Just a quick newbie question: I'm doing a console application and whenever I open a .exe using system(), the console pauses and does not run the next instruction until I close this exe. For example when I do this:
...
system("mspaint.exe");
printf("Hello");
It opens ms paint and does not print the next message until Paint is closed. How do I work around this?
I am trying to display a messagebox, but instead of just displaying the message box, there is also a console the pops up. How do I prevent the console from popping up? I'm using Visual Studio 2012. I'm creating an empty project, but I still get the console.
View 1 Replies View RelatedI'm working on an assignment that's supposed to open a text file, read the first 24 lines and then pause if there are more than 24 lines and wait for the user to hit enter for every 24 lines. my text file has 27 lines and each line is just a number 1-27. My program isn't stopping at 24, it's going all the way to 27 but when it's displaying the contents it shows the numbers 5-27.
#include <iostream>
#include <string>
#include <fstream>
using namespace std;
int main() {
fstream file;
string name;
[code]....
I've got the cin.get(); in hopes to pause the program after 24 but it's obviously not working
Are there anyways to pause the loop until either LCONTROL or RCONTROL is pressed?
Are there alternatives to stopping LCONTROL and RCONTROL increasing or decreasing counter by a huge sum?
#include <iostream>
#include <windows.h>
#include <conio.h>
[Code].....
I'm trying to create a connect four game, but I've ran into a problem while trying to read valid user input.
Code:
// 4 in a row game
#include <stdio.h>
#include <stdlib.h>
#define NUM_ROWS 6
#define NUM_COLS 7
}
[code]...
If the user enters an invalid number, then it works fine, but if a character is entered, the while loop in main never breaks.
I'm new to C programming and am quiet lost, create a program that takes a start and stop time and prints the passed, or show me where I can find the information needed?
View 6 Replies View RelatedI found a for loop in an example that I don't understund fully.
Code:
for (i=64; i; i/=2)
printf("i: %d
",i);
Now this is dividing by 2 until it reach '1' and stops. But why does this stops?
I'm trying to get the user to input their 5 digit zip code but my problem is how can I stop the program after the five characters have been inputted. Also how can I only get numbers to be inputted cause so far letters can be inputted as well.
View 2 Replies View RelatedMy program is designed to take a reading from a glucometer and light an LED if the reading is too high or too low. I used this code:
#include <eHealth.h>
void setup(){
pinMode(13, OUTPUT); //first alert
pinMode(12, OUTPUT); //green LED
[Code]....
The glucometer stores the tests and the code just pulls the most recent one, however even if I unplug the glucometer the first if statement keeps repeating and the LED in pin 12 keeps shining. Is it the if statement itself malfunctioning, or is the Arduino storing the data it pulled and just repeatedly plugging that into the if loop?
Having multiple issue trying to get a while loop to stop after the third entry.
#include <iostream>
#include <fstream>
#include <string>
#include <stdlib.h>
using namespace std;
int mySequentialSearch() {
int value;
string line;
[Code] .....
I am just working on the while loop right now. Then the function after I get the loop working.
double number = 10000000;
int range;//the length of the string result
string result;//holds the number in a string
ostringstream convert; //stream used for the conversion
convert << number;
result = convert.str();
range = result.length();
I'm trying to convert a double to a string and when the number goes to ten million it goes to scientific notation and it shows it in the string. How do I stop it from do that?
Code:
#include <stdio.h>
#include <stdlib.h>
/*
You are to write a program that will allow a user to enter infinite numbers (greater than zero)(One number at a time). You must have a way for the user to stop entering values.Once the user stops, you will immediately display the following:
The lowest number entered is:
The highest number entered is:
The number of values entered is:
The average of the numbers entered is:
*/
main() {
int userNumber=0, sum=0, count=0, highNum=0, lowNum=0, lastNum = 0;
double average;
printf("Enter a number greater than 0: (Enter -1 to stop)
");
scanf_s("%i", &userNumber);
[Code]....
My program outputs everything I need except the lowest number entered. I have been spending a considerable amount of time on this problem to no avail.
If I wanted to add an emergency stop to an elevator programme should I use an if statement? Eg
Code:
If (emergency stop) {
Void main();
}
Or something along those lines?
This is a problem I have been having with every program I write since I started using SDL 2. Whenever I compile my code and run my program, everything works perfectly fine until at some point (usually after 3-8 minutes of running), the program will stop responding completely and I will have to exit out of the console to close it. The code I believe is relevant is:
void MainLoop()
{
InitLoop();
while ( !QuitMain )
[Code].....
