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When do we count ++(post increment or pre increment) as 1 or 2 token ?
in Compiler Design by Loyal (8.5k points)
edited by | 593 views
0
All operators are tokens, hence ++ is taken as a single token.

Correct if wrong.
+6

This will be consider as single token, because compiler apply greedy approach for performing tokenization, it will try to include more character in token if possible.

https://gcc.gnu.org/onlinedocs/cpp/Tokenization.html

0

@ Shubhanshu Thanks it helps :) /\

3 Answers

+7 votes
Best answer
Preprocessor follows greedy approach to tokenize,  This can be done in TWO steps

1 => Separated by space (A+ ++B) => Have 4 tokens(Space is not considered as token), here tokens are A, +, ++, B(In order of expression). If any operator doesn't have any space ,

then ,

2 => Greedy approach (A+++B) for longest possible valid operator in the Grammer => Like we have '++' > '+' so here the tokens are A,++, +, B(In order of expression).
by Junior (819 points)
selected by
0

@Harish Kumar 2 I think there is nothing like priority in tokenization.

While performing tokenization, Compiler use greedy approach and tries to include more token if possible,

ref https://gcc.gnu.org/onlinedocs/cpp/Tokenization.html

0
Ok, so after Lexical analysis then it checks then the compiler checks which operator to be formed?

Like for "A+++B"(if used some where) then, how many tokens? and how will it use this. Or space is used to seperate these now. Kindly clarify.
0

Then tokens are :-

A

++

+

B

Even you can run on ide

int main()
{
    int a,b;
    a = 10;
    b = 10;
    printf("Hello World %d \nval of a = %d\nval of b = %d", (a+++b), a, b);
    return 0;
}

Output

Hello World 20

val of a = 11

val of b = 10

0
So then priority plays role in getting tokens, right? Because if not, then by greedy approach it should have considered it as '+' only, Shouldn't it???
0
For tokenization, greedy is used and while solving the expression,

like a+++b

priority of operators are used because unary operator have higher precedence than binary operator.
0
Ok thanks Shubhansu, now I have got it.

And i have updated the answer, can you please look into that now and provide feedback.
0

@Harish Kumar 2 now it is correct.

0
Can anyone explain no. of tokens in case of unary operator such as the below line

int a = -3;

Is -3 is taken as single token or two tokens?
+2 votes
Always when we count the token complier always see next symbol

Eg-

++= 1token

a+++++b =5tokens

&&= 1token
by Active (1.7k points)
edited by
0 votes
I think 2 tokens because count is an identifier and ++ is an operator.
by (299 points)
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