How do I write an algorithm to convert an equation from infix to prefix notation?
Q. How do I write an algorithm to convert an equation from infix to prefix notation?
Asked by Taz - Fri Sep 19 08:15:45 2008 - - 1 Answers - 0 Comments

A. Hi, take a look at this link: It's fully commented so it should be easy to figure it out from there :D
Answered by Bleak - Fri Sep 19 08:24:04 2008

Please tell me how do you convert infix expression to prefix expression in C++ programming?
Q. With and without using stacks.
Asked by RJ - Thu Mar 15 15:18:38 2007 - - 1 Answers - 0 Comments

A. The Shunting Yard Algorithm does this using a stack. I'm not sure how to do it without a stack.
Answered by Bill Lumbergh - Fri Mar 16 03:22:24 2007

How do you convert this postfix expression to infix ?
Q. How do you convert this postfix expression to infix ? Expression: A B + C + D E * F + \ (Please explain)
Asked by swat535 - Tue Mar 10 15:23:22 2009 - - 2 Answers - 0 Comments

A. Imagine a little stack. Read from left to right and put the values on the "stack" A --> A B --> A, B When you get to an operation, perform this operation on the two items on the top of the "stack" and make this the new top item. + --> (A + B) C --> (A + B), C + --> (A + B) + C D --> (A + B) + C, D E --> (A + B) + C, D, E * --> (A + B) + C, (D * E) F --> (A + B) + C, (D * E), F + --> (A + B) + C, (D + E) + F \ --> [(A + B) + C] / [(D + E) + F] Answer: [(A + B) + C] / [(D + E) + F]
Answered by Puzzling - Tue Mar 10 15:32:51 2009

convert infix to postfix expression and show trace of algorithm?
Q. Convert the infix expression (A - B) * C + D to postfix. Show the trace of the algorithm, i.e., the stack, the infix expression and postfix expression.
Asked by bc050400307 - Wed Aug 15 00:29:21 2007 - - 1 Answers - 0 Comments

A. a b - c * d + stack trace. [a-b] means a single element that represents the value of a-b. a a b a b - [a-b] [a-b] c [a-b] c * [(a-b)*c] [(a-b)*c] d [(a-b)*c] d + [(a-b)*c + d]
Answered by anonymous - Wed Aug 15 01:12:10 2007

How can i convert infix to prefix in a C++ code?
Q. How can i convert infix to prefix in a C++ code?
Asked by khak - Thu May 15 08:03:43 2008 - - 1 Answers - 0 Comments

A. You need to explain more for any help... It sounds like a class homework. How is the values and operators stored? I remember doing something like this with Parse trees. Look up Parse trees and the different notations on Google. Here is at least an explanation of each in different forms.
Answered by spudz000 - Thu May 15 08:29:08 2008

write a program in java using stack and array to convert infix expression to posfix expresion?
Q. comp guys plz help. i want exact code. thanks in advance
Asked by Zeratul - Tue Aug 25 14:13:49 2009 - - 3 Answers - 0 Comments

A. i know only reverse
Answered by Reema - Tue Aug 25 14:30:01 2009

What is the postfix form of the infix expression A*B+(C-D/F)?
Q. I got the answer to be ABC*+D-E/F*, i am confused when the scanning reaches the ..D/F.. PART
Asked by gejopal - Fri Mar 13 14:43:02 2009 - - 1 Answers - 0 Comments

How to convert an infix expression into a prefix expression using stack class in C++.?
Q. How to convert an infix expression into a prefix expression using stack class in C++.?
Asked by ADS - Thu Nov 2 16:42:51 2006 - - 2 Answers - 0 Comments

A. 1.21 Gigawatts is way too much. You'll destabilize the dilithium crystals. This is an interesting question, most likely a well thought out assignment for a class. So, I think you'll benefit from doing it yourself. I just wrote you a big long answer that suggested a solution, but I'm deleting it, because I think it might be right. Instead, I'll give you a hint: Can you use more than one stack? And should we apply operands to the top two values in a stack of numbers? Hold on, you want to convert to prefix, not evaluate. Okay, but I think what I have in mind still holds. Also, you might think about drawing a parse tree for the expression. It might help you to see what's going on.
Answered by arbeit - Thu Nov 2 20:21:08 2006

HOW CAN I CONVERT INFIX TO POSFIX EQUATION AND AFTER THAT GET THE ANSWER OF POSTFIX EQUATION USING TURBO C?
Q. using stack.
Asked by Dionalyn G - Thu Aug 28 23:09:29 2008 - - 1 Answers - 0 Comments

Could anyone please kindly check my answer (regarding the conversion from infix notation to postfix notation)?
Q. So: Infix Notation: A/(B+C*D)*E-F My answer (in Postfix Notation) : AB+CD*/EF*- Is my answer right? If not, can you please help me? Please only answer if you can really help me... Thanks a lot!
Asked by europhile - Sat Feb 9 03:36:53 2008 - - 1 Answers - 0 Comments

A. The layout of the infix statement does not make it clear if the factor E is in the numerator or the denominator.
Answered by openmath - Sat Feb 9 05:11:52 2008

Write a program that inputs infix expression the outputs postfix expression?
Q. please help
Asked by litol_twilky - Sun Aug 12 20:16:47 2007 - - 2 Answers - 0 Comments

A. in which language?
Answered by i_am_the_next_best_one - Mon Aug 13 05:56:36 2007

can someone give me a program in converting infix to post-fix using linked list?i need the real program?
Q. can someone give me a program in converting infix to post-fix using linked list?i need the real program?
Asked by pink angel - Wed Aug 27 03:49:36 2008 - - 1 Answers - 0 Comments

