Strtol implementation different behaviour on 32 and 64 bit machine - 32bit-64bit

Question

#include <ctype.h>
#include <string.h>
#include <stdio.h>
#include <tgmath.h>
#include <limits.h>
#include <stdbool.h>
#include <errno.h>
#define NEGATIVE -1
#define POSITIVE 1
#define OCTAL 8
#define DECIMAL 10
#define HEXADECIMAL 16
#define BASE_MIN 2
#define BASE_MAX 36
long int strtol (const char * str, char ** endPtr, int base)
{
if(base < 0 || base == 1 || base > BASE_MAX)
{
errno = EINVAL;
return 0L;
}
else
{
bool conversion = true;
int i = 0, sign = POSITIVE, save;
while(isspace(*(str + i)))
i++;
if(*(str + i) == '\0')
{
conversion = false;
save = i;
}
if(*(str + i) == '-')
{
sign = NEGATIVE;
i++;
}
else if(*(str + i) == '+')
i++;
if(base == 0) // find out base
{
if(*(str + i) == '0')
{
if(toupper(*(str + i + 1)) == 'X')
{
base = HEXADECIMAL;
i++;
}
else
base = OCTAL;
i++;
}
else
base = DECIMAL;
}
else if(base == OCTAL)
{
if(*(str + i) == '0')
i++;
}
else if(base == HEXADECIMAL)
{
if(*(str + i) == '0')
if(*(str + i + 1) == 'x' || *(str + i + 1) == 'X')
i += 2;
}
int start = i, end, exp, check = i;
long int long_int, sum, multiplier;
if(conversion) // find out the correct part of the string corresponding to the number
{
if(base < DECIMAL)
{
while(*(str + i) >= '0' && *(str + i) < base + '0') // numbers from 0 to base - 1
i++;
}
else if(base == DECIMAL)
{
while(*(str + i) >= '0' && *(str + i) <= '9') // numbers from 0 to 9
i++;
}
else
{
while((*(str + i) >= '0' && *(str + i) <= '9') || (toupper(*(str + i)) >= 'A' && toupper(*(str + i)) < 'A' + base - 10))
i++;// numbers from 0 to 9 and uper and lowercase letters from a to a + base - 11
}
}
if(i == check && conversion) //no digits at all
{
conversion = false;
save = i;
}
else if(endPtr != NULL && conversion) // assign pointer
*endPtr = (char *) (str + i);
if(conversion)
{
for(end = i - 1, exp = 0, long_int = 0L; end >= start; end--, exp++)
{
multiplier = pow(base, exp);
sum = 0L;
if(*(str + end) >= '0' && *(str + end) <= '9')
sum = (*(str + end) - '0') * multiplier;
else if(*(str + end) >= 'A' && *(str + i) <= (base == BASE_MAX ? 'Z' : 'F'))
sum = (*(str + end) - 'A' + 10) * multiplier;
else if(*(str + end) >= 'a' && *(str + i) <= (base == BASE_MAX ? 'z' : 'f'))
sum = (*(str + end) - 'a' + 10) * multiplier;
if(long_int <= LONG_MIN + sum)
{
errno = ERANGE;
return LONG_MIN;
}
if(long_int >= LONG_MAX - sum)
{
errno = ERANGE;
return LONG_MAX;
}
else
long_int += sum;
}
return sign * long_int;
}
else
{
if(endPtr != NULL)
{// if base is 16 we check if the string given is not in the form 0xIncorrect string in that way we need to return xIncorrect part of the string
if(base == HEXADECIMAL && save >= 2 && toupper(*(str + save - 1)) == 'X' && *(str + save - 2) == '0')
*endPtr = (char *) str + save - 1;
else if(base == OCTAL && save >= 1 && *(str + save - 1) == '0')
*endPtr = (char *) str + save;// if the string is of base 8 and in the form 0incorrect string
else //then we return everything after the 0 as the endptr string
*endPtr = (char *) str;//in other cases no conversion was done so we return original pointer
}
return 0L;
}
}
}
I've got problem with writing implementation of strtol() function. The thing is i compiled it on 64 bit machine and the output was correct but today i checked it on another machine that is 32-bit and something got wrong. 32-bit machine showed the result that for example string "7FFFFFFF" is out of range when on 64-bits the results is that strtol succeded which is the same as for th standard function. I also checked errno value and for 32-bit machine it's set to ERANGE which shouldn't be and it's not not on 64-bit. I have program that checks if your implementation gives the same output as the standard one for different strings. I spent few hours looking for possible bug but i'm out of ideas? Any tips?