How2 convert long to sring in C++ ?

[IL96]

How2 convert long to sring in C++ ?
long A = 12345678;
string B =A; // < won't compile obviously.

Does any1 know a function that converts long to string ?

Thanks.

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There are no stupid questions, there are stupid answers.

[namgor]

Originally posted by IL96:
How2 convert long to sring in C++ ?
long A = 12345678;
string B =A; // < won't compile obviously.

Does any1 know a function that converts long to string ?

Thanks.

string B = (string) A;

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[IL96]

Originally posted by namgor:
string B = (string) A;

No, you see A is not a string, it's long.
Anywhoo, i think i found a function taht sorta kinda does that, it works like this (for future reference)

int radix = 10; // base (2,8,10,16, whatever)
char buffer[33];// temp buffer
char *strptr;// pointer to an aray of characters (string)
strptr = ltoa(12345678,buffer,radix);
printf("The value is: %s\n",strptr);

------------------
There are no stupid questions, there are stupid answers.

[This message has been edited by IL96 (edited July 16, 2001).]

[biosx]

What he did was called a type cast.

string B = (string) A;

What this does is that it turns A into a string and then assigns its value to B.

Check in a book or online for 'Type Casts' or just 'Casts'

Namgor knows his stuff


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[Strogian]

Wow, you can cast a long into a string? Man, this C++ stuff never ceases to amaze me!

[namgor]

I know there is another to cast it, (called dynamic cast??) Anyway, I believe this method can do the trick so i didnt bother look up the other method.

------------------
DHAHL3seasons GP:73 G:121 A:55 Pts:176 GWG:12 +/-:184
UWSWA1season GP:9 G:12 A:8 Pts:20 GWG:3 +/-:-3

uwcdc.com or namgor.com

[dighn]

string a = (string)an_int; does not work because std::string does not provide a conversion operator of type int


try this

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

int main()
{
int i = 123;
std::stringstream buf;
buf<<i;

std::string thestring = buf.str();

std::cout << thestring.c_str();
return 0;
}

sorry my first version of code was incorrect because std::string does not provide any overloaded operators for conversion so what you need is a string stream

and dynamic_cast is for casting between hirarchily related objects (eg. base to derived) using runtime type identification so it has no use here

BTW the includes dont have .h in them because that is the C++ style and puts the classes in the namespace "std" hence the use of std::

[This message has been edited by dighn (edited July 17, 2001).]

[biosx]

Thank you for the correction.

I'm just an amateur C programmer. So I only know the basics of casts and such. I don't really use them much except once in a while when I have to for functions I write or what not.

Thanks again, learn something new everyday.

[This message has been edited by biosx (edited July 17, 2001).]