A. You need to use binary search tree not a link list for doing that.
Answered by Rina S - Wed Aug 27 05:21:26 2008

How to scan and store a character string of numerical infix expression e.g. 30 * ( 56 - 2 ) / 7?
Q. and there is one space between each operater and operand like between 30 and * ,* and ( by the way in c language
Asked by jasmine - Wed Mar 4 10:02:51 2009 - - 1 Answers - 0 Comments

A. include int main() { int bytes_read; int nbytes = 100; char *expression; printf("Enter an infix expression.\n"); expression = (char *) malloc (nbytes + 1); bytes_read = getline(&expression, &nbytes, stdin); if (bytes_read == -1) { perror("read error"); } else { printf("You entered %s\n", expression); } return 0; }
Answered by tyrannikal - Wed Mar 4 11:25:28 2009

Convert the following infix expressions into Polish notation?
Q. i. 1 + ( 2 * ( 3 + 4 ) ) ii. 1 + ( ( 2 * 3 ) + ( 4 * 5 ) )
Asked by Sajjad H - Mon Feb 11 23:57:07 2008 - - 1 Answers - 0 Comments

A. Polish notation is also known as prefix notation (the previous answer was given in postfix notation). So the operator goes in front of its operands, which eliminates the need for grouping (i.e. parentheses): i. 1 + (2 * (3 + 4)) = + 1 (2 * (3 + 4)) = + 1 * 2 (3 + 4) = + 1 * 2 + 3 4 ii. 1 + ( (2 * 3) + (4 * 5) ) = + 1 ( (2 * 3) + (4 * 5)) = + 1 ( + (2 * 3) (4 * 5)) = + 1 + (2 * 3) (4 * 5) = + 1 + * 2 3 * 4 5
Answered by Dread Naught - Tue Feb 12 00:13:07 2008

what is infix and postfix in c programming??
Q. what is infix and postfix in c?? pls. give me some example of it. pls. show me the source code (T_T) thnx..
Asked by Ron - Wed Sep 3 02:34:20 2008 - - 1 Answers - 0 Comments

A. Infix: int a = (7 + 3) / 2; That line declares an integer named "a" and sets it equal to 2. Nothing major here. The infix notations are the "+" and "/". They are in-fix because they are in the midst of the formula itself. It requires arguments on either side of it. You couldn't just say "43 +". There has to be something after the "+" or it makes no sense. We are used to this with everyday algebra. Postfix: int b = 5;//declares an integer and sets it to 5 int a = b++; //declares another integer, sets it equal to b. Then, it increments b by one. The ++ is postfix notation. A very common way to see this is in counting; something we do a lot of in programming. To count, we often increment a variable by one. In many languages you… [cont.]
Answered by Michael F - Wed Sep 3 03:41:52 2008

infix to post fix conversion we need ,wheather stack ,or queue, or structure or union?
Q. infix to post fix conversion we need ,wheather stack ,or queue, or structure or union?
Asked by mita k - Tue Jul 7 02:35:02 2009 - - 1 Answers - 0 Comments

A. complete source code is here at:
Answered by Hassan - Tue Jul 7 03:33:05 2009

Can 2 Reverse Polish Expression (PostFix ) Exist For A Single Infix Expression?
Q. I have a expression as (A+B) * [ C * ( D+E)+F] This can be converted to postfix as AB+ DE+C*F+* And also this one seems to be correct AB+ CDE+*F+* So can there be 2 expression for same infix expression or one is wrong in this case? Please answer as soon as possible, with correct explanation and if possible reference too.
Asked by Blackalchemy - Mon May 14 03:04:59 2007 - - 2 Answers - 0 Comments

A. Only trivially, as by reversing the order of commuting operands. Which is what has happened in the second expression.
Answered by rhsaunders - Mon May 14 03:11:15 2007

algorithm of converting infix notation to prefix notation?
Q. algorithm of converting infix notation to prefix notation?
Asked by claire10jagrcs - Thu Aug 17 04:57:13 2006 - - 2 Answers - 0 Comments

A. Usually you would use a stack or a recursive function. Try:
Answered by send_jim - Thu Aug 17 06:17:57 2006

how to convert infix to postfix?
Q. in turbo C programming, how do you convert infix to postfix???thx!!
Asked by dan_abogado - Sat Sep 6 00:28:53 2008 - - 1 Answers - 0 Comments

A. Heres the code: #include #include #include #define oper(x) (x=='+' || x=='-' || x=='*' || x=='/') char in[30], post[30], stack[30]; int top=-1; void push(char x) { stack[++top]=x; } char pop() { return stack[top--]; } int precedence(char c) { if (c=='+' || c=='-') return 1; if (c=='*' || c=='/') return 2; if (c=='(') return 3; } main() { char c; int l,i,j=0,st1[20],k,h,f,eva l,s,N; printf("Enter the infix expression : "); scanf("%s",&in); l=strlen(in); for(i=0;i<=l;i++) { if(oper(in[i])) { post[j++]=' '; while(precedence(in[i])

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Answered by Emperor Harsh, the great - Sat Sep 6 10:00:53 2008

How do you get a file name from a user in C?
Q. I have an assignment where I need to request a user for a file name and then read the characters from the file into a character array for converting an infix expression to postfix. However, I am not well versed in C's input/output syntax. Any help would be appreciated
Asked by Shelly - Sat Oct 24 20:03:26 2009 - - 1 Answers - 0 Comments

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