Programming Forums > Application Development > C++

Cool1Net6 · Nov 27th, 2006, 5:21 AM

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-

kruptof · Nov 27th, 2006, 5:43 AM

you could of done string.c_str() which would return you the string in an const char* format then you could loop through that new constant array of characters and convert each one to an number use atoi or something else, some else reccommended.

Soulstorm · Nov 27th, 2006, 7:10 AM

The easiest way to do it is to use a stringstream

example:

 cpp Syntax (Toggle Plain Text) 
#include <iostream>
#include <sstream>
 
int main (int argc, char * const argv[]){
	std::stringstream s("2903");
	int i;
 
	s >> i;
	std::cout << i+100;
 
 
	return 0;
}
#include <iostream>
#include <sstream>

int main (int argc, char * const argv[]){
	std::stringstream s("2903");
	int i;
	
	s >> i;
	std::cout << i+100;
	
	
	return 0;
}

Probably a bad example, but it should give you an idea on what to do. Give to an std::stringstream the number you want as a constant character, or an std::string and extract it to an integer or enything else you want.

You could also use atoi(), but using stringstreams will be a lot easier for you and a lot more comprehensive.

The STL has many things you should read when you encounter a problem. In many occasions, the STL solves the problem. Check this out for a reference.

Lesliect6 · Nov 27th, 2006, 7:51 AM

Hello Coll1Net6,

Converting a string to an int is very easy, having in mind that a string is only a representation of a number in the ASCII chart. Having also in mind that a string is nothing more than an array of chars, you can easily manipulate each individual character. So, if your string is composed only of numbers in string format (like '4' or '6'), all you have to do is :

(Toggle Plain Text)

int number = string[x] - 48;

where x is the index of the char you want converted to int.

And the job is done.

DaWei · Nov 27th, 2006, 8:11 AM

Leslie, your method is contingent upon the characters being in an ASCII representation. Such is not always the case. One should use string streams, as recommended, or perhaps strtol or similar. Atoi is not a good choice because errors are indistinguishable from converting the value, zero (and other badnasty results).

pegasus001 · Dec 7th, 2006, 12:47 PM

int string_to_int(string number)
{
int num = 0;
int temp = 1;

for (int i = number.length; i >= 0; i--)
{
num += atoi(number[i]) * temp;
temp *= 10;
}

return num;
}

Narue · Dec 7th, 2006, 1:38 PM

>for (int i = number.length; i >= 0; i--)
That's an off-by-one error and a syntax error. You want to start at number.length() - 1.

>num += atoi(number[i]) * temp;
Okay, number is a string, so number[i] is a character. That alone is a type mismatch. You could take the address of that character because atoi expects a C-style string (any number of characters terminated with '\0'), but the string class doesn't guarantee that internal representation.

The good news is that your algorithm only expects single characters at a time, so you can use the subtraction trick to turn something like '5' into 5:

(Toggle Plain Text)

num += (number[i] - '0') * temp;

Then your only problem is when the string doesn't represent a valid integer.

DaWei · Dec 7th, 2006, 1:58 PM

Are you people reading the responses? ATOI is not a good thang. Here's a quote from cplusplus (emphases mine):

Quote:

Return Value.

The converted integer value of the input string.
On overflow the result is undefined.
If an error occurs 0 is returned. (How do you know whether it's an error, or a conversion of zero?)

Using this is essentially dereliction, like using 'gets'.

I have modified Soulstorm's code to present some alternatives. All are easy. It's certainly preferable to looping through a set of characters and calling atoi, when there's no guarantee that each of the characters is convertible.

(Toggle Plain Text)

#include <iostream>
#include <sstream>
#include <string>

int main (int argc, char * const argv[])
{
    int myInt;
    std::string aString = "3092";
    char cString [] = "2903";

    // Use a C string
    std::stringstream withCString (aString.c_str ());
    // Use another C string
    std::stringstream withCString2 (cString);
    // Use a string-class string
    std::stringstream withString (aString);

    withCString >> myInt;
    std::cout << myInt << std::endl;
    withCString2 >> myInt;
    std::cout << myInt << std::endl;
    withString >> myInt;
    std::cout << myInt << std::endl;;
   
    return 0;
}

Output:

Quote:

3092
2903
3092

EDIT: Woops, was off keying in the code while Narue posted.

pegasus001 · Dec 9th, 2006, 12:42 PM

Quote:

Originally Posted by Narue

>for (int i = number.length; i >= 0; i--)
That's an off-by-one error and a syntax error. You want to start at number.length() - 1.

