Programming Forums > Application Development > C++

bl00dninja · Dec 10th, 2006, 1:13 AM

off-topic slightly...what about forgetting the string thing.

depending on your I/O scheme it may differ in emplementation but...
what if you enter the number didgit by digit first into a lifo container like a stack, then insert digit by digit into a fifo container like a queue, then pop() each and add the result cumulatively as each pops by mutiplying it by pow(10, i) where i is a loop index starting at zero?

example:

123 is your num

fifo pops like 1, 2, 3

lifo pops like 3, 2, 1

1*10^0 = 1; 3*10^0 = 3; add them you get 4.

2*10^1 = 20; 2*10^1 = 20; add them you get 40.

3*10^2 = 300; 1*10^2 = 100; add them you get 400.

you know the end value of i by the size() of the container.

add all three you get 444 which = 123 + 321.

or if this sounds like gibberish then fuck me, nevermind. :p

bl00dninja · Dec 10th, 2006, 2:17 AM

ignoring crap like I/O (this prog only accepts 3-digit numbers) here's the basic idea:

(Toggle Plain Text)

#include <iostream>
#include <stack>
#include <queue>
#include <cmath>  //for pow()
using namespace std;

int main()
{
	//declare data structures
	stack<float> lifo;
	queue<float> fifo;

	//value to be printed at end
	int final = 0;

	//populate the containers...
	lifo.push(7);
	lifo.push(6);
	lifo.push(8);

	fifo.push(7);
	fifo.push(6);
	fifo.push(8);

	for(float i = 0; i < 3; i++)//run three times
	{
		//run algorithm
		final += ( lifo.top() * pow(10, i) ) + (fifo.front() * pow(10, i) );
		//remove those values for next iteration
		lifo.pop();
		fifo.pop();
	}

	//show result
	cout<<final<<endl;

	//pause for output
	system("PAUSE");
	return 0;
}

magnus.therning · Dec 10th, 2006, 2:59 AM

Quote:

Originally Posted by Cool1Net6

In a recent contest, the problem was to reverse the digits of a number and add that new number to the old number to come up with a newer number. I had the knowledge to reverse the digits of a number by converting the original number to a string and swapping the positions of each character, but to add this new number to the original number, I needed to convert the new number back to an int. I lost the competition because I lacked this knowledge.

How do you convert a String to an Int?

-Cool-

You can always use sprintf or one of its family members.

DaWei · Dec 10th, 2006, 5:05 AM

In C, sprintf is a very good option. However, this is the C++ forum. C++ has its own mechanisms available that do these things. An OP is not generally looking to change languages as a solution, but is attempting to learn the language associated with the forum in which the question is posted.

It's also somewhat confounding to me that some respondents are still recommending 'atoi', which is a baddie, and recommending that the OP process each character independently. That's what the intended tools are for, and they cover the job more thoroughly than your average coder will be able to do. Simply consider a number represented in scientific notation to get an idea of what I mean.

Compare the code in posts #3 and #8 to that in #12, bearing in mind that the code in #8 is doing it three times, not just once, as an example of handling input in three different forms.

If I were the OP, I'd have grabbed the solution long ago, run with it, and be done with it.

Jessehk · Dec 10th, 2006, 10:12 AM

I've wrapped DaWei's and Soulstorm's recommendation of using stringstreams into two (what I think are) useful functions, and a sample of their use in main().

 cpp Syntax (Toggle Plain Text) 
#include <iostream>
#include <sstream>
#include <string>
#include <stdexcept>
 
template <typename T>
T read_string( const std::string &input ) throw ( std::invalid_argument ) {
    std::istringstream source( input );
 
    T result;
 
    if ( !(source >> result) )
        // If the operation fails,
        // one possible course of action 
        // is to throw an exception.
        throw std::invalid_argument( "Could not parse string!" );
 
    return result;
}
 
template <typename T>
std::string to_string( const T &input ) {
    std::ostringstream result;
 
    result << input;
    return result.str();
}
 
int main() {
    std::cout << read_string<int>( "12" ) << std::endl;
    std::cout << read_string<float>( "3.14159" ) << std::endl;
    std::cout << to_string( 12 ) << std::endl;
    std::cout << to_string( 3.14159 ) << std::endl;
}
#include <iostream>
#include <sstream>
#include <string>
#include <stdexcept>

template <typename T>
T read_string( const std::string &input ) throw ( std::invalid_argument ) {
    std::istringstream source( input );

