GATE CSE 2000 | Question: 1.18, ISRO2015-25

Question

The number of tokens in the following C statement is

printf("i=%d, &i=%x", i, &i);

• A. $3$
• B. $26$
• C. $10$
• D. $21$

Comments

WhiteSpaces will be removed by the lexical analyser from the source code, this is one of the jobs of LA.

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Note to self: 

Token is the atomic thing recognized by the lexical analyser like keywords, strings, operators, etc.

Lexeme is the sequence of the input character that can be associated with the token. Eg. int is the lexeme of the keyword token.

We use pattern matching (Regular language) to identify the token class of a lexeme.

Whitespace is not token in C lang.

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Detailed Video Explanation: https://www.youtube.com/watch?v=f-yJP8KTNrk&t=6466s

Answer 1

answer - C

Tokens are:

1. printf 
2. (
3. "i=%d, &i=%x"
4. ,
5. i
6. ,
7. &
8. i
9. )
10. ;

Comments

There is line of declaration and line of usage column in symbol table. There just the line number where occurrence of i is used/declared can be given.

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http://arantxa.ii.uam.es/~modonnel/Compilers/04_SymbolTablesI.pdf

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NOTE: In the above answer, from $1$ to $10,$ these are Lexemes, not Tokens. For each lexeme, a token is generated by the lexical analyser. So, number of tokens will be $10$.

Lexeme: The piece of the original program from which we make the token is called a Lexeme.

A lexeme is a sequence of characters in the source program that matches the pattern for a token and is identified by the lexical analyzer as an instance of that token.

As the first phase of a compiler, the main task of the lexical analyzer is to read the input characters of the source program, group them into lexemes, and produce as output a sequence of tokens for each lexeme in the source program.

Detailed Video Explanation: https://www.youtube.com/watch?v=f-yJP8KTNrk&t=6466s 

Answer 2

printf("i=%d,&i=%x",i&i);  it has 9 token but option are not matching 

the same question came in gate 2001  printf("i=%d,&i=%x",i,&i); with this statement it has 10 token

1.printf 
2.(
3."i=%d,&i=%x"
4.,
5.i
6.,
7.&
8.i
9.)
10.;

Comments

C tokens are of six types. They are,

1. Keywords               (int, while,etc),
2. Identifiers               ( main, total,etc),
3. Constants              ( 10, 20),
4. Strings                    ( “total”, “hello”),
5. Special symbols  ( (,), {,} etc),
6. Operators              ( +, /,-,*,etc)

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printf("i=%d,&i=%x",i&i); --- comma is missing, that is why geetting 9 as count.

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The question asked about the number of tokens in printf("i=%d,&i=%x",i,&i);

1. printf

2. (

3. "i=%d,&i=%x"

4. ,

5. i

6. ,

7. &

8. i

9.  )

10. ;

Therefore, the number of tokens is 10.

Answer 3

print =  token1

(   =  token2

"i=%d%x"  =  token3 ////anything inside "  " is considered as 1 token

,   =   token4

i = token5

, = token6

&  = token7

i = token8

) = token9

; = token10

so total 10 tokens

Answer 4

1.printf
2.(
3."i=%d, &i=%x"
4.,
5.i
6.,
7.&
8.i
9.)
10.;
option C
 

Comments

it is an operator- returns the address of operand.

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Okey

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Arjun Sir, In the complete reference it has discussed three types of reference-VARIABLE,NUMBER and DELIMITER.Is it correct or the above types of tokens is right?

Answer 5

if it is  i++   in place of &i   then answer will be 10 ??

Comments

yes bcoz both(i++ and &i ) have 2 tokens.

Answer 6

Number of tokens(i.e valid lexemes that matches with their pattern ) are

 1)printf   

2). (

3)"i=%d, &i=%x"

4.) ,

5) i

6) ,

7) &

8) i

9) )

10) ;

Hence ,total no of tokens =10  ,Answer C is correct.