Pop works only once in specific function - c

Question

I implemented stack as linked list and I wanted to make function which tells if brackets are in good order for example : (()) is good ())( is bad . Logic of function isn't good right now but I don't know why pop() works only once.Here is my code(stog is stack and sljedeci is next):
struct stog {
int x;
stog *sljedeci;
};
typedef struct stog stog;
stog *top;
char pop() {
stog *temp;
temp = (stog*)malloc(sizeof(stog));
temp = top;
char n = temp->x;
top = temp->sljedeci;
top->x = temp->sljedeci->x;
free(temp);
return n;
}
void init() {
top = NULL;
}
void push(char x) {
stog *temp;
temp=(stog*)malloc(sizeof(stog));
temp->x = x;
temp->sljedeci = top;
top = temp;
}
void Brackets(const char* msg) {
char z;
for (int i = 0; i < strlen(msg); i++) {
if (msg[i] == '(') {
push('(');
}
if (msg[i]==')'){
z = pop();
printf("Bad\n");
}
}
}
int main() {
const char* msg = "(())";
init();
Brackets(msg);
return 0;
}
Output is:
Bad
It should be:
Bad
Bad
EDIT: Added init() and push() functions

Answer 1

This line in pop doesn't make sense:
top->x = temp->sljedeci->x;
In the prior line, you assign temp->sljedeci to top. So the x member referenced here is actually the same one on both sides, so assuming both top and temp->sljedeci are not null it does nothing. If either one is NULL, you invokes undefined behavior because you derefrence a null pointer. So get rid of this line.
You also have a memory leak here in pop:
temp = (stog*)malloc(sizeof(stog));
temp = top;
You allocate memory and assign its address to temp, but then you immediately overwrite that address with the value of top.
There's no need to allocate more memory here, so remove the malloc call.

Answer 2

We, beginners, should help each other.:)
I am doing your assignment the first time in my life.:)
For starters always use English words for identifiers. Otherwise a program text can be unreadable.
It is unclear why the data member x has the type int instead of char though the stack deals with characters of a string.
struct stog {
int x;
stog *sljedeci;
};
You did not show all your stack implementation nevertheless for example the function pop is invalid. It produces a memory leak.
At first you allocated memory and its address assigned to the pointer temp and at once in the next line you reassigned the pointer. So the allocated memory will not be freed.
temp = (stog*)malloc(sizeof(stog));
temp = top;
This statement
top->x = temp->sljedeci->x;
can invoke undefined behavior if top is equal to NULL.
Also in the function Brackets this if statement
if (msg[i]==')'){
z = pop();
printf("Bad\n");
}
does not make sense. The function will always output "Bad" as soon as the character ')' is encountered.
Here is a solution I have done.
#include <stdio.h>
#include <stdlib.h>
struct stack
{
char c;
struct stack *next;
};
char * top( struct stack **stack )
{
return *stack == NULL ? NULL : &( *stack )->c;
}
int push( struct stack **stack, char c )
{
struct stack *current = malloc( sizeof( struct stack ) );
int success = current != NULL;
if ( success )
{
current->c = c;
current->next = *stack;
*stack = current;
}
return success;
}
void pop( struct stack **stack )
{
if ( *stack )
{
struct stack *current = *stack;
*stack = ( *stack )->next;
free( current );
}
}
int empty( struct stack **stack )
{
return *stack == NULL;
}
void clear( struct stack **stack )
{
while ( *stack ) pop( stack );
}
int bracket_balance( const char *s )
{
struct stack *stack = NULL;
int balanced = 1;
for ( ; *s && balanced; ++s )
{
if ( *s == '(' )
{
push( &stack, *s );
}
else if ( *s == ')' )
{
if ( ( balanced = !empty( &stack ) && *top( &stack ) == '(' ) )
{
pop( &stack );
}
}
}
balanced = balanced && empty( &stack );
clear( &stack );
return balanced;
}
int main(void)
{
const char * s[] =
{
"", "(", ")", "()", ")(", "( ( ) )", "( )( )", "( ) ) (", "Hello"
};
for ( size_t i = 0; i < sizeof( s ) / sizeof( *s ); i++ )
{
if ( bracket_balance( s[i] ) )
{
printf( "\"%s\" has balanced brackets\n", s[i] );
}
else
{
printf( "\"%s\" does not have balanced brackets\n", s[i] );
}
}
return 0;
}
The program output is
"" has balanced brackets
"(" does not have balanced brackets
")" does not have balanced brackets
"()" has balanced brackets
")(" does not have balanced brackets
"( ( ) )" has balanced brackets
"( )( )" has balanced brackets
"( ) ) (" does not have balanced brackets
"Hello" has balanced brackets