    T result;

    if ( !(source >> result) )
        // If the operation fails,
        // one possible course of action 
        // is to throw an exception.
        throw std::invalid_argument( "Could not parse string!" );

    return result;
}

template <typename T>
std::string to_string( const T &input ) {
    std::ostringstream result;

    result << input;
    return result.str();
}

int main() {
    std::cout << read_string<int>( "12" ) << std::endl;
    std::cout << read_string<float>( "3.14159" ) << std::endl;
    std::cout << to_string( 12 ) << std::endl;
    std::cout << to_string( 3.14159 ) << std::endl;
}

Klipt · Dec 10th, 2006, 3:48 PM

You can also reverse a positive number without any string conversions, like so:

(Toggle Plain Text)

int reverse_decimal(int number)
{
	int rebmun = 0;
	while (number > 0)
	{
		rebmun *= 10;
		rebmun += (number % 10);
		number /= 10;
	}
	return rebmun;
}

(Note that if you give it 100 it will give you 1, indistinguishable from 001 - if you give it 1, it'll give you 1 back. So reversing twice doesn't give you back the original number.

If you change the 10's to 2's it'll reverse it in binary.)

Dec 10th, 2006, 1:13 AM	#11
bl00dninja Programming Guru Join Date: Oct 2004 Location: namespace std Posts: 1,246 Rep Power: 5	off-topic slightly...what about forgetting the string thing. depending on your I/O scheme it may differ in emplementation but... what if you enter the number didgit by digit first into a lifo container like a stack, then insert digit by digit into a fifo container like a queue, then pop() each and add the result cumulatively as each pops by mutiplying it by pow(10, i) where i is a loop index starting at zero? example: 123 is your num fifo pops like 1, 2, 3 lifo pops like 3, 2, 1 110^0 = 1; 310^0 = 3; add them you get 4. 210^1 = 20; 210^1 = 20; add them you get 40. 310^2 = 300; 110^2 = 100; add them you get 400. you know the end value of i by the size() of the container. add all three you get 444 which = 123 + 321. or if this sounds like gibberish then fuck me, nevermind. :p __________________ i put on my robe and wizard hat... Have you ever heard of Plato, Aristotle, Socrates?...Morons.

Dec 10th, 2006, 5:05 AM	#14
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	In C, sprintf is a very good option. However, this is the C++ forum. C++ has its own mechanisms available that do these things. An OP is not generally looking to change languages as a solution, but is attempting to learn the language associated with the forum in which the question is posted. It's also somewhat confounding to me that some respondents are still recommending 'atoi', which is a baddie, and recommending that the OP process each character independently. That's what the intended tools are for, and they cover the job more thoroughly than your average coder will be able to do. Simply consider a number represented in scientific notation to get an idea of what I mean. Compare the code in posts #3 and #8 to that in #12, bearing in mind that the code in #8 is doing it three times, not just once, as an example of handling input in three different forms. If I were the OP, I'd have grabbed the solution long ago, run with it, and be done with it. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Dec 10th, 2006, 10:12 AM	#15
Jessehk The Oblivious One Join Date: May 2005 Location: Ontario, Canada Posts: 641 Rep Power: 4	I've wrapped DaWei's and Soulstorm's recommendation of using stringstreams into two (what I think are) useful functions, and a sample of their use in main(). cpp Syntax (Toggle Plain Text) #include <iostream> #include <sstream> #include <string> #include <stdexcept> template <typename T> T read_string( const std::string &input ) throw ( std::invalid_argument ) { std::istringstream source( input ); T result; if ( !(source >> result) ) // If the operation fails, // one possible course of action // is to throw an exception. throw std::invalid_argument( "Could not parse string!" ); return result; } template <typename T> std::string to_string( const T &input ) { std::ostringstream result; result << input; return result.str(); } int main() { std::cout << read_string<int>( "12" ) << std::endl; std::cout << read_string<float>( "3.14159" ) << std::endl; std::cout << to_string( 12 ) << std::endl; std::cout << to_string( 3.14159 ) << std::endl; } #include <iostream> #include <sstream> #include <string> #include <stdexcept> template <typename T> T read_string( const std::string &input ) throw ( std::invalid_argument ) { std::istringstream source( input ); T result; if ( !(source >> result) ) // If the operation fails, // one possible course of action // is to throw an exception. throw std::invalid_argument( "Could not parse string!" ); return result; } template <typename T> std::string to_string( const T &input ) { std::ostringstream result; result << input; return result.str(); } int main() { std::cout << read_string<int>( "12" ) << std::endl; std::cout << read_string<float>( "3.14159" ) << std::endl; std::cout << to_string( 12 ) << std::endl; std::cout << to_string( 3.14159 ) << std::endl; } __________________ Dr. Zoidberg: [ecstatic] I'm going to a movie... with FRIENDS!

