Original Post
is there any function to do this? i'd rather not have to write my own lol.
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converting an int to a string
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Assuming C++, since that seems to be what you've used in your other posts, you can either use boost::lexical_cast or a std::stringstream.
std::stringstream sstr;sstr << my_int;std::string str1 = sstr.str();std::string str2 = boost::lexical_cast<std::string>(my_int);
sprintf works.
int sprintf( char *buffer, const char *format, ... );
like so
char buf[5]
int value = 5555;
sprintf(buf,"%i",value);
int sprintf( char *buffer, const char *format, ... );
like so
char buf[5]
int value = 5555;
sprintf(buf,"%i",value);
Quote:
Original post by Gink
sprintf works.
int sprintf( char *buffer, const char *format, ... );
like so
char buf[5]
int value = 5555;
sprintf(buf,"%i",value);
If you're going to use C, you might as well use something like itoa.
If you care about portable code I highly recommend avoiding itoa(). It is severly non-standard. Not only is it not part of the C standard library, but in non-standard implementations I've seen at least five different function signatures for itoa(), with different memory management conventions:
char * itoa(int, char *, int); // you supply the bufferchar * itoa(int, int, char *); // you supply the bufferchar * itoa(int, char *); // you supply the bufferchar * itoa(int, int); // you need to free the pointerchar * itoa(int); // you need to free the pointerQuote:
Original post by SiCrane
If you care about portable code I highly recommend avoiding itoa(). It is severly non-standard. Not only is it not part of the C standard library, but in non-standard implementations I've seen at least five different function signatures for itoa(), with different memory management conventions:char * itoa(int, char *, int); // you supply the bufferchar * itoa(int, int, char *); // you supply the bufferchar * itoa(int, char *); // you supply the bufferchar * itoa(int, int); // you need to free the pointerchar * itoa(int); // you need to free the pointer
Hmm, didn't know that. Nevermind, then.
alright, for anyone who cares about this at all, here is my little function to convert a int to a string. on vc++ 6, this works up to 999,999,999. It doesn't do any formating or anything, but all my tests show it works just fine. let me know if you see any problems with it.
string int_to_string(int num){ //cout << "number inputed = " << num << endl; int num_size = 0; //holds the size of the number string str; //holds the actual string we will return //find out how large the number is int temp = num/10; //used to find how large the number is num_size++; while(temp != 0){ temp = temp / 10; num_size++; } //cout << "num size = " << num_size << endl; int one_place = 0; int div = 10; //holds the number num should be divided by for(int i = 0; i < num_size-2; i++){ div = div * 10; } //cout << "div = " << div << endl << endl; for(i = num_size; i > 0; i--){ one_place = num/div; //add this number to the string if(one_place == 1){ str = str + "1"; } else if(one_place == 2){ str = str + "2"; } else if(one_place == 3){ str = str + "3"; } else if(one_place == 4){ str = str + "4"; } else if(one_place == 5){ str = str + "5"; } else if(one_place == 6){ str = str + "6"; } else if(one_place == 7){ str = str + "7"; } else if(one_place == 8){ str = str + "8"; } else if(one_place == 9){ str = str + "9"; } else{ str = str + "0"; } //find what remains of the number num = num - (div*one_place); //set up div for the next pass div = div / 10; //output info (for testing only) //cout << "This pass info" << endl; //cout << "div = " << div << endl; //cout << "num = " << num << endl; //cout << "one_place = " << one_place << endl; } return str;}
Quote:
Original post by CJWR
alright, for anyone who cares about this at all, here is my little function to convert a int to a string. on vc++ 6, this works up to 999,999,999. It doesn't do any formating or anything, but all my tests show it works just fine. let me know if you see any problems with it.
Well, it doesn't work with negative numbers, for starters. And, well, SiCrane told you how to do it, why not use what he said?
Quote:
Original post by Fruny Quote:
Original post by CJWR
alright, for anyone who cares about this at all, here is my little function to convert a int to a string. on vc++ 6, this works up to 999,999,999. It doesn't do any formating or anything, but all my tests show it works just fine. let me know if you see any problems with it.
