Programming Forums > Application Development > C++

mikaoj · May 27th, 2006, 5:15 PM

It will check a token on the symbol table, and then next token, if this token is legal for what the previous token, it will look at next token. If it's not legal, it will report an error.

The tokens will be added straight forward, so: I am a boy
Will be just added:
I
am
a
boy

programmingnoob · May 27th, 2006, 5:27 PM

Quote:

Originally Posted by mikaoj

It will check a token on the symbol table, and then next token, if this token is legal for what the previous token, it will look at next token. If it's not legal, it will report an error.

The tokens will be added straight forward, so: I am a boy
Will be just added:
I
am
a
boy

suppose here is the text:
I am a boy
the scanner will break it into I, am, a, boy tokens, right? but if you take a look at the link i copied pasted .. it says the parser does not start working until the scanner is finished working .... and after the scanner is finished working, you would have 4 tokens in the symbol table, right.... but how is the the parser supposed to know that they were received in the order of I am a boy... not in the order of am a boy I? do you see my doubt?
I suspect I have a huge misunderstanding about either parser or scanner, coz this doesnt seem to be working in my head

mikaoj · May 27th, 2006, 5:34 PM

The scanner will break the source down to tokens, right.
While it's scanning, when one token is assembled, it will add it to the table, so it vill assemble next token, and so on.

I have sent you a PM

programmingnoob · May 27th, 2006, 5:44 PM

Quote:

Originally Posted by mikaoj

The scanner will break the source down to tokens, right.
While it's scanning, when one token is assembled, it will add it to the table, so it vill assemble next token, and so on.

I have sent you a PM

check your pm

programmingnoob · May 27th, 2006, 8:31 PM

Quote:

Originally Posted by The Dark

If the assignment is to write a parse, I am not sure that getting a program to generate one for you would be good enough.

You pretty much seem on track, programmingnoob. The parser takes tokens from the scanner and then processes them to see if there is a valid expression. IMO "Recursive Descent Parsers" are the easiest to write.

is the scanner supposed to preserve the order of the tokens?

Harakim · May 27th, 2006, 10:02 PM

If you don't mind me asking, how do you know what you are compiling from and what you are compiling to? Are you making a general purpose compiler?

The Dark · May 28th, 2006, 9:06 AM

Quote:

Originally Posted by programmingnoob

is the scanner supposed to preserve the order of the tokens?

Yes, otherwise the parser won't be able to make any sense of the symbols.

programmingnoob · May 28th, 2006, 10:43 PM

Quote:

Originally Posted by The Dark

Yes, otherwise the parser won't be able to make any sense of the symbols.

exactly

but what if you pass the token to parser as soon as it is generated by the scanner?
here's the pseudo-code given by my professor, which doesnt make much sense to me except they scan the text stream only when it's needed

the code is for the function for the rule: <id-list> ::= $id | <id-list> , $id

(Toggle Plain Text)

if NT->attr =! id, then return NULL.
i = cons (NULL, NT, NULL); 
NT = Sc.get(); 
while (NT->attr == ,) do
j = NT
NT = Sc.get();
if NT->attr =!id, then return Null. 
cons (i, j, cons (NULL) NT, NULL);
NT->attr =! id, then return NULL
cons (i, j, cons (NULL, NT, NULL);
NT = Sc.get();
return (i);

NT means next token
if you have any idea what i = cons... etc mean, please let me know

programmingnoob · Jul 11th, 2006, 3:10 AM

I'm having a hard time coding this particular rule for my yes/no parser ...
<proc/stmt> ::= <compound> ; | procedure <proc> <proc/stmt>

Note: NT means Next Token

(Toggle Plain Text)

item<string>*     procstmt()
	 {
		  if (NT->theStruct.type != procedure)
		 {
		     if(compound()==NULL) return NULL; 
 
			 if (NT->theStruct.type != scln){  return NULL;}
			 NT = Si.cget();
		 }
		 else{ NT = Si.cget();
			y = proc();
			if (y != NULL) {
			while (y != NULL )  
			{if (NT->theStruct.type != procedure) 
			{if(compound() == NULL) return NULL;
			if (NT->theStruct.type != scln){ return NULL;}
			else return NT;}
			else {NT = Si.cget();
			y = proc(); }}}
			else return NULL; 
		 }
	 }

i know i didnt make myself very clear .... it's really late, so i am gonna hit the sack ... i'll get up tomorrow and post more details if needed

