Parsing a string into an array of integers in C - c

Question

So now that I've figured out how to get what I want, I'm just hoping somebody can let me know a cleaner, less ridiculous way of achieving the same thing. I'm just learning C. Here was my approach.
int main()
{
// String of positive and negative integer values
// Numbers are never more than 2 digits
char TEMPS[256] = "1 -22 -8 14 5";
int N = 5;
int ints[N];
int i = 0;
int mult;
// Arbitrary number to identify that num is not yet in use
int num = 999;
int c = 0;
mult = (TEMPS[c] != 45) ? -1 : 1;
while(strcmp(&TEMPS[c], "\0") != 0)
{
if(TEMPS[c] == 32)
{
ints[i] = mult * num;
i++;
num = 999;
mult = (TEMPS[c + 1] == 45) ? -1 : 1;
}
else if((TEMPS[c] != 45) && (TEMPS[c] != 32))
{
if(num == 999)
{
num = TEMPS[c] - '0';
}
else
{
num = num * 10 + (TEMPS[c] - '0');
}
}
c++;
}
ints[i] = mult * num;
}

Answer 1

I would use strtol - here's a good site for how it works and examples
http://www.tutorialspoint.com/c_standard_library/c_function_strtol.htm
long int strtol(const char *str, char **endptr, int base)
Parameters
str -- This is the string containing the representation of an integral number.
endptr -- This is the reference to an object of type char*, whose value is set by the function to the next character in str after the numerical value.
base -- This is the base, which must be between 2 and 36 inclusive, or be the special value 0.
Return Value
This function returns the converted integral number as a long int value, else zero value is r
I've included one example from the site.
#include <stdio.h>
#include <stdlib.h>
int main()
{
char str[30] = "2030300 This is test";
char *ptr;
long ret;
ret = strtol(str, &ptr, 10);
printf("The number(unsigned long integer) is %ld\n", ret);
printf("String part is |%s|", ptr);
return(0);
}