please ....i need help with this program

Write a program that read an expression and tell the user whether the bracket are correct or not (brackets checker)
for example:
( 3 * 4 + 2 ) / ( 8 - 2 ) is correct

while the expreesion ( 3 * 4 + 2 ( / (8-2) is incorrect

i wrote this incomplete prog...... please help me to complete it...


the code....


#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
struct Node;
typedef struct Node * PtrToNode;
struct Node
{
int element;
PtrToNode Next;
};

PtrToNode MakeEmpty(PtrToNode L)
{
L= new(Node);
L->Next=NULL;
return L;
}

void Push(PtrToNode L,int x)
{
PtrToNode S;
S= new(Node);
S->element=x;
S->Next=L->Next;
L->Next=S;
}

int Pop(PtrToNode L)
{
PtrToNode P;
P=L->Next;
int x=P->element;
L->Next=P->Next;
free(P);
return x;
}

For some reason I have a feeling that you have just dumped your homework to us. It smells like your teacher had given the code to you, and asked you to complete it. I doubt that you would've written that if your goal is what you described. Students tend to start by writing a main, not by trying to implement a list...

So please, tell us more specifically waht your problem is. Simple "help me to write a program" does not quite cut it...

you are thinking this too complicated...

think it this way... for every "(" there has to be a ")" otherwise the expression is wrong...

why all this pushing and poping and use of struct? unless you are simulating push-down automata

Quote:

---

Originally Posted by creeping death

you are thinking this too complicated...

think it this way... for every "(" there has to be a ")" otherwise the expression is wrong...

why all this pushing and poping and use of struct? unless you are simulating push-down automata

---

So this expression is correct then:

Code:

---

())(()

---

since it contains the right number of left and right match.

--
Mats

i wrote the main but i am completely sure that it is incorrect......if you can correct it....please help :(



main()
{
PtrToNode L;
int choise;
L = MakeEmpty(NULL);


printf("please enter your numbers without space");
scanf("%d",&choise);
while(L->Next!=NULL)
{
if(L->element=='+')
break;

else if(L->element=='-')
break;

else if(L->element=='*')
break;

else if(L->element=='/')
break;

else if(L->element=='(')
{
Push(L,choise);
}
else if(L->element==')')
{
Pop(L);
}


getch();
}

No! if my "flag" variable if it becomes -1, the expression is wrong(it becomes -1 and does not check any further)...hehe...you almost got me on that occasion

Code:

---

[c_d@localhost c scratchpad]$ ./a.out
enter expression())(()
expr is wrong

---

like i said earlier for every "(" there got to be a ")" ... since ")" came before "(" the expression became wrong...

hi angel_copra,

Code:

---

printf("please enter your numbers without space");
scanf("%d",&choise);

---

what is this?

lol sorry....x input from the user eg, ( 3 + 3 ) / 8

main()
{
PtrToNode L;
int x;
L = MakeEmpty(NULL);


printf("please enter your numbers without space");
scanf("%d",&x);
while(L->Next!=NULL)
{
if(L->element=='+')
break;

else if(L->element=='-')
break;

else if(L->element=='*')
break;

else if(L->element=='/')
break;

else if(L->element=='(')
{
Push(L,choise);
}
else if(L->element==')')
{
Pop(L);
}


getch();
}

you cant store an expression in an integer.

you need to store the expression as a string.
and then consider each character of the string, one at a time.

also are you supposed to only check if the brackets are balanced or if the expression is correct also?

example: (a +b + ) * ( a / ) here the brackets are balanced but expression is wrong.

if you have to check for brackets only , then push the character whenever you find ( and pop whenever you find ) ... you are doing that ...except you should handle the situation when you try to pop() when the stack is empty....(like in the example matsp gave)

but why break whenever you encounter a binary operator?

i am check for brackets only... i have 2 remove this ...right

if(L->element=='+')
break;

else if(L->element=='-')
break;

else if(L->element=='*')
break;

else if(L->element=='/')
break;

what about this code....


for (int i=0;i<sizeof b;i++)
{
if (b[i]=='(')
sum++;
if (b[i]==')')
sum--;
if (sum<0)
break;
}

It's clear you don't have a clue what you're doing and this problem is too far above your difficulty level at this stage. All you've got are a bunch of random bits of code thrown together and you're just hoping that it in some way does something useful. Even then I'd guess that those bits were copied from someone else.
So there are two problems here:

1. You don't understand how to fully solve the problem
2. You aren't yet a proficient C Programmer

Which one of those would you like help with first?
How long have you got left for this assignment?

sum=0;

Quote:

---

Originally Posted by angel_copra

what about this code....

