arranging a statement

[sa-al]

hello i have wrote a program that converts a statement to sth like the example below please help me:
(9*(1+(3*8)))+(1+(4*5))
into:
9 1 3 8 * + * 1 4 5 * + +

while((c=getch())!='\0'){   /*i dont know whether it is true or not*/

	if(c=='('){
	co[k]=1;
   	k++;
   	break;
		}

	if(c==')'){
	while(k>=0&&k<=20){
   	co[k]=0;
      	k--;
         }
      	break;
		}

	if(c==" "){
		break;
		}  /*this ode may be useful for converting a usual statemnt to poland model*/

	if(c==n){
 	while(n>=1){
   	rep[i+n]=rep[i+n-1];
      	n--;
   		}
	rep[i]=c;
   	i++;
   	n=counter;
 		}

	if(c==o){
	if(co[k]=0){
   	rep[i+n]=c;
      	i=i+n;
      	n=1;
   		}
   	else{
       	while(n>=1){
       	rep[i+n]=rep[i+n-1];
         	n--;
       		}
   	rep[i]=c;
   	n=counter;
   	n++;
   	counter=n;
   	}

}

please tell me where is the problem.

[invisal]

while((c=getch())!='\0') {

The getch() get a single character from keyboard. There is no NULL key button in our keyboard, so the condition is always true.

[Salem]

Your code is poorly indented.
The variables are undeclared.
The variable names are meaningless.

> if(c==" ")
This is meaningless in a C program, you can't compare arrays of chars like this.

> /*this ode may be useful for converting a usual statemnt to poland model*/
Huh?
It's the only comment, and it just adds to the confusion.

I assume by "poland", you mean Reverse Polish Notation (RPN).
There is an algorithm, which I believe is called the "shunting yard algorithm", for converting infix expressions into RPN expressions.

>please tell me where is the problem
You need to go back to basics I think.

[sa-al]

hello i have changed my previous code about converting a statement to reverse polish notation but still i think it has serious problems .can any one help me solve it? i have just a little time!

#include <stdio.h>
#include <conio.h>

main(){

int k=0;
int co[20];
int n=0;
char c;
char rep[256]={0};
int i=0;

while((c=getchar())!='\n'){  

	if(c=='('){
		co[k]=1;
   	k++;
   	break;
		}

	if(c==')'){
		while(k>=0&&k<=20){
   		co[k]=0;
      	k--;
         }
      	break;
		}

	if(c==' '){
		break;
		}  



	 if(c=='*'||c=='-'||c=='+'||c=='/'||c=='s'||c=='c'||c=='t'||c=='C'){
		if(co[k]=0){
   		rep[i+n]=c;
         //printf("%c",rep[0]);
         //getch();
      	i=i+n;
      	n=1;
   		}
   	else{
       	while(n>=1){
       		rep[i+n]=rep[i+n-1];
         	n--;
       		}
   	if(rep[i]!='s'||'c'||'t'||'C') rep[i]=c;
   	n++;
   	}
  }
     	 if(c>47&&c<58){
 			while(n>=1){

   		rep[i+n]=rep[i+n-1];
      	n--;
   		}
		rep[i]=c;
   	i++;
 	
}
}

There is no point creating a separate thread again. It just wastes our time.

As I said before I think you need to go back to basics, i.e figuring out basic i/o basic parsing etc.
Have you considered how you are going to represent the stack. Are you going to use a linked list la-la-la-de-da.

[WaltP]

hello i have changed my previous code about converting a statement to reverse polish notation but still i think it has serious problems .can any one help me solve it? i have just a little time!

You think you have a problem? What makes you think so? Have you read the post Read Me: Read This Before Posting yet? The title says it all.

Also, you need to format your code. See this...

[invisal]

Let me interpret how is your code really works.

#include <stdio.h>
#include <conio.h>

// A bad main() function because it doesn't have
// the return type. The standard one is int main()
// Next time learn how to indent your code, because
// I am too lazy to help you indent your code.
main()
{
   // Learn how to name your variable because
   // I couldn't understand what each variable
   // is used for. A nice comment for each
   // variable would be a very good habit
   int k=0;
   int co[20];
   int n=0;
   char c;
   char rep[256]={0};
   int i=0;

   // Waiting the user to press any key and
   // assign it to variable c. If the Enter
   // key were pressed, skip the loop.
   while((c=getchar())!='\n'){  

      if(c=='('){ // If the key was '(' then
         co[k]=1; // Assign 1 to co[k]
         k++; // Increase k by 1
         break; // Skip the whole loop and end the program
      }

      if(c==')') { // If the key was ')' then
         while(k>=0&&k<=20) { // Loop while k value is in 0-20
            co[k]=0; // Assign 0 to co[k]
            k--; // Decrease the k by 1
         }
         break; // Skip the whole loop and end the program
      }

      if(c==' ') { // If the input was a space
         break; // Skip the whole loop and end the program
      }  

      // If the input was *, -, /, s, c, t or C, then
      if(c=='*'||c=='-'||c=='+'||c=='/'||c=='s'
          ||c=='c'||c=='t'||c=='C'){
         // Assign 0 to co[k], which make this condtion 
         // always false
         if(co[k]=0) {
            rep[i+n]=c;
            i=i+n;
            n=1;
         // Because the first condition always false
         // It always proccess here
         } else {
            while(n>=1) { // Loop until n is smaller than 1
               // Assign the rep[i+n] with the rep[i+n-1]
               rep[i+n]=rep[i+n-1];
               n--; // Decrease n by 1
            }
            // If the rep[i] is sometimes besides s, c, t, and C
            if(rep[i]!='s'||'c'||'t'||'C') 
               rep[i]=c; // Assign the input to rep[i]
            n++; // Increase n by 1
         }
      }

      // If c is in between 48-57 (which is a number)
      if(c>47&&c<58) {
         while(n>=1) { // Loop until the n smaller than 1
            // Assign the rep[i+n] with the rep[i+n-1]
            rep[i+n]=rep[i+n-1]; 
            n--; // Decrease n by 1
         }
      // Missing close brace here
      rep[i]=c; // Assign the input to rep[i]
      i++; // Increase i by 1
   }
}