I forgot, about that.

Quote:

Originally Posted by Narue

>num += atoi(number[i]) * temp;
Okay, number is a string, so number[i] is a character. That alone is a type mismatch. You could take the address of that character because atoi expects a C-style string (any number of characters terminated with '\0'), but the string class doesn't guarantee that internal representation.

The good news is that your algorithm only expects single characters at a time, so you can use the subtraction trick to turn something like '5' into 5:

(Toggle Plain Text)

num += (number[i] - '0') * temp;

Then your only problem is when the string doesn't represent a valid integer.

Well we can arrange that with a switch statement, what do you think????:beard:

DaWei · Dec 9th, 2006, 1:50 PM

I think you're not paying attention to the responses. Next question?

Nov 27th, 2006, 5:21 AM	#1
Cool1Net6 Newbie Join Date: Aug 2005 Location: Florida, U.S.A. Posts: 10 Rep Power: 0	Convert String to Int 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- __________________ "Give a programmer enough pizza and coke they'll do anything you want."

Nov 27th, 2006, 7:10 AM	#3
Soulstorm Hobbyist Programmer Join Date: Jan 2006 Location: Menidi, Athens, Greece Posts: 243 Rep Power: 3	The easiest way to do it is to use a stringstream example: cpp Syntax (Toggle Plain Text) #include <iostream> #include <sstream> int main (int argc, char * const argv[]){ std::stringstream s("2903"); int i; s >> i; std::cout << i+100; return 0; } #include <iostream> #include <sstream> int main (int argc, char * const argv[]){ std::stringstream s("2903"); int i; s >> i; std::cout << i+100; return 0; } Probably a bad example, but it should give you an idea on what to do. Give to an std::stringstream the number you want as a constant character, or an std::string and extract it to an integer or enything else you want. You could also use atoi(), but using stringstreams will be a lot easier for you and a lot more comprehensive. The STL has many things you should read when you encounter a problem. In many occasions, the STL solves the problem. Check this out for a reference. __________________ Project::Soulstorm (personal homepage)

Nov 27th, 2006, 7:51 AM	#4
Lesliect6 Programmer Join Date: Aug 2005 Posts: 68 Rep Power: 4	Hello Coll1Net6, Converting a string to an int is very easy, having in mind that a string is only a representation of a number in the ASCII chart. Having also in mind that a string is nothing more than an array of chars, you can easily manipulate each individual character. So, if your string is composed only of numbers in string format (like '4' or '6'), all you have to do is : (Toggle Plain Text) int number = string[x] - 48; int number = string[x] - 48; where x is the index of the char you want converted to int. And the job is done.

Nov 27th, 2006, 8:11 AM	#5
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	Leslie, your method is contingent upon the characters being in an ASCII representation. Such is not always the case. One should use string streams, as recommended, or perhaps strtol or similar. Atoi is not a good choice because errors are indistinguishable from converting the value, zero (and other badnasty results). __________________ 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 7th, 2006, 1:38 PM	#7
Narue Professional Programmer Join Date: Sep 2005 Posts: 419 Rep Power: 4	>for (int i = number.length; i >= 0; i--) That's an off-by-one error and a syntax error. You want to start at number.length() - 1. >num += atoi(number[i]) * temp; Okay, number is a string, so number[i] is a character. That alone is a type mismatch. You could take the address of that character because atoi expects a C-style string (any number of characters terminated with '\0'), but the string class doesn't guarantee that internal representation. The good news is that your algorithm only expects single characters at a time, so you can use the subtraction trick to turn something like '5' into 5: (Toggle Plain Text) num += (number[i] - '0') * temp; num += (number[i] - '0') * temp; Then your only problem is when the string doesn't represent a valid integer. __________________ Even if the voices aren't real, they have some pretty good ideas.

Dec 7th, 2006, 12:47 PM	#6
pegasus001 Hobbyist Programmer Join Date: Nov 2006 Location: 163H Posts: 215 Rep Power: 3	int string_to_int(string number) { int num = 0; int temp = 1; for (int i = number.length; i >= 0; i--) { num += atoi(number[i]) * temp; temp *= 10; } return num; }

Dec 9th, 2006, 1:50 PM	#10
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	I think you're not paying attention to the responses. Next question? __________________ 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

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Programming Forums > Application Development > C++

Convert String to Int