How to extract multi-digit numbers from a string? - c

Question

Firstly, I know similar questions have been asked before but I believe my case is different.
My input string is:
(5,7) (1,6) (2,4) (10,14) (8,9)
I wrote the following code for extraction into an array.
main(){
char s[100];
int i=0,x,n=0;
int a[20];
printf("Enter the sets:");
gets(s);
x=strlen(s);
while(i<x){
if((s[i]=='(' && s[i+2]==',') || (s[i]==',' && s[i+2]==')'))
{
a[n]=s[i+1]-'0';
n++;
}
i++;
}
for(i=0;i<n;i++){
printf("%d\n",a[i]);
}
}
The output I get is:
5 7 1 6 2 4 8 9
I understand why my code will skip numbers having 2 or more digits.
Please suggest some minor changes to the present code to fix this limitation.
P.S.- I'm looking for a solution which doesn't depend on length of the number.

Answer 1

Since you only care about the numbers and not any of the delimiters, you can use strtok, which allows for a set of delimiters.
Use the following in place of you existing while loop:
char *p = strtok(s, "(), ");
while (p) {
a[n++] = atoi(p);
p = strtok(NULL, "(), ");
}
Output:
5
7
1
6
2
4
10
14
8
9
If on the other hand you are particular about the format, you can do the following:
char *start = s, *p1 = NULL, *p2 = NULL, *p3 = NULL;
if (start) p1 = strchr(start, '(');
if (p1) p2 = strchr(p1+1, ',');
if (p2) p3 = strchr(p2+1, ')');
while (p1 && p2 && p3) {
a[n++] = atoi(p1+1);
a[n++] = atoi(p2+1);
start = p3+1;
if (start) p1 = strchr(start, '(');
if (p1) p2 = strchr(p1+1, ',');
if (p2) p3 = strchr(p2+1, ')');
}

Answer 2

I have used a different approach to the problem, but I have solved it and it works. Consider trying this. Btw I have used char *s as a string literal but you can keep it like yours.
main(){
char *s="(5,7) (1,6) (2,4) (10,14) (8,9)";
int i=0,x,n=0;
char a[20];
x=strlen(s);
while(i<x){
if (isdigit(s[i])) {
a[n]=s[i];
if (s[i+1]==',' || s[i+1]==')') {
a[n+1]=' ';
n++;
}
n++;
}
i++;
}
printf("%s\n", a);
}
output:
tenshi#mashiro:~/projects/test$ ./test
5 7 1 6 2 4 10 14 8 9

Answer 3

#include <stdio.h>
int main(void) {
// your code goes here
char s[100];
int i=0,x,n=0;
int a[20];
printf("Enter the sets:");
gets(s);
x=strlen(s);
while(i<x-1){
if(isdigit(s[i]))
{
if(isdigit(s[i+1]))
{
a[n]=(s[i]-'0')*10 +(s[i+1]-'0');
i++;
}
else
{
a[n]=s[i]-'0';
}
n++;
}
i++;
}
printf("\n");
for(i=0;i<n;i++){
printf("%d\n",a[i]);
}
return 0;
}
What about the above code, unfortunately C doesn't have simple string functions like split with Regex(it has split function but i didn't understand well). Alternatively, here is ideone for it https://ideone.com/eRKTbD

Answer 4

If the input is in the exact format as in the question, then you can add two loops inside the main while loop to read one set at a time.
while (i < x)
{
if (s[i] == '(')
{
// temporary var to store number
int num = 0;
// read first number
while (s[++i] != ',')
num = num*10 + s[i]-'0';
a[n++] = num;
num = 0;
// read second number
while (s[++i] != ')')
num = num*10 + s[i]-'0';
a[n++] = num;
}
i++;
}

Answer 5

If you always have the same format (a,b)(c,d)...(y,z) and the same number of values then this solution works :
char * arr = "(5,7)(1,6)(2,4)(10,14)(8,9)";
int a,b,c,d,e,f,g,h,i,j;
sscanf(arr,"(%d,%d)(%d,%d)(%d,%d)(%d,%d)(%d,%d)",&a,&b,&c,&d,&e,&f,&g,&h,&i,&j);
printf("%d %d %d %d %d %d %d %d %d %d\n", a, b, c, d, e, f, g, h, i, j);