My program basically converts an infix expression to a postfix expression, although so far my program only accepts single digits. Anyway when I try to compile, right after inputting my infix expression, the program crashes almost immediately. My code:
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <stdlib.h>
int priority(char x); // Determines priority of incoming operator.
void push(char x); // Pushes element to stack.
char pop(); // Pops element from stack.
char stack[10];
int top = -1;
int main() {
char init[20];
printf("Enter an expression: ");
fgets(init, 20, stdin);
int x = 0, y, z = 0;
static char result[20];
while (init[x++] != '\0') {
if (isalnum(init[x]))
result[z++] = init[x]; // Operand printed out immediately.
else if (init[x] == '(')
push(init[x]); // '(' character pushed.
else if (init[x] == ')') {
while ((y = pop()) != '(')// Popping elements from stack until reaching '('
result[z++] = y;
} else if (init[x] == ' ') {
z++;
else {
while (priority(init[x]) <= priority(stack[top])) // If expression operator has higher precedence than stack operator, expression operator is pushed onto stack. Else stack operator is popped and printed out.
result[z++] = pop();
push(init[x]);
}
}
while (top != -1)
result[z++] = pop(); // Remaining operators printed out.
printf("Final expression is %s.\n", result);
}
int priority(char x) {
int precedence = 0;
if(x == '(')
precedence = 0;
if(x == '+' || x == '-')
precedence = 1;
if(x == '*' || x == '/')
precedence = 2;
if(x == '^')
precedence = 3;
return precedence;
}
void push(char x) {
stack[++top] = x;
}
char pop() {
return stack[top--];
}
I had a version of this that worked but when I look at this version, nothing seems to be any different. Can someone please tell me what I'm missing?
A major issue that I found is:
while (init[x++] != '\0')
While you increment the value of x in the condition check of loop, you again try to access it in call to function :
isalnum(init[x])
the first number is never evaluated this way. So if you enter "5+2", only "+2" will be evaluated, which is an invalid infix expression.
Related
The question is:
WAP to convert a given Prefix expression into its equivalent Postfix
expression and evaluate it using stack.
I have written this code, it takes the input but doesn't display the output.
//header files
#include<stdio.h>
#include<string.h>
//global variables
char stack[50];
int top = -1;
void push(char s)
{
stack[++top]=s;
}
//function to pop an element
char pop()
{
return stack[top--];
}
// funtion to check if the character is operator or not
int is_operator(char x)
{
switch (x)
{
case '+':
case '-':
case '/':
case '*':
return 1;
}
return 0;
}
//function to Convert prefix to Postfix
void convert()
{
int i,l;
char op1,op2,tmp;
char exp[50];
printf("Enter the prefix expression: ");
gets(exp);
//length of expression
l = strlen(exp);
//scanning from right to left
for(i = l - 1; i >= 0; i--)
{
//checking if the symbol is an operator
if (is_operator(exp[i]))
{
//popping two operands from stack
op1 = stack[top];
pop();
op2 = stack[top];
pop();
//concating the operands and operator
tmp = op1 + op2 + exp[i];
//Pushing the temporary string to stack
push(tmp);
}
//if it is an operand
else
{
//push the operand to the stack
push((exp[i]));
}
}
//printf("The postfix expression is: %s",stack[top].c_str());
printf("%s ",stack[top]);
}
//main function
int main()
{
convert();
return 0;
}
when i run the code it takes the prefix expression but immediately after that it crashes and returns to the terminal?
Can anyone please help me with the code i am having difficulty in understanding what mistake i have made.
Also if possible give a little explanation what mistake i did in the code.
In this algorithm, a stack of strings is required. But you are using an array of char.
The exp[i] returns a char. So just can not expect tmp = op1 + op2 + exp[i] to concat them into a string in c.
In c strcpy() can be used to copy a string and strcat() can be used to concat strings.
There are several ways to convert a char to a string of one char tailing NULL. In this code (char[2]){(char)exp[i], '\0'} is used to get a string containing exp[i] and NULL char.
In the solution code there is a memory limitation. The input expression should be in 50 chars (including '\0') .
Solution code :
#include<stdio.h>
#include<string.h>
char stack[50][50];
int top = -1;
void clear_stack() {
top = -1 ;
}
void push(char *s)
{
strcpy(stack[++top], s) ;
}
char* pop()
{
return stack[top--];
}
int is_operator(char x)
{
if(x == '+' ||x == '-'||x == '*'||x == '/'){
return 1;
}
else{
return 0;
}
}
//function to Convert prefix to Postfix
void convert(char *exp)
{
clear_stack() ;
int i,l;
char op1[50],op2[50];
l = strlen(exp);
//scanning from right to left
for(i = l - 1; i >= 0; i--)
{
//checking if the symbol is an operator
if (is_operator(exp[i]))
{
//popping two operands from stack
strcpy(op1, pop()) ;
strcpy(op2, pop()) ;;
//concating the operands and operator
strcat(op1 , strcat(op2 , (char[2]) {(char)exp[i], '\0'})) ;
//Pushing the temporary string to stack
push(op1);
}
//if it is an operand
else
{
//push the operand to the stack
push((char[2]){(char)exp[i], '\0'});
}
}
//printf("The postfix expression is: %s",stack[top].c_str());
printf("%s\n",stack[top]);
}
//main function
int main()
{
convert("*-A/BC-/AKL");
convert("+ab");
convert("*+abc");
convert("*a+bc");
convert("+/ab/cd");
convert("*+ab+cd");
convert("-*+abcd");
return 0;
}
Output :
ABC/-AK/L-*
ab+
ab+c*
abc+*
ab/cd/+
ab+cd+*
ab+c*d-
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.
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 new to the C programming language. I am creating a simple calculator program, but my function for some reason isn't returning the right result. Here's my program:
#include <stdio.h>
#include <math.h>
#include <complex.h>
#include <string.h>
int calculator(int x, char operator, int y);
int main()
{
int x;
char operator;
int y;
printf("Enter an arithmetic experession: ");
scanf("%d%s%d", &x, &operator, &y);
int result = calculator(x, operator, y);
if(result == -1)
{
printf("Error! Try again!");
}
else
{
printf("%d", result);
}
return 0;
}
int calculator(int x, char operator, int y)
{
int result = 0;
if(y = 0)
{
return -1;
}
if(operator == '+')
{
result = x + y;
}
else if(operator == '-')
{
result = x - y;
}
else if(operator == '*')
{
result = x * y;
}
else if(operator == '/')
{
result = x / y;
}
else if(operator == '%')
{
result = x % y;
}
else
{
return -1;
}
return result;
}
So when I run this program it asks for an arithmetic expression. If I put in 5 + 3, it only returns 5! After a few tests it seems as though it only returns the first operand no matter what. I imagine it's something very minor, but I don't see what I'm missing.
Two problems. The first is where you're reading in the expression:
scanf("%d%s%d", &x, &operator, &y);
operator is a char but you're using the %s format specifier which is for a string. This ends up putting the operator in operator, but then a NULL terminator is added to the next byte in memory. Since this byte is not part of operator, this causes undefined behavior.
You want to use %c instead to read a single character.
The second is where you're doing the zero check on y:
if(y = 0)
This is an assignment, not a comparison. The value 0 is assigned to y, then y is evaluated as a boolean, which ends up being false. This is why all of your expressions evaluate as if y is 0.
Change this to an assignment:
if(y == 0)