not. The code is exiting without printing anything. However when I debug using breakpoints, the value of c is getting stored in it.
I'll take a wild guess and say you run this from your IDE.
It does print the output. But the program then ends. And it ends so fast, you cannot see the output. If you want your program to wait for you to see it, you will need to find a way to make it wait. For example with another scanf line.
As there is no standard, environment agnostic way to say "Press any key to continue", have a look here for some alternatives.
Add \n to the end of the last printf. It is implementation-defined whether the last line of output to stdout actually gets flushed , if it doesn't end in a newline.
When you use
scanf("%d %d", &a, &b);
then you should give 2 inputs separated by space or enter
scanf("%d,%d", &a, &b);
then you should give 2 input separated by comma(,).
and at last use
printf("Press any key to continue.."); getche(); instead of return;
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;
}
I wrote program in C , but after compiling and running it the output console does not stay on after I enter anything. My program requires to enter distance and time.
My code is here:
#include <stdio.h>
int spd(int x , int y);
int main() {
int x,y;
printf("enter the distance first then time in their SI units :\n");
scanf("%d",&x);
scanf("%d",&y);
printf("the speed required is ",spd(x,y));
getch();
return 0;
}
int spd(int x , int y) {
return x/y;
}
It seems that the getch feeds on your input-buffer. When you enter your value in scanf(), the '\n' remains.
Try to put flush(stdin) after your scanf() or for now put getch () twice.
If you want to run a console program, you should run it from a console. You have written a console program -- run it from a console.
Also, you need to either flush standard out or write a newline before the getch call. Otherwise, you're waiting for a keypress before you've actually written anything.
Try including this header file : -
#include conio.h