How to find if all brackets are balanced (C) [closed] - c

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So I have an array full of brackets, for example :
1) (()())())((())(())
2) ()((()()))
Any open bracket ( '(' ) should also be closed by another one (')')
So for example 1) ->
(()())())((())(()) -> (....)..)((..)(..) -> ())(()() -> .)(.. , so the answer is no because from here we can see that not all of the brackets are balanced
For example 2) ->
()((()())) -> .((..)) -> (()) -> (..) -> () -> .. , so here the answer is yes because all brackets are balanced. In this case, I would also like to print the positions of all couples of brackets that are balanced, for example :
1,2 & 5,6 & 7,8 & 3,10 & 4,9
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Answer 1

In your case, it's as simple using a counter. Increment for (, and decrement for ). It shouldn't go under 0, and should be 0 in the end if balanced.
You may consider using a stack, if you're creating some syntax parser like the compilers and interpreters do.
EDIT: you need to use a stack to print out the pairs. You need to implement a stack by hand in C, so the following is reference code in C++.
std::stack<int> s;
switch(string[i]) {
case '(':
s.push(i);
break;
case ')':
if(!s.empty()) {
printf("%d, %d\n", s.top(), i);
s.pop();
} else {
// Fail here
}
break;
}
if(!s.empty()) // Fail here

Answer 2

Count opening and closing brackets:
const char* it;
int open_n = 0, close_n = 0;
/* Assume array is zero terminated.
* Otherwise condition: 'it != &array[ArraySize]' instead '*it' */
for(it = &array[0]; *it ; ++it) {
switch( *it ) {
case '(': open_n += 1; break;
case ')': if( open_n ) { close_n += 1; break; }
else { return false; /* ErrorCloseWithoutOpen */ }
default: break;
}
}
return open_n ? false : true;