Algorithm to parse an int from a string in one pass - c

Question

I'm trying to write a function that parses an integer from a string representation.
My problem is that I don't know how to do this with one pass through the string. If I knew ahead of time that the input contained only characters in the range '0', '1', ..., '9' and that the string was of length n, I could of course calculate
character_1 * 10^(n-1) + character_2 * 10^(n-2) + .... + character_n * 10^0
but I want to deal with the general scenario as I've presented it.
I'm not looking for a library function, but an algorithm to achieve this in "pure C".
Here's the code I started from:
int parse_int (const char * c1, const char * c2, int * i)
{
/*
[c1, c2]: Range of characters in the string
i: Integer whose string representnation will be converted
Returns the number of characters parsed.
Exs. "2342kjsd32" returns 4, since the first 4 characters were parsed.
"hhsd3b23" returns 0
*/
int n = 0;
*i = 0;
while (c1!= c2)
{
char c = *c1;
if (c >= '0' && c <= '9')
{
}
}
return n;
}

Answer 1

Just as some of the comments and answers suggested, maybe a bit clearer: You have to "shift" the result "left" by multiplying it by 10 in every iteration before the addition of the new digit.
Indeed, this should remind us of Horner's method. As you have recognized, the result can be written like a polynomial:
result = c1 * 10^(n-1) + c2 * 10^(n-2) + ... + cn * 10^0
And this equation can be rewritten as this:
result = cn + 10*(... + 10*(c2 + 10*c1))
Which is the form this approach is based on. From the formula you can already see, that you don't need to know the power of 10 the first digit is to be multiplied by, directly from the start.
Here's an example:
#include <stdio.h>
int parse_int(const char * begin, const char * end, int * result) {
int d = 0;
for (*result = 0; begin != end; d++, begin++) {
int digit = *begin - '0';
if (digit >= 0 && digit < 10) {
*result *= 10;
*result += digit;
}
else break;
}
return d;
}
int main() {
char arr[] = "2342kjsd32";
int result;
int ndigits = parse_int(arr, arr+sizeof(arr), &result);
printf("%d digits parsed, got: %d\n", ndigits, result);
return 0;
}
The same can be achieved using sscanf(), for everyone that is fine with using the C standard library (can also handle negative numbers):
#include <stdio.h>
int main() {
char arr[] = "2342kjsd32";
int result, ndigits;
sscanf(arr, "%d%n", &result, &ndigits);
printf("%d digits parsed, got: %d\n", ndigits, result);
return 0;
}
The output is (both implementations):
$ gcc test.c && ./a.out
4 digits parsed, got: 2342

Answer 2

I think this is good solution to count parse character
int parse(char *str)
{
int k = 0;
while(*str)
{
if((*str >= '0') & (*str <= '9'))
break;
str++;
k++;
}
return k;
}

Answer 3

Here's a working version:
#include <stdio.h>
int parse_int (const char * c1, const char * c2, int * i)
{
/*
[c1, c2]: Range of characters in the string
i: Integer whose string representnation will be converted
Returns the number of characters parsed.
Exs. "2342kjsd32" returns 4, since the first 4 characters were parsed.
"hhsd3b23" returns 0
*/
int n = 0;
*i = 0;
for (; c1 != c2; c1++)
{
char c = *c1;
if (c >= '0' && c <= '9')
{
++n;
*i = *i * 10 + c - '0';
}
else
{
break;
}
}
return n;
}
int main()
{
int i;
char const* c1 = "2342kjsd32";
int n = parse_int(c1, c1+10, &i);
printf("n: %d, i: %d\n", n, i);
return 0;
}
Output:
n: 4, i: 2342