The Dark · Jul 11th, 2006, 3:19 AM

Try something like:

(Toggle Plain Text)

tem<string>*     procstmt()
	 {
		  if (NT->theStruct.type != procedure)
		 {
		     if(compound()==NULL) return NULL; 
 
			 if (NT->theStruct.type != scln){  return NULL;}
			 NT = Si.cget();
		 }
		 else{ NT = Si.cget();
			y = proc();
			if (y != NULL) {
                          if (procstmt() == NULL)
                            return NULL;
                        }
			else return NULL; 
		 }
                 // Return something not NULL
	 }

May 27th, 2006, 5:15 PM	#21
mikaoj Programmer Join Date: Aug 2005 Location: Norway Posts: 56 Rep Power: 0	It will check a token on the symbol table, and then next token, if this token is legal for what the previous token, it will look at next token. If it's not legal, it will report an error. The tokens will be added straight forward, so: I am a boy Will be just added: I am a boy __________________ Heh.

May 27th, 2006, 5:34 PM	#23
mikaoj Programmer Join Date: Aug 2005 Location: Norway Posts: 56 Rep Power: 0	The scanner will break the source down to tokens, right. While it's scanning, when one token is assembled, it will add it to the table, so it vill assemble next token, and so on. I have sent you a PM __________________ Heh.

Jul 11th, 2006, 3:10 AM	#29
programmingnoob Hobbyist Programmer Join Date: Feb 2006 Posts: 155 Rep Power: 3	I'm having a hard time coding this particular rule for my yes/no parser ... <proc/stmt> ::= <compound> ; \| procedure <proc> <proc/stmt> Note: NT means Next Token (Toggle Plain Text) item<string>* procstmt() { if (NT->theStruct.type != procedure) { if(compound()==NULL) return NULL; if (NT->theStruct.type != scln){ return NULL;} NT = Si.cget(); } else{ NT = Si.cget(); y = proc(); if (y != NULL) { while (y != NULL ) {if (NT->theStruct.type != procedure) {if(compound() == NULL) return NULL; if (NT->theStruct.type != scln){ return NULL;} else return NT;} else {NT = Si.cget(); y = proc(); }}} else return NULL; } } item<string>* procstmt() { if (NT->theStruct.type != procedure) { if(compound()==NULL) return NULL; if (NT->theStruct.type != scln){ return NULL;} NT = Si.cget(); } else{ NT = Si.cget(); y = proc(); if (y != NULL) { while (y != NULL ) {if (NT->theStruct.type != procedure) {if(compound() == NULL) return NULL; if (NT->theStruct.type != scln){ return NULL;} else return NT;} else {NT = Si.cget(); y = proc(); }}} else return NULL; } } i know i didnt make myself very clear .... it's really late, so i am gonna hit the sack ... i'll get up tomorrow and post more details if needed

Jul 11th, 2006, 3:19 AM	#30
The Dark Expert Programmer Join Date: Jun 2005 Posts: 893 Rep Power: 4	Try something like: (Toggle Plain Text) tem<string>* procstmt() { if (NT->theStruct.type != procedure) { if(compound()==NULL) return NULL; if (NT->theStruct.type != scln){ return NULL;} NT = Si.cget(); } else{ NT = Si.cget(); y = proc(); if (y != NULL) { if (procstmt() == NULL) return NULL; } else return NULL; } // Return something not NULL } tem<string>* procstmt() { if (NT->theStruct.type != procedure) { if(compound()==NULL) return NULL; if (NT->theStruct.type != scln){ return NULL;} NT = Si.cget(); } else{ NT = Si.cget(); y = proc(); if (y != NULL) { if (procstmt() == NULL) return NULL; } else return NULL; } // Return something not NULL }

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May 27th, 2006, 10:02 PM	#26
Harakim Hobbyist Programmer Join Date: May 2006 Location: West Jordan, Utah, United States Posts: 176 Rep Power: 3	If you don't mind me asking, how do you know what you are compiling from and what you are compiling to? Are you making a general purpose compiler?

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Programming Forums > Application Development > C++

writing a parser in c++