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Simple c++ help needed

Posts 1–15 of 21 · Page 1 of 2

I'm pretty sure there is a very simple way to do this but I haven't found a good explanation for how to do it. How would I go about getting an input from a user that I can verify is an integer?

One other question I have is related to code similar to this:

Code:

int factor(long innum)
{
	int prev = 0;
	ofstream factors;
	factors.open("factors.txt");
	if (factors.is_open())
	{
		factors << "Factors of " << innum << ":" << endl << endl;
		for (long loops = 1; loops <= innum; loops++)
		{
			if (innum % loops == 0)
			{
				factors << loops << " ";
				prev++;
			}
			if (prev == 7)
			{
				factors << endl;
				prev = 0;
			}
		}
	}
	else
	{
		clearscreen
		wait
		cout << "Error opening file!!!" << endl << "Exiting...";
		return(1);
	}
	factors.close();
	return(0);
}

Now thats the code I have for finding factors of a number. I would want to add a percent type thing to it but I'm not sure how I would do that. I could divide 'loops' by 'innum' but how could I control how many decimal places are kept if i wanted to display something like xx.xx%

Thanks for any help in advance

char f;
cin >> f;

Oh Wait You Want Integer so It Would Be int f;
cin >> f;

Yea but now how do I now take it so it can be manipulated as a number and how do I verify there are no letters in it?

Edit: I wrote that code myself I do know what I am doing to this level I am asking how do I get input from the user and extract a number from it but also making sure that nothing but numbers were typed,

Originally Posted by hobosrock696

Yea but now how do I now take it so it can be manipulated as a number and how do I verify there are no letters in it?

Edit: I wrote that code myself I do know what I am doing to this level I am asking how do I get input from the user and extract a number from it but also making sure that nothing but numbers were typed,

Soon As They Type In The Integer For f Its Goin to Be Saved In So Yea

I think he want to make sure that the user don't type in a letter or smth.

One Solution:

Code:

 char input;
cin>>input;
testinputaboutascii();
Convert char to int (sprintf should work)

i think this is ONE solution but there should be millions of ways you could do it

ASCII codes allows would be 0x41 - 0x5A (A-Z) and 0x61 - 0x7A (a-z)

check the char input with a pointer the the char and then go one step forward
if(*char==......)
char++;

Originally Posted by doofbla

I think he want to make sure that the user don't type in a letter or smth.

One Solution:

Code:

char input; cin>>input; testinputaboutascii(); Convert char to int (sprintf should work)

i think this is ONE solution but there should be millions of ways you could do it

ASCII codes allows would be 0x41 - 0x5A (A-Z) and 0x61 - 0x7A (a-z)

check the char input with a pointer the the char and then go one step forward
if(*char==......)
char++;

He Said He Was No0b He Isnt Goin To Know How Too Do All That...Atleast I Think

Originally Posted by whit

He Said He Was No0b He Isnt Goin To Know How Too Do All That...Atleast I Think

Ah ok I didn't think about that so.. ty.

Anyhow I've a new idea how to do:

Maybe you can just disable all keys instead of 0-9 for the time of input

There are tons of ways

I have done pointers and memory all that jazz im not great with it but this I can handle. Yes I consider myself a noob and will continue to until I know c++ very well and at least a bit of the windows api + direct x

Question:

Would I check for ascii? Or is it unicode? Is it different for linux or for macs?

Ok so I tried writing some code but obviosly im doing something wrong since it doesnt work. Here it is....

Code:

// testingchar.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
using namespace std;


int main()
{
	char * input = "a";
	int counter = 0;
	cin.getline(input, 20);
	while (input != "\0")
	{
		cout << input[counter] << endl;
		counter++;
	}
	return 0;
}

Oh my noob self forgot to add the * before the pointer. Problem is It still wont compile... Anyone willing to write this into a WORKING example XD and thanks for any help once again!

Originally Posted by hobosrock696

Ok so I tried writing some code but obviosly im doing something wrong since it doesnt work. Here it is....

Code:

// testingchar.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
using namespace std;


int main()
{
	char * input = "a";
	int counter = 0;
	cin.getline(input, 20);
	while (input != "\0")
	{
		cout << input[counter] << endl;
		counter++;
	}
	return 0;
}

Oh my noob self forgot to add the * before the pointer. Problem is It still wont compile... Anyone willing to write this into a WORKING example XD and thanks for any help once again!

define input as an array.
char only holds 1 character

Ok i made it for you -.- but i think next time you should show more self engagement:

Code:

#include "windows.h"
#include "stdio.h"
#include "iostream"

using namespace std;

char input[20];

bool checkchar(char * in)
{
    while(*in != 0)
    {
        if(*in < 0x30 || *in > 0x39)
        {
            return false;
        }
        in++;
    }
    return true;
}

int main()
{
    cin.getline(input, 20);

    if(checkchar(input))
    {
        cout<<"You entered only numbers - OK"<<endl;
    }
    else
    {
        cout<<"You entered NOT only numbers - FAIL"<<endl;
    }

    system("PAUSE");
    return 0;
}

with this you can proof if the user entered only numbers -> in ASCII it works

Thanks and well this is why I am a noob. Also yes thats true I didnt think about that much I apologize for being.... a sort of leecher in this case

Edit: Shouldnt you also not be checking for 0 but "\0" which signifies the end of a string of chars? Or is that just me being wrong?

just make a for cicle and check if the variable of the loop is equal to your imput, like:

int f;

cin>>f;

for(int j=0; j<10; j++)
if(f==j)
//it's a number

Yea I was looking at the code. I'm assuming the compiler would optimize that into the same code in the end but I still prefer a for over a while loop in which a variable is incremented.

