My code runs well to my surprise!!
The only problem is that whenever I use a bracketed infix input it comes out a 'J' at the end of postfix expression !! Any suggestions??
Here the algorithm is the basic one all expressions are getting converted and all is right but the tailing 'J' is i just cann't understand !! Suggestions??
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<ctype.h>
#define MAX 50
char stack[MAX];
int top = -1;
void push(char element)
{
stack[++top] = element;
}
char pop()
{
return(stack[top--]);
}
char tope()
{
return(stack[top]);
}
int prec(char c)
{
switch(c){
case '+':
case '-' : return 1;
break;
case '*' :
case '/' : return 2;
break;
default:
return 0;
break;
}
}
int main()
{
char post[MAX],in[MAX],ch,element;
printf("Infix expression : ");
scanf("%s",in);
int i=0,k=0;
in[strlen(in)] = ')';
push('(');
while((ch = in[i++]) != '\0')
{
if(isalnum(ch))
post[k++] = ch;
if(ch == '(')
push(ch);
if(ch == ')')
{
while(tope() != '(')
{
post[k++] = pop();
}
pop();
}
if(ch == '+' || ch =='-' || ch == '*' || ch == '/')
{
while(prec(ch) <= prec(tope()))
{
post[k++] = pop();
}
push(ch);
}
}
post[k] = '\0';
printf("%s",post);
return 0;
}
in[strlen(in)] = ')';
overwrites the nul-terminating character, which explains strange chars when printing (printing stops only when meeting another nul char by luck: undefined behaviour which may even lead to a crash if no nul char is found in the in 50-byte buffer)
You have to shift it, for instance like this:
int l = strlen(in);
in[l] = ')';
in[l+1] = '\0';
Note: you have to store the length of your string in l instead of calling strlen(in) twice not only because of the performance loss but, because putting the parenthesis means that strlen doesn't work properly until you null-terminate.
(also you probably want to protect your scanf like this: scanf("%48s",in); so you're sure to have enough room for the extra parenthesis for your 50-size buffer, even compatible with your macro definition, see scanf: template with macro (#define constant) inside)
Related
What I'm trying to obtain is a calculator that will take infix notation, ignore insignificant whitespace characters like " " or '#', then convert that infix notaion into postfix notation and do simple calculations like addition, subtraction etc. So far the code is taking input in infix notation trimming it in a way that ignores insignificant whitespace characters and outputs the postfix notation.
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>;
#include <ctype.h>;
#define MAX_LENGTH 100
//Functions
void push(char x);
char pop();
void trimString(char string[], char newString[]);
void inputToRPN(char trimmedExp[], char rpnExp[]);
int calculateRPN(char rpnExp[]);
char stack[MAX_LENGTH];
char resStack[MAX_LENGTH];
int top = -1;
int resTop = -1;
int index = 0;
int main() {
int res;
char exp[MAX_LENGTH] = "10 +2";
char trimmedExpression[MAX_LENGTH];
char rpnExpression[MAX_LENGTH];
// Input commented out as per suggestion in comments
//printf("Enter expression : ");
//fgets(exp, 100, stdin);
printf("Infix expression: %s \n", exp);
trimString(exp, trimmedExpression);
printf("\n");
inputToRPN(trimmedExpression, rpnExpression);
res = calculateRPN(rpnExpression);
//printf("Result of calculation: %d", res);
return 0;
}
void push(char x) {
stack[++top] = x;
}
char pop() {
if (top == -1)
return -1;
else
return stack[top--];
}
int priority(char x) {
if (x == '(')
return 0;
if (x == '+' || x == '-')
return 1;
if (x == '*' || x == '/')
return 2;
return 0;
}
void trimString(char string[], char newString[]) {
int i = 0, j = 0;
while (string[i] != '\0' && string[i] != 10) {
// Range of significant characters
if (string[i] >= '(' && string[i] <= '9') {
newString[j] = string[i];
i++, j++;
}
else {
i++;
}
}
newString[j] = 0;
}
void inputToRPN(char trimmedExp[], char rpnExp[]) {
char* e, x;
e = trimmedExp;
while (*e != '\0') {
// Add to RPN if character is alphanumeric
if (isalnum(*e)) {
rpnExp[index] = *e;
index++;
}
// Add to stack if is an open brace
else if (*e == '(')
push(*e);
// Add all operators to the expression until finding open braces
else if (*e == ')') {
while ((x = pop()) != '(') {
rpnExp[index] = x;
index++;
}
}
// If none of the above, that is an operator - check it's priority.
