Application of Stacks (Infix to Postfix) - Part 8

Data Structures: Application of Stacks (Infix to Postfix) - Part 8
Topics discussed:
1) Writing a C program to evaluate the postfix expression.

Music:
Axol x Alex Skrindo - You [NCS Release]

#DataStructuresByNeso #DataStructures #Stacks #ApplicationOfStacks

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[AzreenRajieveMANANGGOLO]

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[bisalranjanswain]

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[shivajichate]

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[ankitxalxotoo]

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[-anuhyajanjam]

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[MDSagor-tuxw]

Does it work for multi digits infix expressions ? What will happen if I use a multi digit infix expression ?

[lokendrasinghrao]

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[galihtanu]

Cool sir. i hope can converted to arduino code.

[mohamedsamir]

For entering more than one digit number i made some modifications as:
void InToPost()
{
int i, j=0, x;
char symbol, next;
for(i=0; i<strlen(infix); i++)
{
symbol = infix[i];
switch(symbol)
{
case ' ':
case '\t':
break;
case '(':
push(symbol);
break;
case ')':
while( (next = pop()) != '(')
postfix[j++] = next;
break;
case '^':
case '*':
case '/':
case '+':
case '-':
while(top!=-1 && precedence(symbol) <= precedence(stack[top]))
postfix[j++] = pop();
push(symbol);
break;
default:
while(infix[i+1] >= '0' && infix[i+1] <= '9')
{
x = ((symbol-'0') * 10) + (infix[++i]-'0');
symbol = (x+'0');
}
postfix[j++] = symbol;
break;
}
}
while(top != -1)
postfix[j++] = pop();
postfix[j++] = '\0';
}

int precedence(char s)
{
switch(s)
{
case '^': return 3;
break;
case '*':
case '/': return 2;
break;
case '+':
case '-': return 1;
break;
default: return 0;
break;
}
}

int post_evaluation()
{
int i, a, b, n;
for(i=0; i<strlen(postfix); i++)
{
if(postfix[i]!='+' && postfix[i]!='-' && postfix[i]!='*'
&& postfix[i]!='/' && postfix[i]!='^')
push(postfix[i]-'0');
else
{
a = pop();
b = pop();
switch(postfix[i])
{
case '+': push(b+a);
break;
case '-': push(b-a);
break;
case '*': push(b*a);
break;
case '/': push(b/a);
break;
case '^': push(pow(b, a));
break;
}
}
}
return pop();
}

void print(char *s)
{
int i=0;
printf("Postfix expression: ");
while(s[i])
{
if(postfix[i]!='+' && postfix[i]!='-' && postfix[i]!='*'
&& postfix[i]!='/' && postfix[i]!='^')
printf("(%d)", (s[i]-'0'));
else
printf("%c", s[i]);
i++;
}
printf("\n");
}

I hope i done well.❤

[kycideas]

Sir tell the computer organisation course sir, please

[anonyanonymous]

Problem: :(
The algorithm will not work If the operand is of multiple digits.
Example:
Infix exp : (31+2)*2/3
Postfix exp: 312+2*3/
Array: |3| |1| |2| |+| |2| |*| |3| |/|
Now when we evaluate the above expression we will push 3 1 2
And the moment we encounter an operator (+) we will pop 2 and 1. And will perform 1+2 which yeilds 3 then we will push this into the stack now the stack contains
3 3

Push 2

3 3 2

Encounter *

Do 3*2 = 6

Push 6

3 6

Push 3

3 6 3

Encounter /

Do 6/3 = 2

Push 2

3 2

Scanning finishes
Now pop 2

So as per the algorithm the result of the above expression is 2 and that's obviously wrong.
The result must be 22