How to handle negative integer when converting them to strings? [closed] - c

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I need to convert integer to a string, be it negative or positive.
So far i able to convert positive integers to strings using the following code.
But not the negative ones. How can i handle them properly to convert them to strings.
Here is the code that i was using.
Thanks
Rajat!
/*
* C Program which Converts an Integer to String & vice-versa
*/
#include <stdio.h>
#include <string.h>
#include <math.h>
void tostring(char [], int);
int main()
{
char str[10];
int num, result;
printf("Enter a number: ");
scanf("%d", &num);
tostring(str, num);
printf("Number converted to string: %s\n", str);
return 0;
}
void tostring(char str[], int num)
{
int i, rem, len = 0, n;
if(num<0)
{
n=(-1)*(num);
}
else
{
n=num;
}
while (n != 0)
{
len++;
n /= 10;
}
for (i = 0; i < len; i++)
{
rem = num % 10;
num = num / 10;
str[len - (i + 1)] = rem + '0';
}
str[len] = '\0';
}

Answer 1

Before doing anything else, see if n is negative. If it is, start the output with the sign - (make sure len is also one more than it would be otherwise, and that your stringification starts one character later), make n positive, and continue as you were doing it before.

Answer 2

You have to be really wary of "corner-case" like many other have said :
You can't have the positive value of a negative int in an int
The range value of an int is [MIN; +MIN -1], like –2,147,483,648 to 2,147,483,647 for a 4 byte int.
So, if you have –2,147,483,648 and you *-1, you will not have 2,147,483,648 since it will overflow the int capacity
Your loop for finding the len of int is "bad", because you don't take negative number in count (but I suppose this is the purpose of this post) and you don't take car of the corner case 0 ("len" value is 0 but must be 1).
So, how can you make this function ?
1 : Fix the int len calculation
size_t len = 1;
for (int n = number; n <= -10 || 10 <= n; n /= 10) {
++len;
}
2 : Fix the "digit to char" loop
for (size_t i = 0; i < len; ++i) {
str[len - i - 1] = abs(number % 10) + '0';
number /= 10;
}
str[len] = '\0';
There. The only thing that left is to put "-" in the beginning and have an offset for negative case number.
You can have an offset variable (fix the "digit to char loop" if you use it) or you can simply do "++str".
There, you can do the function on your own now, I pratically gived you the answer.
On a funny note, you can skip the "abs" function if you simply do the following :
void tostring(char *str, int number)
{
char *digit = "9876543210123456789" + 9;
size_t len = 1;
if (number < 0) {
// TODO : Put '-' in the first case
// TODO : Make str to be *str[1]
}
for (int n = number; n <= -10 || 10 <= n; n /= 10) {
++len;
}
for (size_t i = 0; i < len; ++i) {
str[len - i - 1] = digit[number % 10];
number /= 10;
}
str[len] = '\0';
}
If you don't understand how this work, take your time and read how pointer work (especially pointer arithmetic).

Answer 3

I have solved the issue by the help from #Amadan .
Here is the code that i am using. Please feel free to tell me a more better way to solve this issue.
Thanks
Rajat
/*
* C Program which Converts an Integer to String & vice-versa
*/
#include <stdio.h>
#include <string.h>
#include <math.h>
void tostring(char [], int);
int main()
{
char str[10];
int num, result;
printf("Enter a number: ");
scanf("%d", &num);
tostring(str, num);
printf("Number converted to string: %s\n", str);
return 0;
}
void tostring(char str[], int num)
{
int i, rem, len = 0, n;
bool flag=0;
if(num<0)
{
n=-num;
flag=1;
}
else
{
n=num;
flag=0;
}
while (n != 0)
{
len++;
n /= 10;
}
if(flag==1)
{
num=-1*num;
str[0]='-';
}
for (i = 0; i < len; i++)
{
rem = num % 10;
num = num / 10;
if(flag==1)
{
str[len - (i)] = rem + '0';
}
else
{
str[len - (i + 1)] = rem + '0';
}
}
if(flag==1)
{
str[len+1] = '\0';
}
else
{
str[len] = '\0';
}
}

Answer 4

Just use sprintf function as below,
sprintf(str,"%i",num);
So the modified code will be,
#include <stdio.h>
#include <string.h>
#include <math.h>
void tostring(char [], int);
int main()
{
char str[10];
int num, result;
printf("Enter a number: ");
scanf("%d", &num);
//tostring(str, num);
sprintf(str,"%i",num);
printf("Number converted to string: %s\n", str);
return 0;
}
void tostring(char str[], int num)
{
int i, rem, len = 0, n;
n =num;
while (n != 0)
{
len++;
n /= 10;
}
for (i = 0; i < len; i++)
{
rem = num % 10;
num = num / 10;
str[len - (i + 1)] = rem + '0';
}
str[len] = '\0';
}
Hope this helps.

Answer 5

Here you have the function which works with any base (you need just to find enough chars to represent digits) but it will not work in one corner case (can you find the case?)
static const char Digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUWXYZ";
static char *reverse(char *s, size_t len)
{
size_t pos;
for(pos = 0; pos < len / 2; pos++)
{
char tmp = s[pos];
s[pos] = s[len - pos - 1];
s[len - pos - 1] = tmp;
}
return s;
}
char *toString(char *buff, int n, unsigned base)
{
int saved = n;
char *savedbuff = buff;
n = n < 0 ? -n : n; // <= there is one number which will cause problems. Which one?
do
{
*buff++ = Digits[n % base];
n /= base;
}while(n);
if(saved < 0) *buff++ = '-';
*buff = 0;
return reverse(savedbuff, buff - savedbuff);
}
int main()
{
char buff[50];
printf("%s\n", toString(buff, -255,12)); // 12 base :)
printf("%s\n", toString(buff, -976,10)); // 10 base
printf("%s\n", toString(buff, -976,8)); // or maybe octal ?
return 0;
}