Grab all integers from irregular strings in C - c

Question

I am looking for a (relatively) simple way to parse a random string and extract all of the integers from it and put them into an Array - this differs from some of the other questions which are similar because my strings have no standard format.
Example:
pt112parah salin10n m5:isstupid::42$%&%^*%7first3
I would need to eventually get an array with these contents:
112 10 5 42 7 3
And I would like a method more efficient then going character by character through a string.
Thanks for your help

Answer 1

A quick solution. I'm assuming that there are no numbers that exceed the range of long, and that there are no minus signs to worry about. If those are problems, then you need to do a lot more work analyzing the results of strtol() and you need to detect '-' followed by a digit.
The code does loop over all characters; I don't think you can avoid that. But it does use strtol() to process each sequence of digits (once the first digit is found), and resumes where strtol() left off (and strtol() is kind enough to tell us exactly where it stopped its conversion).
#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>
int main(void)
{
const char data[] = "pt112parah salin10n m5:isstupid::42$%&%^*%7first3";
long results[100];
int nresult = 0;
const char *s = data;
char c;
while ((c = *s++) != '\0')
{
if (isdigit(c))
{
char *end;
results[nresult++] = strtol(s-1, &end, 10);
s = end;
}
}
for (int i = 0; i < nresult; i++)
printf("%d: %ld\n", i, results[i]);
return 0;
}
Output:
0: 112
1: 10
2: 5
3: 42
4: 7
5: 3

Answer 2

More efficient than going through character by character?
Not possible, because you must look at every character to know that it is not an integer.
Now, given that you have to go though the string character by character, I would recommend simply casting each character as an int and checking that:
//string tmp = ""; declared outside of loop.
//pseudocode for inner loop:
int intVal = (int)c;
if(intVal >=48 && intVal <= 57){ //0-9 are 48-57 when char casted to int.
tmp += c;
}
else if(tmp.length > 0){
array[?] = (int)tmp; // ? is where to add the int to the array.
tmp = "";
}
array will contain your solution.

Answer 3

Just because I've been writing Python all day and I want a break. Declaring an array will be tricky. Either you have to run it twice to work out how many numbers you have (and then allocate the array) or just use the numbers one by one as in this example.
NB the ASCII characters for '0' to '9' are 48 to 57 (i.e. consecutive).
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
int main(int argc, char **argv)
{
char *input = "pt112par0ah salin10n m5:isstupid::42$%&%^*%7first3";
int length = strlen(input);
int value = 0;
int i;
bool gotnumber = false;
for (i = 0; i < length; i++)
{
if (input[i] >= '0' && input[i] <= '9')
{
gotnumber = true;
value = value * 10; // shift up a column
value += input[i] - '0'; // casting the char to an int
}
else if (gotnumber) // we hit this the first time we encounter a non-number after we've had numbers
{
printf("Value: %d \n", value);
value = 0;
gotnumber = false;
}
}
return 0;
}
EDIT: the previous verison didn't deal with 0

Answer 4

Another solution is to use the strtok function
/* strtok example */
#include <stdio.h>
#include <string.h>
int main ()
{
char str[] = "pt112parah salin10n m5:isstupid::42$%&%^*%7first3";
char * pch;
printf ("Splitting string \"%s\" into tokens:\n",str);
pch = strtok (str," abcdefghijklmnopqrstuvwxyz:$%&^*");
while (pch != NULL)
{
printf ("%s\n",pch);
pch = strtok (NULL, " abcdefghijklmnopqrstuvwxyz:$%&^*");
}
return 0;
}
Gives:
112
10
5
42
7
3
Perhaps not the best solution for this task, since you need to specify all characters that will be treated as a token. But it is an alternative to the other solutions.

Answer 5

And if you don't mind using C++ instead of C (usually there isn't a good reason why not), then you can reduce your solution to just two lines of code (using AXE parser generator):
vector<int> numbers;
auto number_rule = *(*(axe::r_any() - axe::r_num())
& *axe::r_num() >> axe::e_push_back(numbers));
now test it:
std::string str = "pt112parah salin10n m5:isstupid::42$%&%^*%7first3";
number_rule(str.begin(), str.end());
std::for_each(numbers.begin(), numbers.end(), [](int i) { std::cout << "\ni=" << i; });
and sure enough, you got your numbers back.
And as a bonus, you don't need to change anything when parsing unicode wide strings:
std::wstring str = L"pt112parah salin10n m5:isstupid::42$%&%^*%7first3";
number_rule(str.begin(), str.end());
std::for_each(numbers.begin(), numbers.end(), [](int i) { std::cout << "\ni=" << i; });
and sure enough, you got the same numbers back.

Answer 6

#include <stdio.h>
#include <string.h>
#include <math.h>
int main(void)
{
char *input = "pt112par0ah salin10n m5:isstupid::42$%&%^*%7first3";
char *pos = input;
int integers[strlen(input) / 2]; // The maximum possible number of integers is half the length of the string, due to the smallest number of digits possible per integer being 1 and the smallest number of characters between two different integers also being 1
unsigned int numInts= 0;
while ((pos = strpbrk(pos, "0123456789")) != NULL) // strpbrk() prototype in string.h
{
sscanf(pos, "%u", &(integers[numInts]));
if (integers[numInts] == 0)
pos++;
else
pos += (int) log10(integers[numInts]) + 1; // requires math.h
numInts++;
}
for (int i = 0; i < numInts; i++)
printf("%d ", integers[i]);
return 0;
}
Finding the integers is accomplished via repeated calls to strpbrk() on the offset pointer, with the pointer being offset again by an amount equaling the number of digits in the integer, calculated by finding the base-10 logarithm of the integer and adding 1 (with a special case for when the integer is 0). No need to use abs() on the integer when calculating the logarithm, as you stated the integers will be non-negative. If you wanted to be more space-efficient, you could use unsigned char integers[] rather than int integers[], as you stated the integers will all be <256, but that isn't a necessity.