What is a output?. How will Come?.

main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}

The values are i=0,j=0,k=1,l=3,m=1.its very simple here we r using post increments and logical and,or operations.

I think m=0

l=0,j=0,k=1,l=3,m=1;
according to relatonal operator presedence.
&& , ||,!
T&&T=T
T&&F=f
T||t=t
t||f=t
F||f=f

I think may be you got the error "m is not initialized before using".....

-----------------------
suresh

You will not get any error.
Its just application of logical operators.
Start evaluating from Left to right. All the values get post incremented.
m=i++&&j++&&k++||l++;
m = -1 && -1 && 0 || 2
since i, j, k ,l are post increments, they will increment after the complete statement evaluation.
m = 1 && 0 || 2
m = 0 || 2 , m = 1
Here, since left side is 0, it will evaluate aven the right side also.
Finally, i = 0, j = 0, k = 1, l = 3, m = 1.

main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}

0 0 1 3 true

It will not show any errors.
The out put of your program is :
[B]i=0;j=0;k=1;l=3;m=1(true).[/B]
m=i++&&j++&&k++||l++;

[COLOR=black][B][FONT=arial black]execution is as follows :[/FONT][/B][/COLOR]
==> m= (i++&&j++)&&k++||l++; i=0 ; j=0
==> m=(-1 && -1)&&k++||l++;
==> m= (1) && k++||l++;
==> m= (1 && 0)|| l++; k= 1;
==> m= (0 || 2); ==> m= 1 l=3;
[B][U]


and (&&) operator truth table:[/U][/B] [B]

true(1) && true(1) -------> true(1)
true(1) && false(0) -------> false(0)
false(0) && true(1) -------> false(0)
false(0) && false(0) -------> true(1)
[/B] [U][B]

or(||) operator truth table:[/B][/U]

[B] true(1) || true(1) -------> true(1)
true(1) || false(0) -------> true(1)
false(0) || true(1) -------> true(1)
false(0) || false(0) -------> false(0)[/B] [COLOR=blue][B]
::remember these while doing the logical operator programs ::[/B][/COLOR][B]

*** vishnu ***[/B]

[QUOTE=Golda;18013]main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}[/QUOTE]

ans:
0 0 1 3 0

main()
{
int i=5;
printf("%d%d%d%d%d",i++, i--, ++i, --i,i);
}
I got the output [B][COLOR="Blue"]45545[/COLOR][/B]. How it will come?...:confused:

ya 45545 is correct

it will check from left to right
initially i=5

so for output value i it ll print as 5

for o/p --i is a prefix it will decrease and assign value to i(5-1) so for --i its 4


for o/p ++i is a prefix it will increase and assign value to i(4+1) so for ++i its 5

for i-- is post fix so it will assign previous valuue then it will decrease by 1.

so o/p will be 5 only but after printing value will be 4

for i++ is post fix so it will assign previous valuue then it will increase by 1.

so o/p will be 4 only but after printing value will be 5

so finally value of i=5

what about this?:confused:
void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u ",gotit=-giveit);
}

hmm no idea abt unsigned int

[QUOTE=rupesh kumar singh;18188]l=0,j=0,k=1,l=3,m=1;
according to relatonal operator presedence.
&& , ||,!
T&&T=T
T&&F=f
T||t=t
t||f=t
F||f=f[/QUOTE]

A good explaination with truth values m=1 cause at bit level
00000001 || 00000011=00000001
Nice job rupesh

i think u have declared giveit as unsigned and assigned "-1"

Ya i declared giveit as unsigned and assigned -1 only

the o/p of this program is 00131
in given expression (m=i++&&j++&&k++||l++) operater play an imp role
i.e. we will first evaluted incre. op. that has higher predency after that we will evaluted logical AND op(&&) and after that logical OR(||) op.and at the end as.op(=) will be evaluted . note that it is a logical exp. so m have value either 1 or 0. while in this exp it is true so has 1 as o/p
and the value of i,j,k and l is incremented when write exp , so it print these incremented value by 1 so these value become 0,0,1,and 3 and the value of m is evaluted in exp ,1
dilip agrawal mca indore mp

00131 is the answer.
m=i++ && j++ && k++ || l++;
let us leave all the postfix increments because they will change the values only in the next statement of their respective variables. so m=i && j&& k|| l
-1 && -1=0
0 && 0=0
0 || 2=1
so m is 1. this statement is over then all the variables will be incremented by 1 because of that postfix increments. so output is 00131

out put vil b
-1,-1,0,2,1

[QUOTE=Golda;18013]main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}[/QUOTE]

i=0,j=0,k=1,l=3,m=0

0 0 1 2 -1

[QUOTE=Golda;18866]main()
{
int i=5;
printf("%d%d%d%d%d",i++, i--, ++i, --i,i);
}
I got the output [B][COLOR="Blue"]45545[/COLOR][/B]. How it will come?...:confused:[/QUOTE]
the output is 5 6 6 5 5

0,0,1,3,0
is this correct

the output will be 00131
All the effect will be occur at next statement because of post increment
the || operator is the last element and the value of l is 2 so that any value of the previous generation will be ommited and the value of m will be true i.e. 1
m=0||1 is 1
i=-1 --> 0
j=-1 --> 0
k=0 --> 1
l=2 --> 3
m = 1

[QUOTE=kharade Sagar M;46668]the output is 5 6 6 5 5[/QUOTE]

The output for this expression is [B]ambiguous[/B]. The ANSI C standard does not define the order of evaluation of arguments to a function. Hence, the output may vary depending on the compiler you use.

[QUOTE=Golda;18960]what about this?:confused:
void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u ",gotit=-giveit);
}[/QUOTE]

"giveit" has a bit pattern
[CODE]11111111 11111111 11111111 11111111[/CODE]

Incrementing it by [B]1[/B] gives you the following bit pattern
[CODE]1 00000000 00000000 00000000 00000000[/CODE]
This [B]1[/B] results from overflow, and is ignored. So, you effectively get a zero value on return

[QUOTE=Golda;18013]main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}[/QUOTE]
Output : 0 0 1 3 1
EXP : As the expression have 3 && operators at the begin it will check for all the case s ,& will get 0 as resultant wch is || with 1 (i.e.,2) So that m=0||1=1.

i=0,j=0,k=1,l=3,m=0

[QUOTE=Golda;18960]what about this?:confused:
void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u ",gotit=-giveit);
}[/QUOTE]
++giveit=2 n -gotit= -2