Code:

---

for (int i=0;i<sizeof b;i++)
    {
    if (b[i]=='(')
        sum++;
    if (b[i]==')')
        sum--;
    if (sum<0)
        break;
    }

---

---

Ah now that code could actually do something useful. It doesn't solve the whole problem, but it does solve part of it, if you know what comes after it. However, unless you can demonstrate that you understand how that works, I'm inclined to believe that it is someone else's code.

Perhaps you could attempt to describe in your own words, the algorithm you will use to solve the problem. It doesn't matter if you get it wrong or get stuck, just give it a try and we'll give you a nudge.

1.yes i don't understand how to solve the problem :(

my assignment 8/4/2009 about 4 days

Quote:

---

Originally Posted by iMalc

Ah now that code could actually do something useful. It doesn't solve the whole problem, but it does solve part of it, if you know what comes after it. However, unless you can demonstrate that you understand how that works, I'm inclined to believe that it is someone else's code.

Perhaps you could attempt to describe in your own words, the algorithm you will use to solve the problem. It doesn't matter if you get it wrong or get stuck, just give it a try and we'll give you a nudge.

---

You are right it’s not my code; I can’t solve it by myself….

Okay it might be easiest if we start with pinning down what it takes for the expression to be invalid.
Here are a few examples:

Code:

---

1 ( 2 )
+ 3
)

---

The first one is invalid because a bracket should not come directly after a number
The second one is invalid because a plus sign must not be the first thing in the expression.
The Third one is invalid because an expression must not start with a close bracket, or more accurately, the close bracket does not have a corresponding open bracket before it.
See how many more you can come up with.
Then once you think you've got all the rules about when something is invalid, then you've got a good idea on what it takes for an expression to be valid, and after that you might be able to work out pseudocode for solving this and then onto the real code.

One thing though, do you know all of the possible things that can be in an expression? I mean it can have plus, minus, multiply and divide, but can it also have thing like ^ which means to-the-power-of? If you aren't sure, then just assume it only has +, -, *, /, (, ).

Quote:

---

Originally Posted by iMalc

Okay it might be easiest if we start with pinning down what it takes for the expression to be invalid.
Here are a few examples:

Code:

---

1 ( 2 )
+ 3
)

---

The first one is invalid because a bracket should not come directly after a number
The second one is invalid because a plus sign must not be the first thing in the expression.
The Third one is invalid because an expression must not start with a close bracket, or more accurately, the close bracket does not have a corresponding open bracket before it.
See how many more you can come up with.
Then once you think you've got all the rules about when something is invalid, then you've got a good idea on what it takes for an expression to be valid, and after that you might be able to work out pseudocode for solving this and then onto the real code.

One thing though, do you know all of the possible things that can be in an expression? I mean it can have plus, minus, multiply and divide, but can it also have thing like ^ which means to-the-power-of? If you aren't sure, then just assume it only has +, -, *, /, (, ).

---

k what else...?

Quote:

---

Originally Posted by iMalc

Okay it might be easiest if we start with pinning down what it takes for the expression to be invalid.
Here are a few examples:

Code:

---

1 ( 2 )
+ 3
)

---

The first one is invalid because a bracket should not come directly after a number
The second one is invalid because a plus sign must not be the first thing in the expression.
The Third one is invalid because an expression must not start with a close bracket, or more accurately, the close bracket does not have a corresponding open bracket before it.
See how many more you can come up with.
Then once you think you've got all the rules about when something is invalid, then you've got a good idea on what it takes for an expression to be valid, and after that you might be able to work out pseudocode for solving this and then onto the real code.

One thing though, do you know all of the possible things that can be in an expression? I mean it can have plus, minus, multiply and divide, but can it also have thing like ^ which means to-the-power-of? If you aren't sure, then just assume it only has +, -, *, /, (, ).

---

hi, i think the OP just wants to check if the brackets are correct and balanced out...but anyway...please ignore me and continue...you're doing an awesome thing btw.

Quote:

---

Originally Posted by creeping death

hi, i think the OP just wants to check if the brackets are correct and balanced out...but anyway...please ignore me and continue...you're doing an awesome thing btw.

