Elysian Shadows

Posted: **Fri Mar 27, 2009 3:15 pm**

Yes, I know that you can just use atoi(), but I was bored so I didn't care.

Note: this will treat the string as though there are no other characters in the string other than numbers, so if the string was "1a3", it would return 13.

Edit: Now it also works with negative numbers, so if you had "-q1a3", it would return -13

Code: Select all

int stringToInt(string str)
{
    int x = 0;
    for(int i = str.size() - 1, j = 1; i >= 0; --i)
    {
        char current = str[i];
        if(current >= '0' && current <= '9')
        {
            if(x < 0)
                x = abs(x);
            int y = (int)current - '0';
            x += y * j;
            j *= 10;
        }
        else if(current == '-')
            x *= -1;
    }
    return x;
}

I've tested it, so it should work.

Edit2: I made a char* version:

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int stringToInt(char str[])
{
    int ret = 0, multiple = 1;
    char* ptr;
    for(ptr = str; *ptr; ++i)
    {;
        if(*ptr >= '0' && *ptr <= '9')
        {
            if(x < 0)
                x *= -1;
            int y = (int)(*ptr) - '0';
            x += y * multiple;
            multiple *= 10;
        }
        else if(*ptr == '-')
            x *= -1;
    }
    return x;
}

Posted: **Fri Mar 27, 2009 4:02 pm**

Ginto8 wrote:So if the string was "1a3", it would return 13.

Actually, your function is better since atoi("1a3") would terminate at the first invalid (non integer) value. So atoi() would return 1, while yours returns 13. On second though, I retract; "better" would depend on the situation or what you're trying to do. So they both have their uses.

By the way, in Java (I've learning it in my spare time) you can also do something like this:

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Integer.parseInt("1a3");

Posted: **Fri Mar 27, 2009 4:53 pm**

Edit: There, I fixed it. For those curious, I had hit "edit" instead of "quote" and overwrote his post with my reply :)

My reply is,

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        if(!(current < '0' || current > '9'))

is equivalent to the clearer

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if (current >= '0' && current <= '9')

Very nice (assuming it works), btw - gj :D

Posted: **Fri Mar 27, 2009 4:54 pm**

sparda wrote:By the way, in Java (I've learning it in my spare time) you can also do something like this:
Code: Select all
Integer.parseInt("1a3");

No you can't.

Posted: **Fri Mar 27, 2009 6:06 pm**

wearymemory wrote:No you can't.

LULZ!

Here is the OFFICIAL java documentation for the Integer class. Apparently you're too lazy to search Google before posting erroneous remarks.

Posted: **Fri Mar 27, 2009 6:37 pm**

MarauderIIC wrote:Edit: There, I fixed it. For those curious, I had hit "edit" instead of "quote" and overwrote his post with my reply

My reply is,
Code: Select all
        if(!(current < '0' || current > '9'))
is equivalent to the clearer
Code: Select all
if (current > '0' && current < '9')
Very nice (assuming it works), btw - gj :D

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if (current >= '0' && current <= '9')

Fixed.

Posted: **Fri Mar 27, 2009 7:19 pm**

sparda wrote:
wearymemory wrote:No you can't.
LULZ!

Here is the OFFICIAL java documentation for the Integer class. Apparently you're too lazy to search Google before posting erroneous remarks.

I am? It's a shame you feel that way.

The line of code you've provided will throw a NumberFormatException.

I have a lot more experience with Java than you do. Not that you asked.

Posted: **Fri Mar 27, 2009 7:55 pm**

wearymemory wrote:I am? It's a shame you feel that way.

The line of code you've provided will throw a NumberFormatException.

I have a lot more experience with Java than you do. Not that you asked.

LOLOLOLOLOL!!!

Oh my god LOL!

You are absolutely correct

But you're also right that I never asked you for your credentials. In any case, you could have added A LITTLE MORE than simply:

No you can't.

Just for future reference.

Anything less, and I swear you were trying to flame-bait me. But I've learned my lesson, I won't even minutely comment about things I don't fully understand.

Posted: **Sun Mar 29, 2009 12:25 pm**

Made some changes, added negatives. So if you entered "-q1a3", it would return -13, and if you entered "1q-a3", it would return 13.

Posted: **Sun Mar 29, 2009 2:24 pm**

Ginto8 wrote:Made some changes, added negatives. So if you entered "-q1a3", it would return -13, and if you entered "1q-a3", it would return 13.

what if you did "1qa-3"

anyways, this is a total waste of effort, there is not reason you need this.

Posted: **Sun Mar 29, 2009 2:30 pm**

avansc wrote:anyways, this is a total waste of effort, there is not reason you need this.

If he's learning by doing this, it wouldn't be a waste of effort.

Posted: **Sun Mar 29, 2009 3:38 pm**

avansc wrote:
Ginto8 wrote:Made some changes, added negatives. So if you entered "-q1a3", it would return -13, and if you entered "1q-a3", it would return 13.
what if you did "1qa-3"

anyways, this is a total waste of effort, there is not reason you need this.

"1qa-3" would do the same as "1q-a3".

And I think I said in the OP that I was bored when I made this.

Posted: **Tue Jul 07, 2009 8:20 am**

I prefer to use stringstream for converting than atoi, etc.

#include <sstream>

strings to whatever:
stringstream input(string variable);
input >> variable;

whatever to strings:
stringstream output;
output << variable;
output = variable.str();

Posted: **Tue Jul 07, 2009 10:14 am**

I have a complete solution to this li'l problem, I use a little bit of bitmasking to convert my strings to integers, I just knocked this out when I was bored today

Code: Select all

#include<iostream>
#include<string>

using namespace std;



int StringToInt(const char* target); // char* means you can pass any kind of string
// just as long as you use .c_str() on a C++ string if it isnt literal

int main(int argc, char** argv)
{
        cout << StringToInt("cyrax thinks he's a 13x3x7 hax0r");
        return 0;
}

int StringToInt(const char* target)
{
       int multiple = 1; // this will increase as we go through the decimals
       int total = 0; // the number to return

       for(int i = (strlen(target) -1);i >= 0; i--) // go from end to beginning of string
       // added -1 to strlen as not to go past array bounds 
        {
            if(target[i] >= '0' && target[i] <= '9') // check character is digit
            {
                total  += ((target[i] & 0x0F )* multiple);
                // add to total, AND against 0F to get digit,
                // then multiply
                multiple *= 10; // multiply the multiple to increase
                // next number (HTU etc)

            }
        }

        return total; // return the final total, simples :D

}

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Output: 13370

Since I read bitmasking in one of my books a few months back, I've been hooked, so there you go

~~EDIT: bare in mind that with simple adjustment, this would work with C-style string too, you would just use strlen() instead of .length() , actually, scratch that out, i'll stick that in the code :P~~
EDIT 2: the code now takes any kind of string, literal, c-style, and C++ style as long as you use c_str()

EDIT 2.5: btw, I had no problem with your solution Ginto8, I was impressed by how you did it, I've just been mulling over something like this in my head for a while now, and i'm a complete bitmask zealot

Posted: **Tue Jul 07, 2009 2:29 pm**

I added a char* version.

PiP: How do you do that? teach me!

Elysian Shadows

std::string-to-int converter

std::string-to-int converter

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