Please help with bug in my function - converts int to string

[Ivar]

Hi guys - So basically I am trying to implement a function that
converts an int to a string, but it is not working for some reason -
any thoughts? My function, intToStr, is shown below. I'm just trying to
implement this to gain practice with c-style strings.

#include<iostre am>
#include"testSt ring.h"

int main(int argc, char* argv[]) {
char* c2 = new char[];
testString::int ToStr(c2, -254);
cout << c2 << endl;
delete c2;
c2 = NULL;
return 0;*/
}
void testString::int ToStr(char str[], int number) {
int x = number;
if(x < 0)
x = -x;
int order = 0;
while(x > 0) {
x = x/10;
order++;
}
char* tmp = new char[order+2];
tmp[0] = '\0';
int y = number;
for(int i=1; i <= order; i++) {
tmp[i] = (char)(y%10);
y = y/10;
}
if(number < 0)
tmp[order+1] = '-';
else
tmp[order+1] = '+';
testString::rev erseString(tmp) ; /*reverseString works - there is no
bug in that code*/
while(*str++ = *tmp++);
delete tmp;
tmp = NULL;
}

Thanks

[Neelesh Bodas]

Ivar wrote:

testString::rev erseString(tmp) ; /*reverseString works - there is no
bug in that code*/

but there is a bug in your usage. since the first character of tmp is
null character, I suspect whether the string will be reversed (provided
reverseString doesnot do anything unusual.)

[Jonathan Mcdougall]

Ivar wrote:

Hi guys - So basically I am trying to implement a function that
converts an int to a string, but it is not working for some reason -
any thoughts? My function, intToStr, is shown below. I'm just trying to
implement this to gain practice with c-style strings.
Many things don't work here.
#include<iostre am>
#include"testSt ring.h"

int main(int argc, char* argv[]) {
char* c2 = new char[];
Illegal, you must specify an array size here.
testString::int ToStr(c2, -254);
cout << c2 << endl;
delete c2;
Illegal. This should be

delete[] c2;
c2 = NULL;
return 0;*/
Remove that */
}
void testString::int ToStr(char str[], int number) {
int x = number;
if(x < 0)
x = -x;
Use std::abs().
int order = 0;
while(x > 0) {
x = x/10;
order++;
}
char* tmp = new char[order+2];
tmp[0] = '\0';
int y = number;
Watch out! y should be absolute here or you'll get negative values!
for(int i=1; i <= order; i++) {
tmp[i] = (char)(y%10);
This should be

tmp[i] = '0' + (y%10);

if you want characters. This will only work on ASCII machines.
y = y/10;
}
if(number < 0)
tmp[order+1] = '-';
else
tmp[order+1] = '+';
testString::rev erseString(tmp) ; /*reverseString works - there is no
bug in that code*/
while(*str++ = *tmp++);
Nooo! You just lost the pointer to the memory you allocated. Save it
*before*

char *to_delete = tmp;
delete tmp;
This should crash the application because you are not deleting from the
correct address (tmp has moved in your loop).
tmp = NULL;
}
I think there was some other errors as well, but start by fixing these.
By the way, I understand you are doing that for fun, but you should use
std::istringstr eam instead:

# include <sstream>

int main()
{
int i = 0;
std::istringstr eam iss("-254");
iss >> i;
}
Thanks

Jonathan

[Jonathan Mcdougall]

Jonathan Mcdougall wrote:

Ivar wrote:

char* tmp = new char[order+2];

delete tmp;

This should crash the application because you are not deleting from the
correct address (tmp has moved in your loop).

What's more, you should do

delete[] tmp;

as in main().

int *i = new int;
delete i;

int *i = new int[10];
delete[] i;
Jonathan

[Clark S. Cox III]

On 2005-11-17 04:43:01 -0500, "Jonathan Mcdougall"
<jo************ ***@gmail.com> said:

Ivar wrote:

for(int i=1; i <= order; i++) {
tmp[i] = (char)(y%10);

This should be

tmp[i] = '0' + (y%10);

if you want characters. This will only work on ASCII machines.

No, it will work on all machines. The characters for the digits '0'
through '9' are guaranteed to be sequential. That is, the following
will *always* produce the character '5':

putc('0' + 5, stdout);
--
Clark S. Cox, III
cl*******@gmail .com

[shikn]

tmp[i] = (char)(y%10)+ '0'; ?

"Ivar" <ra**********@g mail.com> wrote in message
news:11******** *************@o 13g2000cwo.goog legroups.com...

Hi guys - So basically I am trying to implement a function that
converts an int to a string, but it is not working for some reason -
any thoughts? My function, intToStr, is shown below. I'm just trying to
implement this to gain practice with c-style strings.

#include<iostre am>
#include"testSt ring.h"

int main(int argc, char* argv[]) {
char* c2 = new char[];
testString::int ToStr(c2, -254);
cout << c2 << endl;
delete c2;
c2 = NULL;
return 0;*/
}
void testString::int ToStr(char str[], int number) {
int x = number;
if(x < 0)
x = -x;
int order = 0;
while(x > 0) {
x = x/10;
order++;
}
char* tmp = new char[order+2];
tmp[0] = '\0';
int y = number;
for(int i=1; i <= order; i++) {
tmp[i] = (char)(y%10);
// the above statement maybe need some change? tmp[i] = (char)(y%10)+ '0';
y = y/10;
}
if(number < 0)
tmp[order+1] = '-';
else
tmp[order+1] = '+';
testString::rev erseString(tmp) ; /*reverseString works - there is no
bug in that code*/
while(*str++ = *tmp++);
delete tmp;
tmp = NULL;
}

Thanks

[Jonathan Mcdougall]

Clark S. Cox III wrote:

On 2005-11-17 04:43:01 -0500, "Jonathan Mcdougall"
<jo************ ***@gmail.com> said:

Ivar wrote:

for(int i=1; i <= order; i++) {
tmp[i] = (char)(y%10);

This should be

tmp[i] = '0' + (y%10);

if you want characters. This will only work on ASCII machines.