---

whyyyyyyyyy? please i need your help

How about putting the expression, into a string of char's:

Code:

---

Pseudocode:
char exp[50] = { '\0' };

fgets() can get the expression from the user, and put it into the exp array

then have 2 int variables to count up the left and right tally of brackets:
int left, right, noGood;

then we can scan the exp array, char by char:
left = right = noGood = 0;
for(i = 0; exp[i] != '\0'; i++)  {
  if(exp[i] == '(')
      left++;
  else if(exp[i] == ')')
      right++;


  //there can't be more right brackets then left brackets in the expression, at any time
  if(right > left)
      noGood = 1;
   
}

---

Now if left == right and noGood still equals 0, it's OK.

So print that message

Note that this is pseudocode, not ready-to-run code.

Quote:

---

Originally Posted by creeping death

hi, i think the OP just wants to check if the brackets are correct and balanced out...but anyway...please ignore me and continue...you're doing an awesome thing btw.

---

Oh, that could be the case. Okay I'll assume that only the brackets are to be checked. Though that makes it even stranger that the first bit of code posted would be for linked lists.

In that case, all you need to do is starting from the left hand side of the expression, look at one character at a time, looking for ( and ) characters. Then just make sure that the number of close brackets never exceeds the number open brackets at any point.
Some examples:

Code:

---

(        invalid - there are more open that close
( )        valid
) (        invalid - there is a point at which we have closed more than we have opened
( ( )        invalid - not all brackets are closed (the counts are not equal)
( ) ) ( ( )        invalid - the counts match, but again there is a point at which more have been closed than were previously opened

---

The for-loop with 'sum' in it is the kind of thing that would help here, though there are definitely a few problems with the original code for that. It should not use sizeof for a start. You'll probably want to use strlen.

Okay, now start with a basic Hello World program, and add in something to read in the expression from the user. You'll need something like fgets and sscanf and an array of chars, though there is certainly more than one way of doing it.
See if you can write a program that reads in a sentence and writes it out back to you again.

Quote:

---

Originally Posted by angel_copra

whyyyyyyyyy? please i need your help

---

hi angel_copra,
iMalc is already helping you . I just re-conveyed what your requirement was. What he was about to tell, could have drastically increased the complexity of the program. Dont worry you are in good hands. Keep trying. C is easy once you understand what you are doing.

Quote:

---

Originally Posted by iMalc

Okay, now start with a basic Hello World program, and add in something to read in the expression from the user. You'll need something like fgets and sscanf and an array of chars, though there is certainly more than one way of doing it.
See if you can write a program that reads in a sentence and writes it out back to you again.

---

#include<stdio.h>
#include<conio.h>
void main()
{
int x,y,sum;
printf("please enter your value");
scanf("%d%d",&x,&y);
sum=x+y;
printf("the answer is %d ",z);
getch();
}


loool

you can't use z, since you didn't declare it. sum is what you want, isn't it?

Please use code tags around your code. Click on the # sign after you highlight your code, to do that.

should it fix it or only detect if there its right or not ?
i coded a fix program for it

Code:

---

#include <stdio.h>
void string_fixer(char *buffer)
{
    int i;//loop counter
    for(i=0;buffer[i]!=0;i++){
            if(buffer[0]=='('){
                 
            if(buffer[i]=='/'){
                buffer[i-2]=')';
                }
          }
    }
}
int main(void)
{
    char buffer[100];
    double x;//for atoi
    sprintf(buffer,"( 3 * 4 + 2 ( / (8-2)");
    string_fixer(buffer);
    puts(buffer);
    return getchar();
}

---

you should ident your code lil bit and put instead of z put num coz thats ur equation whish sum = x+y

here you go~~

code not tested!!

Code:

---

main(){

        char *input="(2+3)+((4-6)";
        int flag=0;

        while(*input){
                if(*input=='('){
                        flag++;
                }else if(*input==')'){
                        flag--;
                }
                if(flag<0){
                        printf("Incorrect brackets starting at: %s\n",input);
                        exit(0);
                }
                *input++;       
        }

        if(flag!=0){
                        printf("Probably %d closing bracket(s) missing brackets\n",flag);
        }else{
                printf("Correct brackets!\n");
        }               
}

---

main should be int main(void) - read FAQ
input shoudl be const char*

Quote:

---

*input++;

---

Why do you need * here?

Also your main is missing return 0; at the end

no offence prakash0104 lolguy,but if you notice the OPs post #11 you can see that he/she had already got the solution partly...the point was to help the OP build the basics of C, not to spoon-feed him/her and give ready-made answer...as that would not help in the long run...(see homework policy)