Understanding ++

[yes]

Could someone help me understand this:
{
int i=-3,j=2,k=0,m;
m=++i && ++j || ++k;
}

{
int i=-3,j=2,k=0,m;
m=++i || ++j && ++k;
}



What are the final values of variable in both cases, and
how?

TIA.

 

[Christopher Benson-Manica]

(e-mail address removed) spoke thus:

{
int i=-3,j=2,k=0,m;
m=++i && ++j || ++k;
}

&& has precedence over ||, so this is equivalent to

m=(++i && ++j) || ++k;

Assuming you didn't forget a loop or something, you get

m=(-2 && 3) || 1;

so m=1 (true), i=-2, j=3, and k=1.

{
int i=-3,j=2,k=0,m;
m=++i || ++j && ++k;
}

m=-2 || (3 && 1);

which gives you the same result.

Now, if you meant something like

while( ++i && ++j || ++k );

you get

( -2 && 3 ) || 1
( -1 && 4 ) || 2
( 0 && 5 ) || 3
etc.

This is an infinite loop.

For

while( ++i || ++j && ++k );

you get

-2 || ... <- short circuit evaluation
-1 || ...
0 || ( 3 && 1 )
1 || ...
..etc

Another infinite loop. Notice that if k is -1 instead of 0, you end up with

-2 || ...
-1 || ...
0 ||| ( 3 && 0 ) <- false

so i=0, j=3, and k=0. (I have tested this ^_^)

If I screwed up somewhere, someone please correct me! (And sorry for doing
what seems like a homework problem - I just realized it, and damn if I won't
post after I wrote all this...)

 

[Thomas Matthews]

(e-mail address removed) spoke thus:




&& has precedence over ||, so this is equivalent to

m=(++i && ++j) || ++k;

To Nitpick, the && and || operators have equal precedence.
The && operator is evaluated first because the operators
are evaluated left to right (since they have equal
precedence).

--
Thomas Matthews

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Other sites:
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[me]

(e-mail address removed) spoke thus:


&& has precedence over ||, so this is equivalent to

m=(++i && ++j) || ++k;

Assuming you didn't forget a loop or something, you get

m=(-2 && 3) || 1;

#include <stdio.h>
int main(void)
{
int i=-3,j=2,k=0,m;
m=++i && ++j || ++k;
printf("%d %d %d %d",i,j,k,m);
return 0;
}

so m=1 (true), i=-2, j=3, and k=1.

i, j, k, m
-2 3 0 1

 

[Arthur J. O'Dwyer]

To Nitpick, the && and || operators have equal precedence.

Not in C, they don't.

HAND,
-Arthur

 

[Christopher Benson-Manica]

&& has precedence over ||, so this is equivalent to

m=(++i && ++j) || ++k;

Assuming you didn't forget a loop or something, you get

m=(-2 && 3) || 1;

Even though someone else *already* corrected me, I want to come forward and
admit that I am dumb... it's funny how I only remembered short-circuit
evaluation halfway through this post. Of course this is

m=(-2 && 3) ... -> i=-2,j=3,k=0
m=-2 || ... -> i=-2,j=2,k=0

One day I'll make a totally correct post - until then I'll just have to
pretend I look sexy in this dunce cap...

 

[Christopher Benson-Manica]

To Nitpick, the && and || operators have equal precedence.
The && operator is evaluated first because the operators
are evaluated left to right (since they have equal
precedence).

According to a book I've respected until now, && is just higher than ||. If
the Standard says differently, I may have to revise my opinion of said book.

 

[you]

Even though someone else *already* corrected me, I want to come forward and
admit that I am dumb...

Not at all. Everyone makes mistakes.

it's funny how I only remembered short-circuit
evaluation halfway through this post. Of course this is

m=(-2 && 3) ... -> i=-2,j=3,k=0
m=-2 || ... -> i=-2,j=2,k=0

I don't think the logic is clear. Could you (or someone
else) explain it in more detail?

Thanks.

Also note:
#include <stdio.h>
int main(void)
{
int i=-3,j=2,k=0,m;
m=++i || ++j && ++k;
printf("%d %d %d %d",i,j,k,m);
return 0;
}

-2 2 0 1

 

[Robert Stankowic]

Even though someone else *already* corrected me, I want to come forward and
admit that I am dumb...

Not at all. Everyone makes mistakes.

it's funny how I only remembered short-circuit
evaluation halfway through this post. Of course this is

m=(-2 && 3) ... -> i=-2,j=3,k=0
m=-2 || ... -> i=-2,j=2,k=0

I don't think the logic is clear. Could you (or someone
else) explain it in more detail?

Thanks.

Also note:
#include <stdio.h>
int main(void)
{
int i=-3,j=2,k=0,m;
m=++i || ++j && ++k;

++i yiels -2.
-2 != 0 therefore the whole exprssion yields "true" which is by definition 1
due to lazy evaluation[1] the rest of the expression is skipped and we have
m == 1 (the standars´d value a true logical expression yields)
i == -2 (this part of the expression was evaluated)
j == 2
k == 0
because ++j as well as ++k were never executed

printf("%d %d %d %d",i,j,k,m);
return 0;
}

-2 2 0 1

[1] "lazy evaluation means, that the parts of a logical expression are only
evaluated until the outcome is known:
in
a || b for example b is only evaluated if a yields false
in
a && b b is only evaluated if a yields true

HTH
Robert

 

[Grumble]

To Nitpick, the && and || operators have equal precedence.

The && operator is evaluated first because the operators
are evaluated left to right (since they have equal
precedence).

http://www.difranco.net/cop2220/op-prec.htm

 

[pete]

To Nitpick, the && and || operators have equal precedence.
The && operator is evaluated first because the operators
are evaluated left to right (since they have equal
precedence).

Whta you just wrote, makes it seem as though you believe that
(1 || 1 && 0) is equivalent to ((1 || 1) && 0)
That can't be what you mean, can it ?