how to open file from user entered directory location

[bkara]

Hello,

I would like to write a program that asks the user the location of a file.
example of what the user would enter/input:
C:\Users\Documents\Visual Studio 2017\file.txt

then the program opens the file.

after the file is opened the program then asks the user where they would like to save the file.
the user would enter different location
example:C:\Users\Downloads

Is it possible to write a program that can do this? I tried googling stuff related to this topic but I have yet to find something that can really help me.

(After the text file is opened the program would convert it to a csv file, I already figured out how to convert the file)

[H00G0]

You can get the file location with std::cin as a string e.g. std::string input = "C:\Users\Documents\Visual Studio 2017\file.txt"

EDIT: Additionally if your file is in the same fiolder as your .exe (at least in visual studio) you can name it directly without using C:\.... e.g. std::string input = "file.txt"

Then you can open your file like so:
yourFile.open(input)

Whereas it is for opening or saving your file, the process remains the same.

For more information and examples on file operations you might want to check the following link: http://www.cplusplus.com/doc/tutorial/files/

Hope this helps,

Regards,

Hugo.

[ria1]

@hoogo can you help me in implementing the below hackerrank question ---

Janet is in an Uber on her way to an interview. The driver promises to take her to the venue as soon as possible. The driver is aware
that:-
-There are 'n' junctions in the city of Mumbai, numbered from '1' to 'n'.
-Janet's interview location is at junction 'n'. They are initially at junction '1'.
-There are 'm' bidirectional roads connecting some pairs of junctions, each one requiring some amount of time to pass through it.

At every junction, there are traffic lights denoting whether they are allowed to go further or to wait. Traffic lights have two colors,
'red' and 'green'. The driver can commute through junctions based on these conditions:
1. At any junction, if the traffic signal's light is green, then they can go immediately, otherwise, they have to wait until traffic signal
becomes green.
2. Traffic signal changes its color every 'k' seconds of time at all junctions simultaneously.

Initially, at the '0th' second, all traffic lights have changed to green color at all the junctions. If the cab driver reaches a junction at a second when the traffic light changes its color, then he sees the traffic light after the change.
Can you help the driver determine the least amount of time needed to reach the interview location?
Complete the function leastTimeToInterview which takes in three integers n,k and m and returns the least amount of time
needed to reach the interview location, in seconds. You need to take the information about the roads from the standard input. They
will be specified in 'm' lines, as described in the input format section below.

Input Format:-

-The first line contains an integer 'n', the number of junctions.
-The second line contains an integer 'k' denoting the time taken by a signal to change its color.
-The third line contains an integer 'm' denoting the number of roads.
-The next 'm' lines describe the roads. Each consist of three space-separated integers 'i','j' and 't' where 'i' and 'j' denotes a road between two junctions and 't' denotes time required to travel through it.

Constraints-:
- 1<= n <= pow(10,4)
- 1<= k <= pow(10,2)
- 1<= m <= pow(10,5)
- 1<= i,j <= n
- 1<= t <= pow(10,3)

- There can be self-loops, i.e., roads connecting the same pair of junctions.
- There is at least one path from junction '1' to junction 'n'.

Output Format-:

Print a single integer denoting the shortest amount of time required to reach junction 'n'.

Sample Input 0
7
4
7
1 2 3
2 3 1
1 4 4
4 6 7
7 5 2
3 5 1
4 5 5

Sample Output 0
11


Explanation 0-

- Junction number 1: The cab driver can visit any of the adjacent junctions. He chooses to visit junction 2. Since the traffic signal is
'green' at the '0th' second, He can visit junction 2 which takes 3 seconds.
- Junction number 2: The traffic signal is still 'green' since traffic signals change color every '4' seconds. The cab driver now chooses
to visit junction '3' which takes '1' second, and we are now at the '4th' second.
- Junction number 3: 4 seconds have passed, and the traffic signal has already become 'red'. They have to wait for 4 more seconds
until the signal again becomes 'green'.
- Junction number 5: The cab takes the route to junction 5 which takes 1 second. So far, 9 seconds have passed.
- Junction number 7: The traffic signal is still 'green' and the cab can go to junction 7. It takes 2 seconds to reach junction 7.

In total, it takes 11 seconds to go from junction 1 to junction 7 . It can be shown that this is the minimum amount of time possible.


#include <bits/stdc++.h>
using namespace std;

// Complete the leastTimeToInterview function below.
long leastTimeToInterview(int n, int k, int m) {

// Return the least amount of time needed to reach the interview location in seconds.
}

int main()
{
ofstream fout(getenv("OUTPUT_PATH"));
int n;
cin >> n;

cin.ignore(numeric_limits<streamsize>::max(), '\n');
int k;
cin >> k;

cin.ignore(numeric_limits<streamsize>::max(), '\n');
int m;
cin >> m;

cin.ignore(numeric_limits<streamsize>::max(), '\n');
long result = leastTimeToInterview(n, k, m);
fout << result << "\n";
fout.close();
return 0;
}