No, it will work on all machines. The characters for the digits '0'
through '9' are guaranteed to be sequential. That is, the following
will *always* produce the character '5':

putc('0' + 5, stdout);

Any reference? All I could find is

2.13.2.1 "[...] An ordinary character literal that contains a
single c-char has type char, with value equal to the numerical value of
the encoding of the c-char in the execution character set."
Jonathan

[werasm]

Ivar wrote:

Hi guys - So basically I am trying to implement a function that
converts an int to a string, but it is not working for some reason -
any thoughts? My function, intToStr, is shown below. I'm just trying to
implement this to gain practice with c-style strings.

#include<iostre am>
#include"testSt ring.h"

int main(int argc, char* argv[]) {
char* c2 = new char[];
How large an array should be allocated? Is it really necessary to make
use of the heap in this case. Why not just use:

char resultStr[100];
testString::int ToStr(c2, -254);
cout << c2 << endl;
delete c2;
Bug here. Calling new[] requires you to call delete [], for example:

char* result = new char[x];
....requires...
delete []result;
c2 = NULL;
In these circumstances, this is really not necessary, as c2 is never
accessed again.
return 0;*/
}


For a function like this, I would consider providing a std::string as
argument. If not, I would explicitly indicate the result buffers size
to prevent overflow.
void testString::int ToStr(char str[], int number) {
int x = number;
if(x < 0)
x = -x;
IMO the line here above will not give you the desired effect. Maybe
abs( x ) where abs gives you the absolute value of x, else try x *= -1;

int order = 0;
while(x > 0) {
x = x/10;
order++;
}
char* tmp = new char[order+2];
Why always using memory allocated on the heap?- expensive, you know...
Also, is it at all necessary to create a temporary here?
tmp[0] = '\0'; I'll assume reverseString ignores the fact that the string is
null-terminated. If this is the case, it is not string flavoured
(traditionally) at all. Null terminating your first character doesn't
make sense at all.
int y = number;
for(int i=1; i <= order; i++) {
tmp[i] = (char)(y%10);
y = y/10;
}
if(number < 0)
tmp[order+1] = '-';
else
tmp[order+1] = '+';
testString::rev erseString(tmp) ; /*reverseString works - there is no
bug in that code*/
Yes, terminating your strings first character effectively makes it
empty :-). Reversing an empty strings gives you, well, an empty string
:-).
while(*str++ = *tmp++);
Hmmm, how about std::copy or memcpy here. Functions are there to be
used. Complex functions are made of less complex ones. Why did you not
write or own memcpy then, and use that if you want to have some
practice at doing it right;
delete tmp;
Ooops, delete []tmp; This may crash...
tmp = NULL;
Hmmm, not necessary. }

Thanks

Pleasure,

W

[Krishanu Debnath]

Jonathan Mcdougall wrote:

Clark S. Cox III wrote:

On 2005-11-17 04:43:01 -0500, "Jonathan Mcdougall"
<jo************ ***@gmail.com> said:

Ivar wrote:

for(int i=1; i <= order; i++) {
tmp[i] = (char)(y%10);
This should be

tmp[i] = '0' + (y%10);

if you want characters. This will only work on ASCII machines.

No, it will work on all machines. The characters for the digits '0'
through '9' are guaranteed to be sequential. That is, the following
will *always* produce the character '5':

putc('0' + 5, stdout);

Any reference? All I could find is

2.13.2.1 "[...] An ordinary character literal that contains a
single c-char has type char, with value equal to the numerical value of
the encoding of the c-char in the execution character set."
Jonathan

You are looking in the wrong place, try 2.2 character sets.

Krishanu

[Jonathan Mcdougall]

Krishanu Debnath wrote:

Jonathan Mcdougall wrote:

Clark S. Cox III wrote:

On 2005-11-17 04:43:01 -0500, "Jonathan Mcdougall"
<jo************ ***@gmail.com> said:

Ivar wrote:

> for(int i=1; i <= order; i++) {
> tmp[i] = (char)(y%10);
This should be

tmp[i] = '0' + (y%10);

if you want characters. This will only work on ASCII machines.
No, it will work on all machines. The characters for the digits '0'
through '9' are guaranteed to be sequential. That is, the following
will *always* produce the character '5':

putc('0' + 5, stdout);

Any reference? All I could find is

2.13.2.1 "[...] An ordinary character literal that contains a
single c-char has type char, with value equal to the numerical value of
the encoding of the c-char in the execution character set."

You are looking in the wrong place, try 2.2 character sets.

Well I don't have the standard (I will, someday), but there's nothing
in the draft at 2.2, except the characters accepted in a source file.
If there are more explanations in the standard, would it be possible
for someone to quote it here?

Thank you,
Jonathan