Extracting chars from inside '(' and ')' from char array in C - c

Question

I am stuck on this problem:
A user writes a math expression and the program need to extract from it all subexpressions that are inside ( ) ex. (x+2) or (x+(y-2)).
For example, if the user enters 2-(5+(x-6)-(y+9)), the program should return (x-6), (y+9), (5+(x-6)-(y+9))
This is what I've tried to do.
#include<stdio.h>
int main()
{
char a[100];
int i=0,t,j=0;
printf("enter math expression:\n");
while( (a[i++]=getchar()) != '\n' && i < 100);
a[i] = '\0';
for (i=0; a[i]!='\0'; i++)
{
if (a[i]=='(')
{ printf("found (\n");
j++;
while (a[i] !=')')
printf("%c",a[i++]);
printf("%c",a[i]);

Answer 1

Since you are dealing with nested expressions, you need to keep a stack around in order to match parentheses. Ideally inside the loop you should:
Whenever found a '(', push position within the string into the stack
When a ')' is found, then pop from the stack the position of the matching '('. You have start-end indexes for your expression.
Continue until string is finished
Example (x + (y+2)):
i == 0 -> '(' -> s.push(0);
i == 1 -> 'x'
i == 2 -> '+'
i == 3 -> '(' -> s.push(3);
i == 4 -> 'y'
i == 5 -> '+'
i == 6 -> '2'
i == 7 -> ')' -> s.pop() [3, 7] contains '(y + 2)'
i == 8 -> ')' -> s.pop() [0, 8] contains '(x + (y+2))'

Answer 2

Let, s be a cstring in this context.
Since in valid mathematical expression parenthesis are balanced, we can observe that,
s+1 is balanced if s[0] is not '('
if s[0] is '(' then we are at the start of a parenthesis'ed expression.
At the case of #1 we do not care about the first character. So, we can set s = s+1 and start over again.
At the case of #2 we have to find the matched end for the first character. So, we divide it in two part. The matched expression and the tail. Both are valid expression. So, we start with them over again.
int main()
{
char s[] = "2-(5+(x-6)-(y+9))";// s have to be modifiable.
extractAndPrint(s, 0); // we are at start and 0 indicates balanced
return 0;
}
Now,
void extractAndPrint(char *s, int found)
{
if(s[0] == 0) return;//no more honey
if(!found){ //so far balanced
if(s[0] == '(') extractAndPrint(s+1, 1);// case 2
else extractAndPrint(s+1, 0); //this is the case 1
} else {//start of a expression
char *t;
//separates the end of current expression
//and the remaining part.
mark_end(s, 0, &t);
printf("(%s)\n", s);
extractAndPrint(s, 0);//potentially contain more expression
extractAndPrint(t, 0);//check past the previous exp
}
}
I liked recursion here too. It takes less thinking for me. Can be implemented in more elegant way.
void mark_end(char *s, int c, char **t)
{
if(c == 0){
if(s[0] == ')'){
*t = s+1;
*s = '\0';
} else if(s[0] == '('){
mark_end(s+1, c+1, t);
} else {
mark_end(s+1, c, t);
}
} else {
if(s[0] == ')'){
mark_end(s+1, c-1, t);
} else if(s[0] == '('){
mark_end(s+1, c+1, t);
} else {
mark_end(s+1, c, t);
}
}
}
OUTPUT:
(5+(x-6)-(y+9))
(x-6)
(y+9)

Answer 3

I recommend you to use state machine, in your case this state machine should be suitable :
so your main will be a kind of a switch case, take care of writing your pop and push fuctions and represent your heap with a good structure

Answer 4

You should use recursion to parse the string as a tree of parenthesis. Here is an example showing how to do it :
#include <stdio.h>
#include <string.h>
#define MAX_DATA 100
int mathexp(const char *exp, int start, int end)
{
int i = start;
while(i < (start + end) && exp[i] != ')') {
if(exp[i] == '(') {
i = mathexp(exp, i + 1, end) + 1;
} else {
i++;
}
}
char subexp[i - start];
memcpy(subexp, &exp[start], i - start);
subexp[i - start] = '\0';
printf("%s\n", subexp);
return i;
}
int main(int argc, char *argv[])
{
char input[MAX_DATA];
printf("Enter a math expression: ");
fgets(input, sizeof(input), stdin);
input[strcspn(input, "\r\n")] = '\0'; // remove trailing \r\n
mathexp(input, 0, strlen(input));
return 0;
}
and when you test it, you have :
Enter a math expression: 2-(5+(x-6)-(y+9))
x-6
y+9
5+(x-6)-(y+9)
2-(5+(x-6)-(y+9))