Dec 10th, 2006, 3:48 PM	#16
Klipt Hobbyist Programmer Join Date: Dec 2005 Posts: 118 Rep Power: 0	You can also reverse a positive number without any string conversions, like so: (Toggle Plain Text) int reverse_decimal(int number) { int rebmun = 0; while (number > 0) { rebmun = 10; rebmun += (number % 10); number /= 10; } return rebmun; } int reverse_decimal(int number) { int rebmun = 0; while (number > 0) { rebmun = 10; rebmun += (number % 10); number /= 10; } return rebmun; } (Note that if you give it 100 it will give you 1, indistinguishable from 001 - if you give it 1, it'll give you 1 back. So reversing twice doesn't give you back the original number. If you change the 10's to 2's it'll reverse it in binary.) __________________ Get PLT Scheme\|SB Common Lisp\|Haskell\|wxDev-C++\|Python Last edited by Klipt; Dec 10th, 2006 at 3:59 PM.

Thread Tools
Show Printable Version Email this Page
Display Modes
Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

Dec 10th, 2006, 2:17 AM	#12
bl00dninja Programming Guru Join Date: Oct 2004 Location: namespace std Posts: 1,246 Rep Power: 5	ignoring crap like I/O (this prog only accepts 3-digit numbers) here's the basic idea: (Toggle Plain Text) #include <iostream> #include <stack> #include <queue> #include <cmath> //for pow() using namespace std; int main() { //declare data structures stack<float> lifo; queue<float> fifo; //value to be printed at end int final = 0; //populate the containers... lifo.push(7); lifo.push(6); lifo.push(8); fifo.push(7); fifo.push(6); fifo.push(8); for(float i = 0; i < 3; i++)//run three times { //run algorithm final += ( lifo.top() * pow(10, i) ) + (fifo.front() * pow(10, i) ); //remove those values for next iteration lifo.pop(); fifo.pop(); } //show result cout<<final<<endl; //pause for output system("PAUSE"); return 0; } #include <iostream> #include <stack> #include <queue> #include <cmath> //for pow() using namespace std; int main() { //declare data structures stack<float> lifo; queue<float> fifo; //value to be printed at end int final = 0; //populate the containers... lifo.push(7); lifo.push(6); lifo.push(8); fifo.push(7); fifo.push(6); fifo.push(8); for(float i = 0; i < 3; i++)//run three times { //run algorithm final += ( lifo.top() * pow(10, i) ) + (fifo.front() * pow(10, i) ); //remove those values for next iteration lifo.pop(); fifo.pop(); } //show result cout<<final<<endl; //pause for output system("PAUSE"); return 0; } __________________ i put on my robe and wizard hat... Have you ever heard of Plato, Aristotle, Socrates?...Morons.

Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
C# corruption!!!	Kilo	C++	32	May 21st, 2006 8:44 PM
Array issues :(	Alo Tsum	Java	10	Nov 26th, 2005 5:45 PM
Convert a string to unicode string	sma_soft	Delphi	3	Nov 16th, 2005 5:06 AM
replace space with ' * '	TecBrain	C++	15	Apr 13th, 2005 12:32 PM
function	solomon_13000	Java	6	Apr 2nd, 2005 11:42 PM

Programming Forums > Application Development > C++

Convert String to Int