Originally Posted by hobosrock696

Thanks and well this is why I am a noob. Also yes thats true I didnt think about that much I apologize for being.... a sort of leecher in this case

Edit: Shouldnt you also not be checking for 0 but "\0" which signifies the end of a string of chars? Or is that just me being wrong?

0x48 = 0
0x49 = 1
.
.
.
0x57 = 9

0x0 is a what you get whan you press SPACE

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Home › Forum › Programming › C++/C Programming › Simple c++ help needed

Simple c++ help needed

Posts 1–15 of 21 · Page 1 of 2

Simple c++ help needed

Code:

int factor(long innum)
{
	int prev = 0;
	ofstream factors;
	factors.open("factors.txt");
	if (factors.is_open())
	{
		factors << "Factors of " << innum << ":" << endl << endl;
		for (long loops = 1; loops <= innum; loops++)
		{
			if (innum % loops == 0)
			{
				factors << loops << " ";
				prev++;
			}
			if (prev == 7)
			{
				factors << endl;
				prev = 0;
			}
		}
	}
	else
	{
		clearscreen
		wait
		cout << "Error opening file!!!" << endl << "Exiting...";
		return(1);
	}
	factors.close();
	return(0);
}

char f;
cin >> f;

Oh Wait You Want Integer so It Would Be int f;
cin >> f;

I think he want to make sure that the user don't type in a letter or smth.

One Solution:

Code:

 char input;
cin>>input;
testinputaboutascii();
Convert char to int (sprintf should work)

i think this is ONE solution but there should be millions of ways you could do it

ASCII codes allows would be 0x41 - 0x5A (A-Z) and 0x61 - 0x7A (a-z)

check the char input with a pointer the the char and then go one step forward
if(*char==......)
char++;

Originally Posted by whit

He Said He Was No0b He Isnt Goin To Know How Too Do All That...Atleast I Think

Ah ok I didn't think about that so.. ty.

Anyhow I've a new idea how to do:

Maybe you can just disable all keys instead of 0-9 for the time of input

There are tons of ways

Ok so I tried writing some code but obviosly im doing something wrong since it doesnt work. Here it is....

Code:

// testingchar.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
using namespace std;


int main()
{
	char * input = "a";
	int counter = 0;
	cin.getline(input, 20);
	while (input != "\0")
	{
		cout << input[counter] << endl;
		counter++;
	}
	return 0;
}

Oh my noob self forgot to add the * before the pointer. Problem is It still wont compile... Anyone willing to write this into a WORKING example XD and thanks for any help once again!

Originally Posted by hobosrock696

Ok so I tried writing some code but obviosly im doing something wrong since it doesnt work. Here it is....

Code:

// testingchar.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
using namespace std;


int main()
{
	char * input = "a";
	int counter = 0;
	cin.getline(input, 20);
	while (input != "\0")
	{
		cout << input[counter] << endl;
		counter++;
	}
	return 0;
}

Oh my noob self forgot to add the * before the pointer. Problem is It still wont compile... Anyone willing to write this into a WORKING example XD and thanks for any help once again!

define input as an array.
char only holds 1 character

Ok i made it for you -.- but i think next time you should show more self engagement:

Code:

#include "windows.h"
#include "stdio.h"
#include "iostream"

using namespace std;

char input[20];

bool checkchar(char * in)
{
    while(*in != 0)
    {
        if(*in < 0x30 || *in > 0x39)
        {
            return false;
        }
        in++;
    }
    return true;
}

int main()
{
    cin.getline(input, 20);

    if(checkchar(input))
    {
        cout<<"You entered only numbers - OK"<<endl;
    }
    else
    {
        cout<<"You entered NOT only numbers - FAIL"<<endl;
    }

    system("PAUSE");
    return 0;
}

with this you can proof if the user entered only numbers -> in ASCII it works

Thanks and well this is why I am a noob. Also yes thats true I didnt think about that much I apologize for being.... a sort of leecher in this case

Edit: Shouldnt you also not be checking for 0 but "\0" which signifies the end of a string of chars? Or is that just me being wrong?

just make a for cicle and check if the variable of the loop is equal to your imput, like:

int f;

cin>>f;

for(int j=0; j<10; j++)
if(f==j)
//it's a number

Yea I was looking at the code. I'm assuming the compiler would optimize that into the same code in the end but I still prefer a for over a while loop in which a variable is incremented.

Originally Posted by hobosrock696

Thanks and well this is why I am a noob. Also yes thats true I didnt think about that much I apologize for being.... a sort of leecher in this case

Edit: Shouldnt you also not be checking for 0 but "\0" which signifies the end of a string of chars? Or is that just me being wrong?

0x48 = 0
0x49 = 1
.
.
.
0x57 = 9

0x0 is a what you get whan you press SPACE

Posts 1–15 of 21 · Page 1 of 2

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