// assume input is string starting with '('
void get_string_up_to_matching_parens(char* const string, char* matched) {
int i = 1;
// initialize stack with the open parens, when stack is empty, initial parens
// was matched
char_stack* c_stack = malloc(sizeof(struct char_stack));
push_char_stack('(', c_stack);
while (!char_stack_empty(c_stack) && string[i] != EOF && string[i] != '\0') {
// printf("%c", string[i]);
if (string[i] == '(') {
push_char_stack('(', c_stack);
}
else if (string[i] == ')') {
pop_char_stack(c_stack);
}
// if I have string = (abc) I want to store matched = abc
// so matched[0] = string[1] and so on
matched[i-1] = string[i];
i++;
}
if (!char_stack_empty(c_stack)) {
error(1,0,"no matching close parentheses found");
}
// if I have string = (abc) I want to store matched = abc
// at the end when ) is matched i = 5 and we need to subtract 2 to
// overwrite the the close parentheses
matched[i-2] = '\0';
}
//when popping one char in charstack, two nums will be popped.
void pop_pro(stacknum *list_num, stackchar *list_char)
{
int pop_num1,pop_num2,res;
char pop_char1;
pop_num1=pop_num_stack(list_num);//pop the first elem
pop_num2=pop_num_stack(list_num);//pop the second elem
pop_char1=pop_char_stack(list_char);//pop the top operator
res=jisuan( pop_char1, pop_num1, pop_num2);//calcu the res and push it into the numstack
push_num_stack(list_num, res);
}
