CharAt For C?

xboxrulz

Sep 15 2008, 12:05 AM

Hey guys,

I'm just wondering if there's a CharAt function like there is in Java?

I know in Java, you just do this:

CODE

intArray[0][intCount] = strSIN.charAt(intCount) - '0';

Thanks,
xboxrulz

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faulty.lee

Sep 15 2008, 04:27 AM

QUOTE(xboxrulz @ Sep 15 2008, 08:05 AM)

Hey guys,

I'm just wondering if there's a CharAt function like there is in Java?

I know in Java, you just do this:

CODE

intArray[0][intCount] = strSIN.charAt(intCount) - '0';

Thanks,
xboxrulz

I'm not sure what type did you declare strSIN as. As far as I know, there's no base string type available in C++ (well, I've not touch C++ for a few years now, maybe something new popup). String normally are represented as char[] or *char or CString. There are other representation, but are less popular. For char[] and *char, it should be easy as it's an array, "CharAt" would simply means char[at] (I forgotten how to do it with *char, but you can lookup in google).

As for CString, you need to refer to your class reference, since it could be from ATL, MFC or other lib, they all have different implementation.

EDIT: PS: I'm refering to windows here. If you're talking about linux, then CString might not applies

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magiccode9

Sep 15 2008, 05:26 AM

You may try this to see if it it help

for char*

CODE

int index = 0;
char *message = "Hello, Java";
char achar = *(message + index);

or

char achar2 = message[index];

If I'am not forgot it, there is a class called string on STL.
But I'am not good with this class.

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xboxrulz

Sep 15 2008, 05:35 PM

well, let me clarify that I use the string to store the SIN (Social Insurance Number) and I want to pick out the certain number out to analyze it.

Many thanks,
xboxrulz

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magiccode9

Sep 23 2008, 12:31 PM

Sorry, I haven't see this

Hi, for c++ string reference.
Here have a great web site.

[qutoe]
http://www.cplusplus.com/reference/string/string/
[/quote]

It should not hard because it looks like java and c# string

-----------------------
For quick way
because the message pointer variable was 0-based index array
So, please change the index variable to point to the char you need.
then loop throught it, as,

CODE

#include "stdio.h"
#include "string.h"

int main()
{
    // change the index to start the looping
    int index = 2;

    // change this to the number of char required (+ 1 if including null char)
    int numrequiredchars = 8;

    // build a simple buffer to hold a new data
    char *requiredchars = new char[numrequiredchars];

    // the source text
    char *message = "Hello, Java and long long text";

    memset(requiredchars, 0, numrequiredchars);

    for (int i = index; i < numrequiredchars; i++)
    {
       *requiredchars = *(message + i);
         ++requiredchars;
    }

    printf("%s\n", requiredchars);
    //printf("\n%s\n", message);

}

---
For relatively full story

I'am not good in explaination but I try my best,

CODE

char *message = "Hello, Java" means a char pointer variable that is pointing to a memory location in system that start with char H and contain a string Hello, Java

Because char *message = "Hello, Java" basically equal char[] = "Hello, Java".
So both of it are an char array, except that the formal including a null char at the end of char array ( the string )
And since a char is occupy 1 byte memory, so a pointer address = 1 byte, i.e.

CODE

0001 => H
0002 => e

compared with an integer type that is 4 byte in size, i.e.

CODE

0001 => 1
0005 => 2

That result in,

CODE

*message = H
*(message + 1) = e
*(message + 2) = l
...

If you just print out the message variable with printf,
it will show the address that it contains.
To get the actually value it pointed to,
use the * operator to dereference, as,
*message

Since not very good in English.
So it might be something missed.

Wish this help !

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xboxrulz

Sep 24 2008, 12:15 AM

Alright, I found an alternative.

I'll keep this for reference.

Many thanks,
Adrien

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