// If it's priority is less that that of the one on top of the stack add the operator from the top of the stack to the expression; untill it's priority is higher.
// At the end add current operator to the stack.
else {
while (priority(stack[top]) >= priority(*e)) {
rpnExp[index] = pop();
index++;
}
push(*e);
}
e++;
}
while (top != -1) {
rpnExp[index] = pop();
index++;
}
// Terminating character at the end of the string
rpnExp[index] = 0;
}
void pushRes(char x) {
printf("pushing: %c \n", x);
resStack[++resTop] = x;
}
char popRes() {
printf("poping \n");
if (resTop == -1)
return -1;
else
return resStack[resTop--];
}
int isValidOperator(char c) {
if (c == '/' || c == '*' || c == '+' || c == '-')
return 1;
else
return 0;
}
int calculateRPN(char rpnExp[]) {
// Doesnt do anything yet, just prints out the compiled reverse polish notation
char* c;
int result = 0;
c = rpnExp;
printf("Postfix expression: %s", rpnExp);
return result;
}
The problem I've stumbled upon is when the infix input has multiple digits say 10+2 the code will treat each digit individually. Therefore the whole expression will be invalid when calculating result. I'm almost certain the issue lies in this line of code:
// Add to RPN if character is alphanumeric
if (isalnum(*e)) {
rpnExp[index] = *e;
index++;
}
Despite that I've got no idea how should i treat multiple digits while adding them to the expression, since the input is in form of character and there can be N amount of digits that have coresponding ascii values which range from 0-9. Looking forward to your answears.
Edit: made it so the code compiles and the input is hard coded.
Okay, so thanks to Bodos suggestions I've fixed the issue. Adding one while loop in this section:
if (isalnum(*e)) {
rpnExp[index] = *e;
index++;
}
enabled me to add one character after every number (including the N digit ones).
Thanks to which I was later able to perform calculations in calculateRPN function that would eventually lead to correct answear.
The issue has been resolved.
I am trying to implement an infix-to-postfix conversion program. On executing the code I get to see only alpha-numeric characters. Arithmetic operators are not printing. Upon debugging, I found the operators are not inserting in the stack. I couldn't find out the reason behind this.
Any help is appreciated.
#include<stdio.h>
#include <ctype.h>
#define MAX 50
char st[MAX];
int top = -1;
void push(char);
char pop();
int priority(char);
int main()
{
int i=0;
char x,s[50];
printf("Enter infix expression: ");
gets(s);
while(s[i]!='\0')
{
if(isalnum(s[i]))
printf("%c",s[i]);
else if(s[i] == '(')
push(s[i]);
else if(s[i] == ')')
{
while((x=pop())!='(')
printf("%c",x);
}
else
{
while(priority(st[top])>=priority(s[i]))
printf("%c",pop());
push(st[i]);
}
i++;
}
while(top!=-1)
printf("%c",pop());
}
void push(char x)
{
st[++top] = x;
}
char pop()
{
if(top == -1)
return -1;
else
return (st[top--]);
}
int priority(char x)
{
if(x == '(')
return 0;
if(x == '+' || x == '-')
return 1;
if(x == '*' || x == '/' || x == '%')
return 2;
}
As you correctly detected on your debugging session, you don't see operators in your postfix expression because you never push() them to your stack.
