read only integers from complex string in c

Question

I am trying to read only the integers from a string. The whitespace and non-integer characters are essentially random with each iteration. Ideally my code would run through this and print(or otherwise store) only the integers. So for this test case it would be 6,1,0,16,2,3 etc etc.

char line [ 200 ] ="T06      1/ 0 16/ 2  3/15 41/40 17/36  0/11  0/11  0/11  1/14  1/ 7";
int num=0;
int count=0;
int i=0;
char *str = line;   

while (sscanf(str, "%d%n", &num,&i)){
      printf ("number: %d\n",num);
      str+=i;
      count++;
}

Clearly its not working and I think I need a way for sscanf to skip anything that is not 0-9. Any help?

Thanks

I would suggest to read it as `%c` and then perform a check if the `char` u scanned are between 48 and 57 which are integer numbers in ASCII table — Coldsteel48, Feb 20 '14 at 00:18

score 3 · Accepted Answer · answered Feb 20 '14 at 00:24

3

change to

while (sscanf(str, "%*[^0-9]%d%n", &num, &i)==1){
    printf ("number: %d\n",num);
    str+=i;
    count++;
}

answered Feb 20 '14 at 00:24

BLUEPIXY

39,699
7
33
70

1

The loop will fail if the string starts with a digit :( – chqrlie Feb 21 '21 at 16:22

score 2 · Answer 2 · answered Feb 20 '14 at 00:51

Replace all non-digit characters with whitespace and then just scanf them all. You'll need to make adjustments for negatives but I tested this and it gives the output you want for the given input.

char line [ 200 ] ="T06      1/ 0 16/ 2  3/15 41/40 17/36  0/11  0/11  0/11  1/14  1/ 7";
int num=0;
int count=0;
int i=0;
char *str = line; 

for (int i = 0; i < strlen(line); i++)
    if (!isdigit(line[i]))
        line[i] = ' ';

while (sscanf(str, "%d%n", &num,&i) == 1){
      printf ("number: %d\n",num);
      str+=i;
      count++;
}

Kaz · Answer 3 · 2014-02-20T01:39:50.547

#include <stdio.h>
#include <string.h>
#include <errno.h>
#include <limits.h>
#include <stdlib.h>

enum oflow { OFL_GOOD, OFL_OVER, OFL_UNDER };

int getnums(const char *str, int nums[], enum oflow oflow[], int n)
{
  int i;

  for (i = 0; *str; ) {
    char *junk;
    long val;

    str += strcspn(str, "0123456789+-");

    if (!*str)
      break;

    errno = 0;
    val = strtol(str, &junk, 10);

    if (junk > str) {
      if (i < n) {
        if (((val == LONG_MIN) && errno == ERANGE) || (val < INT_MIN)) {
          nums[i] = 0;
          oflow[i++] = OFL_UNDER;
        } else if (((val == LONG_MAX) && errno == ERANGE) || (val > INT_MAX)) {
          nums[i] = 0;
          oflow[i++] = OFL_OVER;
        } else {
          oflow[i] = OFL_GOOD;
          nums[i++] = val;
        }
      } else {
        i++;
      }
      str = junk;
    } else {
      str++; /* no number was pulled out: skip one char and try again */
    }
  }

  return i;
}

int main(int argc, char **argv)
{
  int nums[256];
  enum oflow oflow[256];

  if (argc > 1) {
    int n = getnums(argv[1], nums, oflow, 256);
    int i;

    if (n > 256) {
      printf("array is too small: we need %d elements\n", n);
      n = 256;
    }

    for (i = 0; i < n; i++) {
      if (oflow[i])
        printf("nums[%d] = %sflow\n", i, oflow[i] == OFL_OVER ? "over" : "under");
      else
        printf("nums[%d] = %d\n", i, nums[i]);
    }
  }

  return 0;
}

tests:

$ ./nums ""
$ ./nums "0"
nums[0] = 0
$ ./nums "-"
$ ./nums "+"
$ ./nums "+1"
nums[0] = 1
$ ./nums "-1"
nums[0] = -1
$ ./nums "abc"
$ ./nums "asd+asdf+1  -0+1-3234abc-10zzz-"
nums[0] = 1
nums[1] = 0
nums[2] = 1
nums[3] = -3234
nums[4] = -10
$ ./nums "+++-1+++"
nums[0] = -1
$ ./nums "++"
$ ./nums "1+11111111111111111111111-111111111111111111111"
nums[0] = 1
nums[1] = overflow
nums[2] = underflow
$ ./nums "$(seq 1 300)" | head -5
array is too small: we need 300 elements
nums[0] = 1
nums[1] = 2
nums[2] = 3
nums[3] = 4
$ ./nums "$(seq 1 300)" | tail -5
nums[251] = 252
nums[252] = 253
nums[253] = 254
nums[254] = 255
nums[255] = 256

Homework exercise: in what situations is junk == str after the strtol, so that the str++ case executes? Why do we need this logic?

score 0 · Answer 4 · answered Feb 21 '21 at 16:27

0

To filter everything but the digits, use strcspn() and sscanf() in a loop:

int i;
while (sscanf(str += strcspn(str, "0123456789"), "%d%n", &num, &i) == 1) {
    printf("number: %d\n", num);
    count++;
    str += i;
}

answered Feb 21 '21 at 16:27

chqrlie

131,814
10
121
189

Luke · Answer 5 · 2014-02-20T00:31:39.840

sscanf is really a pain to use, and it's even worse if your input is not the same every time. You might try regular expressions, or you might try to parse out the integers by hand doing something like this:

int i;
int num = 0;
int num_valid = 0;
int count = 0;
for(i = 0; line[i]; i++) {  // loop through every character
    if(line[i] >= '0' && line[i] <= '9') { // check to see if it's a digit
        num *= 10;                   // scale up num
        num += lin[i] - '0';         // convert ASCII to INT
        num_valid = 1;               // indicate that num contains a real number

    } else if(num_valid) {           // if the number is done
        printf ("number: %d\n",num);
        count++;
        num = 0;
        num_valid = 0;
    }
}

chux - Reinstate Monica · Answer 6 · 2021-02-21T15:38:23.510

OP almost had it!

Add tests for end-of-string and return value from sscanf().

char line[200] = "T06 1/ 0 16/ 2  3/15 41/40 17/36  0/11  0/11  0/11  1/14  1/ 7";
int num;
int count = 0;
char *str = line;   

while (*str) {
  int i; // Place to save count of characters scanned.
  if (1 == sscanf(str, "%d%n", &num, &i)) {
    printf("number: %d\n", num);
    count++;
    str += i;
  } else {
    str++; // Go to the next character
  }
}

read only integers from complex string in c

6 Answers6

Linked