Program crash ?!

**tilex** · 12-21-2004

Hi everyone ! I was bored and decided to make my own version of itoa that would work for a std::string (since I work so often with it and always tell myself how how sad it is it wasn't yet overloaded for strings). Well I coded something but the program crashes when I call my function, could someone tell me where the bug is ? It's not a big function, it's only 6 lines :P

Code:

#include <iostream>
#include <string>
#include <cmath>

template <class int_type>
inline char* itoa_t(std::string& buf, const int_type nb) {
	for(int i=0; pow(10, i)<nb; ++i)
		buf += char((nb/(10*(i-1)))+'0');
	return const_cast<char*>(buf.c_str());
}

int main(void) {
	std::string blah = "";
	itoa_t(blah, 123);
	return 0;
}

Thank you all

**Kybo_Ren** · 12-21-2004

You divide by 0 when i = 1.

**tilex** · 12-21-2004

Well nevermind, I just found out, but could anyone tell me why I get junk with the following code ? It output "½½"

Code:

#include <iostream>
#include <string>
#include <cmath>

#define VALIDATE(a) (((a)==0)?1:a)

template <class int_type>
inline const char* itoa_t(std::string& buf, const int_type nb) {
	for(int i=1; pow(10, i)<nb; ++i)
		buf += char((nb/VALIDATE((10*(i-1))%10))+'0');
	return const_cast<char*>(buf.c_str());
}

int main(void) {
	std::string blah = "";
	itoa_t(blah, 123);
	std::cout << blah;
	return 0;
}

For the most curious ones, the error occurred(spelling?) because it would divide by zero on the first run of the for loop.
*EDIT* Beat me to it lol

**tilex** · 12-21-2004

Now it outputs 326 with the following code, why ?

Code:

template <class int_type>
inline char* itoa_t(std::string& buf, const int_type nb) {
	buf += char(nb%10+'0');
	for(int_type i=1; pow(10, i)<nb; ++i)
		buf += char((nb/(10*i)%10)+'0');
	return const_cast<char*>(buf.c_str());
}

*EDIT* I know that the string will be backwards

but I will take care of that after, it should still output "321".

**Kybo_Ren** · 12-21-2004

I don't want to take the time to look at your code.

Here's my personal itoa function, converted to your template function:

Code:

template <class int_type>
inline char* itoa_t(std::string& buf, const int_type nb) 
{
	std::string buffer;
	int num = abs(nb);
	const unsigned int numdigits = (log((float)num) / log(10.0f)) + 1;
	unsigned int dig_buf;
	
	for(int_type i = 1; i <= numdigits; ++i)
	{
		dig_buf = static_cast<int>(pow(10.0f, i-1));

		buffer = (num % static_cast<int>(pow(10.0f, i))) / dig_buf + '0';
		buf += buffer;
	}

	if(nb < 0)
	{
		buf += '-';
	}

	std::reverse(buf.begin(), buf.end());
	return const_cast<char*>(buf.c_str());
}

Maybe you can get ideas from that?

It's longer and probably takes longer to execute, but I've never had to use itoa in a performace-critical part of code.

**Sebastiani** · 12-21-2004

>> buf += char((nb/(10*i)%10)+'0');

on the first iteration 10 * i == 10, then it's 20, 30, 40, etc. you want to divide nb by 10, then 100, 1000, 10000, etc.

**tilex** · 12-21-2004

How stupid am I, thank you mate.

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