In fact
in the first if you check for alphanumeric characters
in the following else if you check for opening parenthesis
in the second else if you check for closing parenthesis
in the final else you manage pops from stack... but you didn't push anything! (1)
What you need to fix is the last else, where you have at least two distinct problems:
You access st[top] without checking top value. You need to manage the case in which top = -1, that would lead to out of bounds access of the stack array and to undefined behavior. I think that in that scenario you just need to push the operator
You push in the stack st[i]. Probably you meant s[i]
In this way the while analysing the expression becomes
while(s[i]!='\0')
{
if(isalnum(s[i]))
printf("%c ",s[i]);
else if(s[i] == '(' )
push(s[i]);
else if(s[i] == ')')
{
while((x=pop())!='(')
printf("%c ",x);
}
else
{
if( top != -1 )
{
while(priority(st[top])>=priority(s[i]))
printf("%c ",pop());
}
push(s[i]);
}
i++;
}
Input:
4*6+3
Output:
4 6 * 3 +
(I added a further space after each %c in printfs in order to improve output readability).
Note: You still have to fix some problems in operators priority management.
enter code here,
#include <stdio.h>
#include <ctype.h>
#define SIZE 50
char s[SIZE];
int top=-1;
void push(char elem)
{
s[++top]=elem;
}
char pop()
{
return(s[top--]);
}
int pr(char elem)
{
switch(elem)
{
case '#': return 0;
case '(': return 1;
case '+':
case '-': return 2;
case '*':
case '/': return 3;
}
}
void main()
{
char infx[50],pofx[50],ch,elem;
int i=0,k=0,j;
for(j=0;j<=3;j++)
{
printf("\n\nRead the Infix Expression ? ");
scanf("%s",infx);
push('#');
while( (ch=infx[i++]) != '\0')
{
if( ch == '(')
push(ch);
else
if(isalnum(ch))
pofx[k++]=ch;
else
if( ch == ')')
{
while( s[top] != '(')
pofx[k++]=pop();
elem=pop();
}
else
{
while( pr(s[top]) >= pr(ch) )
pofx[k++]=pop();
push(ch);
}
}
while( s[top] != '#')
pofx[k++]=pop();
pofx[k]='\0';
printf("\n\nGiven Infix Expn: %s Postfix Expn: %s\n",infx,pofx);
}
}
When looping through infix to post fix expression,it's not looping beyond 1st loop.
It is producing output of 1st loop and showing segmentation fault for rest of loops
Please can somebody guide me?
Consider an example
given input,
Given Infix Expn: a+b Postfix Expn: ab+
Read the Infix Expression ? a-b
Segmentation fault (core dumped)
you need to reset the index for pofx and infx for every new input. i.e to reset i and k in the for-loop. Fix as below,
void main()
{
char infx[50],pofx[50],ch,elem;
int i=0,k=0,j;
for(j=0;j<=3;j++)
{
i=0;j=0;//here is the update
//rest of your code follows,
}
}
I am stuck on this problem:
A user writes a math expression and the program need to extract from it all subexpressions that are inside ( ) ex. (x+2) or (x+(y-2)).
For example, if the user enters 2-(5+(x-6)-(y+9)), the program should return (x-6), (y+9), (5+(x-6)-(y+9))
This is what I've tried to do.
#include<stdio.h>
int main()
{
char a[100];
int i=0,t,j=0;
printf("enter math expression:\n");
while( (a[i++]=getchar()) != '\n' && i < 100);
a[i] = '\0';
for (i=0; a[i]!='\0'; i++)
{
if (a[i]=='(')
{ printf("found (\n");
j++;
while (a[i] !=')')
printf("%c",a[i++]);
printf("%c",a[i]);
Since you are dealing with nested expressions, you need to keep a stack around in order to match parentheses. Ideally inside the loop you should:
Whenever found a '(', push position within the string into the stack
When a ')' is found, then pop from the stack the position of the matching '('. You have start-end indexes for your expression.