Well, it doesn't work with negative numbers, for starters. And, well, SiCrane told you how to do it, why not use what he said?
i wanted to write my own lol, but the negative thing isn't a problem for what i need this for. but it is something to work on later. :)
Turn stuff into strings, free with your purchase of 1 or more immortality rings.
#include <sstream>#include <string>template <typename T> std::string toString(const T &value) { std::stringstream stream; stream << value; return stream.str(); }Quote:
Original post by smart_idiot
Turn stuff into strings, free with your purchase of 1 or more immortality rings.
*** Source Snippet Removed ***
that dude is a fruitcake lol. not you, the guy that owns that site.
Wholly crap people. They guy is looking for a simple way to do this, not template rocket science!!!
Use sprintf()
Use sprintf()
Quote:
Original post by zealotgi
Wholly crap people. They guy is looking for a simple way to do this, not template rocket science!!!
Use sprintf()
well, i'm happy with my little function, it does all i need to do. and it allows me to get back to creating my game :).
and here is a version that works with negatives as well.
i know this is sort of reinventing the wheel, but it was an interesting little hour project.
string int_to_string(int num){ //cout << "number inputed = " << num << endl; int num_size = 0; //holds the size of the number string str; //holds the actual string we will return //first check if the number is negative (less than 0) if(num < 0){ num = (-num); str = str + '-'; } else; //find out how large the number is int temp = num/10; //used to find how large the number is num_size++; while(temp != 0){ temp = temp / 10; num_size++; } //cout << "num size = " << num_size << endl; int one_place = 0; int div = 1; //holds the number num should be divided by for(int i = 0; i < num_size-1; i++){ div = div * 10; } //cout << "div = " << div << endl << endl; for(i = num_size; i > 0; i--){ one_place = num/div; //add this number to the string if(one_place == 1){ str = str + "1"; } else if(one_place == 2){ str = str + "2"; } else if(one_place == 3){ str = str + "3"; } else if(one_place == 4){ str = str + "4"; } else if(one_place == 5){ str = str + "5"; } else if(one_place == 6){ str = str + "6"; } else if(one_place == 7){ str = str + "7"; } else if(one_place == 8){ str = str + "8"; } else if(one_place == 9){ str = str + "9"; } else{ str = str + "0"; } //find what remains of the number num = num - (div*one_place); //set up div for the next pass div = div / 10; //output info (for testing only) //cout << "This pass info" << endl; //cout << "div = " << div << endl; //cout << "num = " << num << endl; //cout << "one_place = " << one_place << endl; } return str;}i know this is sort of reinventing the wheel, but it was an interesting little hour project.
How about instead of all those if statements, you use str += static_cast('0'+one_place) ?
Edit: Seems to need a cast to work right.
Edit: Seems to need a cast to work right.
Quote:
Original post by smart_idiot
How about instead of all those if statements, you use str += static_cast('0'+one_place) ?
Edit: Seems to need a cast to work right.
well, it did work fine with any number 10 and above (up to however high ints go lol), and i just fixed it so it works on int 1-9 as well. i think it works for any number that fits into an int now. have you found a number that it wouldn't work for?
I'm bored, how about I reinvent the wheel, too?
* head explody *
[Edited by - smart_idiot on July 10, 2005 12:28:40 AM]
std::string toString(int value) { char buffer[32] = {0}, *out = buffer+31; if(!value) return "0"; else if(value >= 0) for(; value > 0; value /= 10) *--out = '0'+value%10; else { for(value *= -1; value > 0; value /= 10) *--out = '0'+value%10; *--out = '-'; } return out; }* head explody *
[Edited by - smart_idiot on July 10, 2005 12:28:40 AM]
Quote:
Original post by smart_idiot
I'm bored, how about I reinvent the wheel, too?
*** Source Snippet Removed ***
you win lol. much better than mine. why the heck didn't i think of that?
Actually, I think mine would print an empty string for 0. Oopsies.
Quote:
Original post by bytecoder
If you're going to use C, you might as well use something like itoa.
itoa() is not ANSIC C, so it's not guaranteed that it's implemented on every compiler.
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