infix to postfix expression string for evalution.

[KidLogik]

Hello!

I am converting an infix expression string into a postfix so that I
will be able to evaluate it easier ->

(5*(((9+8)*(4*6))+7)) == 598+46**7+*

I believe the rule is "Replace all occurances of two operands followed
by an operator by their infix equivalent"

This works fine like so ->

(2*2) == (22*)

but what happens if I need something bigger then 9?

(10*2) == (102*) ->

which would multiple the 0 and the 2 instead of 10*2...
Am I making any sense? Please help me.

Thanks.

 

[John Harrison]

Hello!

I am converting an infix expression string into a postfix so that I
will be able to evaluate it easier ->

(5*(((9+8)*(4*6))+7)) == 598+46**7+*

I believe the rule is "Replace all occurances of two operands followed
by an operator by their infix equivalent"

This works fine like so ->

(2*2) == (22*)

but what happens if I need something bigger then 9?

(10*2) == (102*) ->

which would multiple the 0 and the 2 instead of 10*2...
Am I making any sense? Please help me.

Thanks.

I think you need to add a space between the 10 and the 2.

John

 

[Stewart Gordon]

While it was 2/2/04 9:39 am throughout the UK, KidLogik sprinkled little
black dots on a white screen, and they fell thus:

Hello!

I am converting an infix expression string into a postfix so that I
will be able to evaluate it easier ->

but what happens if I need something bigger then 9?

(10*2) == (102*) ->

<snip>

Surely you should be building the postfixed expression in a binary form?
For example:

struct RPNNode {
enum { NUM, PLUS, MINUS, TIMES, DIVIDE } op;
int value;
};

That way, you won't have any such problem.

Stewart.

 

[David Rubin]

I am converting an infix expression string into a postfix so that I
will be able to evaluate it easier ->

(5*(((9+8)*(4*6))+7)) == 598+46**7+*

I believe the rule is "Replace all occurances of two operands followed
by an operator by their infix equivalent"

Firstly, there is more than one rule! When I wrote this program, I found
that dealing with parentheses and operator precedence was not
straightforward; it took a bit of fiddling around. In any case,
generally speaking, you need to use a stack and a parser...

[snip]

but what happens if I need something bigger then 9?

(10*2) == (102*) ->

which would multiple the 0 and the 2 instead of 10*2...

The parser comes in handy for solving this problem. You can basically
parse the input string as characters using the following loop:

char *p = buf;
while(*p != '\n'){
if(isdigit((int)*p)){
/* operands */
}else
if(strchr("+-*/%^()", *p) == 0){
/* whitespace */
p++;
}else{
/* operators */
}
}

Note that my program does not evaluate the expression (which you seem to
imply as your goal), but rather converts between two strings: infix to
postfix.

HTH,

/david

 

[Thomas Matthews]

KidLogik wrote:

I am converting an infix expression string into a postfix so that I
will be able to evaluate it easier ->

(5*(((9+8)*(4*6))+7)) == 598+46**7+*

I believe the rule is "Replace all occurances of two operands followed
by an operator by their infix equivalent"

Firstly, there is more than one rule! When I wrote this program, I found
that dealing with parentheses and operator precedence was not
straightforward; it took a bit of fiddling around. In any case,
generally speaking, you need to use a stack and a parser...

[snip]

but what happens if I need something bigger then 9?

(10*2) == (102*) ->

which would multiple the 0 and the 2 instead of 10*2...

The parser comes in handy for solving this problem. You can basically
parse the input string as characters using the following loop:

char *p = buf;
while(*p != '\n'){
if(isdigit((int)*p)){
/* operands */
}else
if(strchr("+-*/%^()", *p) == 0){
/* whitespace */
p++;
}else{
/* operators */
}
}

Note that my program does not evaluate the expression (which you seem to
imply as your goal), but rather converts between two strings: infix to
postfix.

HTH,

/david

If one throws the operators and numbers into a binary tree, then
conversion is just a matter of how the tree is traversed. I wrote
the program so long ago, I don't remember how I handled the parens.

--
Thomas Matthews

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http://www.slack.net/~shiva/welcome.txt
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http://www.raos.demon.uk/acllc-c++/faq.html
Other sites:
http://www.josuttis.com -- C++ STL Library book

 

[David Rubin]

Thomas Matthews wrote:

[snip - infix to postfix]

Firstly, there is more than one rule! When I wrote this program, I found
that dealing with parentheses and operator precedence was not
straightforward; it took a bit of fiddling around. In any case,
generally speaking, you need to use a stack and a parser...

[snip]
If one throws the operators and numbers into a binary tree, then
conversion is just a matter of how the tree is traversed. I wrote
the program so long ago, I don't remember how I handled the parens.

I found the stack-based algorithm in the project notes from some random
college intro CS course on the web. I usually just browse college course
web pages when I'm looking for a short "keep your skills sharp" project.
The stack algorithm explicitly described how to handle parens, but it
was apparantly lacking a few details.

/david