Continue until string is finished
Example (x + (y+2)):
i == 0 -> '(' -> s.push(0);
i == 1 -> 'x'
i == 2 -> '+'
i == 3 -> '(' -> s.push(3);
i == 4 -> 'y'
i == 5 -> '+'
i == 6 -> '2'
i == 7 -> ')' -> s.pop() [3, 7] contains '(y + 2)'
i == 8 -> ')' -> s.pop() [0, 8] contains '(x + (y+2))'
Let, s be a cstring in this context.
Since in valid mathematical expression parenthesis are balanced, we can observe that,
s+1 is balanced if s[0] is not '('
if s[0] is '(' then we are at the start of a parenthesis'ed expression.
At the case of #1 we do not care about the first character. So, we can set s = s+1 and start over again.
At the case of #2 we have to find the matched end for the first character. So, we divide it in two part. The matched expression and the tail. Both are valid expression. So, we start with them over again.
int main()
{
char s[] = "2-(5+(x-6)-(y+9))";// s have to be modifiable.
extractAndPrint(s, 0); // we are at start and 0 indicates balanced
return 0;
}
Now,
void extractAndPrint(char *s, int found)
{
if(s[0] == 0) return;//no more honey
if(!found){ //so far balanced
if(s[0] == '(') extractAndPrint(s+1, 1);// case 2
else extractAndPrint(s+1, 0); //this is the case 1
} else {//start of a expression
char *t;
//separates the end of current expression
//and the remaining part.
mark_end(s, 0, &t);
printf("(%s)\n", s);
extractAndPrint(s, 0);//potentially contain more expression
extractAndPrint(t, 0);//check past the previous exp
}
}
I liked recursion here too. It takes less thinking for me. Can be implemented in more elegant way.
void mark_end(char *s, int c, char **t)
{
if(c == 0){
if(s[0] == ')'){
*t = s+1;
*s = '\0';
} else if(s[0] == '('){
mark_end(s+1, c+1, t);
} else {
mark_end(s+1, c, t);
}
} else {
if(s[0] == ')'){
mark_end(s+1, c-1, t);
} else if(s[0] == '('){
mark_end(s+1, c+1, t);
} else {
mark_end(s+1, c, t);
}
}
}
OUTPUT:
(5+(x-6)-(y+9))
(x-6)
(y+9)
I recommend you to use state machine, in your case this state machine should be suitable :
so your main will be a kind of a switch case, take care of writing your pop and push fuctions and represent your heap with a good structure
You should use recursion to parse the string as a tree of parenthesis. Here is an example showing how to do it :
#include <stdio.h>
#include <string.h>
#define MAX_DATA 100
int mathexp(const char *exp, int start, int end)
{
int i = start;
while(i < (start + end) && exp[i] != ')') {
if(exp[i] == '(') {
i = mathexp(exp, i + 1, end) + 1;
} else {
i++;
}
}
char subexp[i - start];
memcpy(subexp, &exp[start], i - start);
subexp[i - start] = '\0';
printf("%s\n", subexp);
return i;
}
int main(int argc, char *argv[])
{
char input[MAX_DATA];
printf("Enter a math expression: ");
fgets(input, sizeof(input), stdin);
input[strcspn(input, "\r\n")] = '\0'; // remove trailing \r\n
mathexp(input, 0, strlen(input));
return 0;
}
and when you test it, you have :
Enter a math expression: 2-(5+(x-6)-(y+9))
x-6
y+9
5+(x-6)-(y+9)
2-(5+(x-6)-(y+9))
I am reading over the K&R book, and am a little stuck.
What is wrong with the following?
void getInput(int* output) {
int c, i;
for(i=0; (c = getchar()) != '\n'; i++)
output[i] = c; // printf("%c", c) prints the c value as expected
output[++i] = '\0';
}
When I run the program it never gets out of the loop and I have to Ctrl+C to exit. However if I replace the fifth line with printf("%c", c);, it prints out all the input just fine after hitting enter and creating the new line.
What is wrong with the following?
1. void getInput(int* output) {
Why is the input argument an int* when what you want to store in an array of characters?
Probably
void getInput(char* output) {
is better.
