Expression evaluation in C

Question

#include <stdio>
int main(){      

       int x = 4;  
       int y = 3;  
       int z;

       z = x---y;
       printf("%d" , z);
       return 0;
}

The gcc compiler in Linux Mandriva evaluates it as (x--)-y. I am confused as to why is it so. It could have been x - (--y).

I know some of the answers would tell me to look at precedence tables. Ihave gone through all of them, still the doubt persists.

Please anybody clarify this.

score 9 · Answer 1 · answered Jul 14 '11 at 09:09

9

The C lexical tokeniser is greedy, so your expression is tokenised as

x -- - y

before precedence rules are applied.

answered Jul 14 '11 at 09:09

Greg Hewgill

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Out of curiosity, is this a implementation defined behaviour or does the standard mandate it to be greedy? Can i rely on this behaviour on different platforms/compilers? – RedX Jul 14 '11 at 09:15
@RedX: I'm fairly certain you can rely on this. I can't find a specific reference, but the behaviour is implied by this lexer for C: http://www.lysator.liu.se/c/ANSI-C-grammar-l.html – Greg Hewgill Jul 14 '11 at 09:21

score 4 · Accepted Answer · answered Jul 14 '11 at 09:11

4

The rule is "when getting the next token, use the longest sequence of characters possible that constitute a valid token". So --- is -- followed by a - and not the other way around. Precedence has actually nothing to do with this.

answered Jul 14 '11 at 09:11

n. m. could be an AI

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score -1 · Answer 3 · edited Jun 08 '12 at 12:46

-1

x-- is stronger than --x , so it is compiled this way. Postfix is stronger than prefix.

See C Operator Precedence Table.

edited Jun 08 '12 at 12:46

Wladimir Palant

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answered Jul 14 '11 at 09:11

powerMicha

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Expression evaluation in C

3 Answers3