If I change SDL_PollEvent(&Event) to SDL_WaitEventTimeout(&Event,100), then the problem goes away (I did that and had the program running for about an hour without it stop responding before I decided that it solved the problem), so I believe that the problem has something to do with event handling. Also, it might be noteworthy to mention that when I use SDL_WaitEventTimeout with the second parameter being a small number (because 100 milliseconds is a long time to wait and makes the program run at like 8 FPS), the problem returns.
My program has a goto label in it and when the program is not instructed to goto the label, it does. When control goes to the line that it is on, even if no where does it say to execute the label's content, it does execute it.
View 6 Replies View RelatedI am new to c++ programming, and I wrote this program. It runs perfectly but the problem is I want the program to stop If I entered letters other than A,B or C
The code as follows:
#include <cstdlib>
#include <iostream>
#include <cstring>
#include <iomanip>
using namespace std ;
int main() {
// Deceleration Statements
[Code] ....
#include <iostream>
using namespace std;
int main() {
[Code].....
When my program is working with a big list, it always connects to the internet and checks information from a website. It is going fast, but when I'm trying to update a label on every connection, it just shows at the end, when everything is checked. Progressbar value is updating. How to solve this?
View 1 Replies View RelatedPHP Code:
using namespace System;
ref class A {
public:
A(){Console::WriteLine(L"A constructor");}
virtual void func(){foo();}
void foo()
[Code] .....
Supposing that class A is the development code; and so for unit-testing purpose I create a mock for it. But I don't know how to override A's constructor such that it will not get executed in my unit-test code.
I have a simple UDP client that should be able to send and receive packets. It does its job, however if there are no packets received for a while it stops listening for any new packets and I'm not sure what's causing it.
This is my code:
#define WIN32_LEAN_AND_MEAN
#include "Client.h"
#include <winsock2.h>
#include <ws2tcpip.h>
#include <vector>
#include <sstream>
#pragma comment(lib, "Ws2_32.lib")
[Code] ....
is there a way to stop the cursor from blinking in the output window(or showing up at all)?
Example:
cout<<"Hello world";
Output:
Hello world_
//where last character keeps blinking
I use DevC++ in Windows 7.
I wondering how to stop a infinite loop. The program clears a file then opens it and refreshes it over and over(the file is going to be modified by another program). I want it to start from the beginning anytime i press enter or escape, it doesn't really matter as long as you can restart it.
Here is the complete code:
#include <cstdlib>
#include <iostream>
#include <fstream>
#include <conio.h>
#include <windows.h>
using namespace std;
void x() {Sleep(1000);}
[Code] .....
It's a tic-tac-toe program. I haven't finished doing everything I need to for the project, I'm just trying to get certain parts working as I go.
Anyway, my problem: The program will print the board and ask the first player which square they want to mark. Then it will print the updated board, print the question for the second player to input their square choice but not actually stop to let the user enter anything. Instead it prints the board again and then prints the question for player one again -- this time stopping to let them type in their choice.
Here's my code:
#include <stdio.h>
// Function prototypes
void printBoard(char board[3][3]);
int main() {
int quit = 0;
// Loop so game continues until player quits
[Code] .....
And it looks like this when I run it, as an example:
-------------
| 1 | 2 | 3 |
-------------
| 4 | 5 | 6 |
-------------
| 7 | 8 | 9 |
-------------
Player 'C' enter a square: 5
-------------
| 1 | 2 | 3 |
-------------
| 4 | C | 6 |
-------------
| 7 | 8 | 9 |
-------------
Player 'U' enter a square:
-------------
| 1 | 2 | 3 |
-------------
| 4 | C | 6 |
-------------
| 7 | 8 | 9 |
-------------
Player 'C' enter a square:
i am tring to update a sql database table, first i select the values from database after it i display them into text boxes and drop dwon lists.and then i make required changes to values but when i click on update button the page is post backed and all values are set as they were already in database.
View 1 Replies View RelatedProgram in c++ that if i ran it the cmd window will stop without using getch statement and conio.h header file, or how ?
View 7 Replies View RelatedMy question is how the button click mull ti times on page load in C# windows form??because in my project i am working on video for LAN based Skype in which i use 2 pictures boxes one for capture images and another for display .. so i want that on page load picture box capture images until i stop it .. following is my code :
image = ((CaptureDevice)cboDevices.SelectedItem).Capture() ;
pictureBox2.Image = image;