Also, how do you know that the output pointer is pointing somewhere where you hold enough memory to write the user's input? Maybe you must have the maximum buffer length as an extra parameter to avoid buffer overflow errors as PW pointed out.
5. output[++i] = '\0';
i has already been incremented an extra time inside the for loop, so you can just do:
output[i] = '\0';
Other than these, the program runs fine and outputs what we input until return.
FWIW, I tested it by calling it like so:
int main(void)
{
char o[100];
getInput(o);
printf("%s", o);
return 0;
}
It looks correct to me as written except that you don't need to increment i outside of the loop. The i gets incremented right before the loop exits, thus it is already where you want it.
Make sure that a '\n' is actually making it into c.
Sometimes an '\n' will get thrown away as a delimiter.
your last code as posted have 3 errors I can see:
char* userInput[MAX_INPUT_SIZE];
Should be:
char userInput[MAX_INPUT_SIZE+1];
(this was already mentioned by Pax Diablo)
getInput(&userInput);
Should be:
getInput( userInput );
This last error means you passed to getInput an address inside your call stack. you have a memory overwrite. probably one of your calls to getchar() returnes to the wrong address.
a simple way to risk buffer overflow, because output's size is never passed/checked
Have you tried using a debugger? You should step through the code in gdb or visual studio or whatever you are using and see what is going on. You said you were a beginner so maybe you hadn't considered that yet - this is a pretty normal debugging technique to use.
Here is the complete program with a couple of updates from your input, but it still won't make it out of the loop. BTW this was exercise 1-24 on pg 34
#include <stdio.h>
#define STACK_SIZE 50
#define MAX_INPUT_SIZE 1000
#define FALSE 0
#define TRUE 1
void getInput();
int validInput();
int main() {
char* userInput[MAX_INPUT_SIZE];
getInput(&userInput);
if (validInput(&userInput) == TRUE)
printf("Compile complete");
else
printf("Error");
}
// Functions
void getInput(char* output) {
int c, i;
for(i=0; (c = getchar()) != '\n' && c != EOF && i <= MAX_INPUT_SIZE; i++)
output[i] = c;
output[i] = '\0';
}
int validInput(char* input) {
char stack[STACK_SIZE];
int c;
int j;
for (j=0; (c = input[j]) != '\0'; ) {
switch(c){
case '[': case '(': case '{':
stack[j++] = c;
break;
case ']': case ')': case '}':
if (c == ']' && stack[j] != '[')
return FALSE;
else if (c == '}' && stack[j] != '{')
return FALSE;
else if (c == ')' && stack[j] != '(')
return FALSE;
// decrement the stack's index
--j;
break;
}
}
return TRUE;
}
Here is the final working code. I must say I picked up quite a bit from doing this one. Thanks for the help and pointers.
Any suggestions on how things could be done better?
#include <stdio.h>
#define STACK_SIZE 50
#define MAX_INPUT_SIZE 1000
#define FALSE 0
#define TRUE !FALSE
void get_input();
int valid_input();
int main() {
char user_input[MAX_INPUT_SIZE + 1]; // +1 for the \0
get_input(user_input);
if (valid_input(user_input))
printf("Success\n");
else
printf("Error\n");
}
// Functions
void get_input(char* output) {
int c, i;
for(i=0; (c = getchar()) != '\n' && c != EOF && i <= MAX_INPUT_SIZE; i++)
output[i] = c;
output[i] = '\0';
}
int valid_input(char* input) {
char stack[STACK_SIZE];
char c;
int i = 0;
int stack_index = -1;
while ((c = input[i]) != '\0' && i < STACK_SIZE) {
switch(c){
case '[': case '(': case '{':
stack_index++;
stack[stack_index] = c;
break;
case ']': case ')': case '}':
if ((c == ']' && stack[stack_index] != '[') ||
(c == '}' && stack[stack_index] != '{') ||
(c == ')' && stack[stack_index] != '('))
return FALSE;
// decrement the stack's index now that the closing bracket is found
stack_index--;
break;
}
i++;
}
// stack index should be back where it started
return (stack_index == -1);
}