Count and get integers from a string using C - c

Question

I am self teaching C programming.
I am trying to count number of int present in given string which are separated by space.
exp:
input str = "1 2 11 84384 0 212"
output should be: 1, 2, 11, 84384, 0, 212
total int = 6
When I try. It gives me all the digits as output which make sense since I am not using a right approach here.
I know in python I can use str.split (" ") function which can do my job very quickly.
But I want to try something similar in C. Trying to create my own split method.
#include <stdio.h>
#include <string.h>
void count_get_ints(const char *data) {
int buf[10000];
int cnt = 0, j=0;
for (int i=0; i<strlen(data); i++) {
if (isspace(data[i] == false)
buf[j] = data[i]-'0';
j++;
}
printf("%d", j);
}
// when I check the buffer it includes all the digits of the numbers.
// i.e for my example.
// buf = {1,2,1,1,8,4,3,8,4,0,2,1,2}
// I want buf to be following
// buf = {1,2,11,84384,0,212}
I know this is not a right approach to solve this problem. One way to keep track of prev and dynamically create a memory using number of non space digits encountered.
But I am not sure if that approach helps.

Answer 1

You want to build your number incrementally until you hit a space, then put that into the array. You can do this by multiplying by 10 then adding the next digit each time.
void count_get_ints(const char *data) {
int buf[10000];
int j = 0;
int current_number = 0;
// Move this outside the loop to eliminate recalculating the length each time
int total_length = strlen(data);
for (int i=0; i <= total_length; i++) {
// Go up to 1 character past the length so you
// capture the last number as well
if (i == total_length || isspace(data[i])) {
// Save the number, and reset it
buf[j++] = current_number;
current_number = 0;
}
else {
current_number *= 10;
current_number += data[i] - '0';
}
}
}

Answer 2

I think strtok will provide a cleaner solution, unless you really want to iterate over every char in the string. It has been a while since I did C, so please excuse any errors in the code below, hopefully it will give you the right idea.
#include <stdio.h>
#include <stdlib.h>
int main() {
char str[19] = "1 2 11 84384 0 212";
const char s[2] = " ";
char *token;
int total;
total = 0;
token = strtok(str, s);
while (token != NULL) {
printf("%s\n", token);
total += atoi(token);
token = strtok(NULL, s);
}
printf("%d\n", total);
return 0;
}

Answer 3

You can check the ascii value of each character by doing c-'0'. If it's between [0,9], then it's an integer. By having a state variable, when you're inside an integer by checking if a given character is a number of space, you can keep track of the count by ignoring white space. Plus you don't need a buffer, what happens if data is larger than 10,000, and you write pass the end of the buffer?, undefined behavior will happen. This solution doesn't require a buffer.
Edit, the solution now prints the integers that are in the string
void count_get_ints(const char *data) {
int count = 0;
int state = 0;
int start = 0;
int end = 0;
for(int i = 0; i<strlen(data); i++){
int ascii = data[i]-'0';
if(ascii >= 0 && ascii <= 9){
if(state == 0){
start = i;
}
state = 1;
}else{
//Detected a whitespace
if(state == 1){
count++;
state = 0;
end = i;
//Print the integer from the start to end spot in data
for(int j = start; j<end; j++){
printf("%c",data[j]);
}
printf(" ");
}
}
}
//Check end
if(state == 1){
count++;
for(int j = start; j<strlen(data); j++){
printf("%c",data[j]);
}
printf(" ");
}
printf("Number of integers %d\n",count);
}

Answer 4

I believe the standard way of doing this would be using sscanf using the %n format specifier to keep track of how much of the string is read.
You can start with a large array to read into -
int array[100];
Then you can keep reading integers from the string till you can't read anymore or you are done reading 100.
int total = 0;
int cont = 0;
int ret = 1;
while(ret == 1 && total < 100) {
ret = sscanf(input, "%d%n", &array[total++], &cont);
input += cont;
}
total--;
printf("Total read = %d\n", total);
and array contains all the numbers read.
Here is the DEMO

Answer 5

Example using strtol
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <errno.h>
#include <ctype.h>
int count_get_ints(int output[], int output_size, const char *input) {
const char *p = input;
int cnt;
for(cnt = 0; cnt < output_size && *p; ++cnt){
char *endp;
long n;
errno = 0;
n = strtol(p, &endp, 10);
if(errno == 0 && (isspace((unsigned char)*endp) || !*endp) && INT_MIN <= n && n <= INT_MAX){
output[cnt] = n;
while(isspace((unsigned char)*endp))
++endp;//skip spaces
p = endp;//next parse point
} else {
fprintf(stderr, "invalid input '%s' in %s\n", p, __func__);
break;
}
}
return cnt;
}
int main(void) {
const char *input = "1 2 11 84384 0 212";
int data[10000];
int n = sizeof(data)/sizeof(*data);//number of elements of data
n = count_get_ints(data, n, input);
for(int i = 0; i < n; ++i){
if(i)
printf(", ");
printf("%d", data[i]);
}
puts("");
}

Answer 6

Assuming you don't have any non-numbers in your string, you can just count the number of spaces + 1 to find the number of integers in the string like so in this pseudo code:
for(i = 0; i < length of string; i++) {
if (string x[i] == " ") {
Add y to the list of strings
string y = "";
counter++;
}
string y += string x[i]
}
numberOfIntegers = counter + 1;
Also, this reads the data between the white spaces. Keep in mind this is pseudo code, so the syntax is different.