Stack Overflow in Balanced Parenthesis - c

Question

I am trying to write a program to check if a string of brackets is balanced or not. If balanced, I do not want any output. But if it is not balanced then I want it to return the brackets needed to balance it. For example '([])' should be balanced and '({}' should be unbalanced and should return "open: )". The code seems to work fine until I have two or more brackets left open. So if I input '({' my code works but when I add a few more characters like '([])([' it fails because of a stack overflow. I do not understand what the error could be or how to fix it.
Here is my code:
void push(struct sNode** top_ref, int new_data){
struct sNode* new_node = (struct sNode*)malloc(sizeof(struct sNode));
if (new_node == NULL) {
printf("Stack overflow \n");
getchar();
exit(0);
}
new_node->data = new_data;
new_node->next = (*top_ref);
(*top_ref) = new_node;
}
int pop(struct sNode** top_ref){
char res;
struct sNode* top;
if (*top_ref == NULL) {
printf("Stack overflow \n");
getchar();
exit(0);
}else {
top = *top_ref;
res = top->data;
*top_ref = top->next;
free(top);
return res;
}
}
// Returns 1 if character1 and character2 are matching left and right Brackets
bool isMatchingPair(char character1, char character2)
{
if (character1 == '(' && character2 == ')')
return 1;
else if (character1 == '{' && character2 == '}')
return 1;
else if (character1 == '[' && character2 == ']')
return 1;
else
return 0;
}
// Return 1 if expression is balanced
int areBracketsBalanced(char exp[]){
int i = 0;
struct sNode* stack = NULL;
while (exp[i]){
if (exp[i] == '{' || exp[i] == '(' || exp[i] == '[')
push(&stack, exp[i]);
if (exp[i] == '}' || exp[i] == ')' || exp[i] == ']') {
if (stack == NULL){
printf("%d:%c\n",i,exp[i]);
return 0;
}else if (!isMatchingPair(pop(&stack), exp[i])){
printf("%d:%c\n",i,exp[i]);
return 0;
}
}
i++;
}
if (stack != NULL){
printf("open: ");
while(stack!=NULL){
if(pop(&stack)=='('){
printf("%c",')');
}else if(pop(&stack)=='['){
printf("%c",']');
}else if(pop(&stack)=='{'){
printf("%c",'}');
}
}
printf("\n");
return 0;
}
return 0;
}
int main(int argc, char* argv[])
{
char *exp = (char *)malloc(strlen(argv[1]) * sizeof(char));
for(int i = 0; i<strlen(argv[1]); i++){
exp[i] = argv[1][i];
}
//char exp[] = "([])([";
areBracketsBalanced(exp);
return 0;
}

Answer 1

As it has been pointed out in comments, your exp in main is wrong as it end up without a zero termination. Actually there is no need for it... just pass argv[1] to the function.
int main(int argc, char* argv[])
{
if (argc > 1)
{
areBracketsBalanced(argv[1]);
}
else
{
puts("Missing argument");
}
return 0;
}
Besides that, you have a problem here:
while(stack!=NULL){
if(pop(&stack)=='('){
printf("%c",')');
}else if(pop(&stack)=='['){
printf("%c",']');
}else if(pop(&stack)=='{'){
printf("%c",'}');
}
}
Each of these pop(&stack) will change stack and maybe cause it to be NULL. At worst you do 3 pop-operations before checking for NULL. That's not good.
Change the code so that you have one pop operation before the if statements. Save the result in a local variable that you use for the if statement. Like:
while(stack!=NULL){
int value = pop(&stack);
if(value == '('){
printf("%c",')');
}else if(value == '['){
printf("%c",']');
}else if(value == '{'){
printf("%c",'}');
}
}