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# c aptitude array

#include

#include

void main()

{

int a[]={0,1,2,3,4,5,6,7,8,9,10,11,12};

int num=0,i=0;

clrscr();

num=a[++i+a[++i]]+a[++i];

printf("%d",num);

getch();

}

how it produce output as 9

can u explain parsing of a[++i+a[++i]]+a[++i]

#include

void main()

{

int a[]={0,1,2,3,4,5,6,7,8,9,10,11,12};

int num=0,i=0;

clrscr();

num=a[++i+a[++i]]+a[++i];

printf("%d",num);

getch();

}

how it produce output as 9

can u explain parsing of a[++i+a[++i]]+a[++i]

Thread moved to CS section.

a[++i+a[++i]]+a[++i]

a[ ++i + a[++i]] + a[++i]

a[ ++i + a[++i]] + a[1]

a[ ++i + a[2]] + a[1]

a[ 3 + a[2]] + a[1]

a[ 3 + 2] + a[1]

a[5] + a[1]

5 + 1

6

This is the way it would parse... Let me try again... 😀

a[ ++i + a[++i]] + a[++i]

a[ ++i + a[++i]] + a[1]

a[ ++i + a[2]] + a[1]

a[ 3 + a[2]] + a[1]

a[ 3 + 2] + a[1]

a[5] + a[1]

5 + 1

6

This is the way it would parse... Let me try again... 😀

I encountered the same kind of problem few days ago. I had a statement like this and it was not showing the expected result. @praveen: i also thought that it should work like this. But the real thing is ++i is not the whole statement so the value of i is not incremented at this place only. First the value of i is incremented in the whole statment i.e. it is incremented 3 times. So i becomes 3. Now wherever the compiler sees pre-incremented part(++x) it puts the final value and wherever it sees the post incremented part(x++), it puts the initial value. Now our statement:

a[++i+a[++i]]+a[++i]

Here all the "i"s are pre incremented, so the final i.e. 3 will be posted.

a[3+a[3]]+a[3]

a[3+3]+3

a[6]+3

6+3

9

This is one of the examples why these kind of statements are avoided in programming.

You can take one more example....

int x=10;

y=x++ + x++;

Logically the output should be 21, but its 20 only.

You can try more examples like y=x++ + x++ + ++x;

P.S.: Correct me if i am wrong.

a[++i+a[++i]]+a[++i]

Here all the "i"s are pre incremented, so the final i.e. 3 will be posted.

a[3+a[3]]+a[3]

a[3+3]+3

a[6]+3

6+3

9

This is one of the examples why these kind of statements are avoided in programming.

You can take one more example....

int x=10;

y=x++ + x++;

Logically the output should be 21, but its 20 only.

You can try more examples like y=x++ + x++ + ++x;

P.S.: Correct me if i am wrong.

The logic is undeniable!!! 😲 But yeah, right!!! 😀nicepeeyushI encountered the same kind of problem few days ago. I had a statement like this and it was not showing the expected result. @praveen: i also thought that it should ork like this. But the real thing is ++i is not the whole statement so the value of i is not incremented at this place only. First the value of i is incremented in the whole statment i.e. it is incremented 3 times. So i becomes 3. Now wherever the compiler sees pre-incremented part(++x) it puts the final value and wherever it sees the post incremented part(x++), it puts the initial value. Now our statement:

a[++i+a[++i]]+a[++i]

Here all the "i"s are post incremented, so the final i.e. 3 will be posted.

a[3+a[3]]+a[3]

a[3+3]+3

a[6]+3

6+3

9

This is one of the examples why these kind of statements are avoided in programming.

You can take one more example....

int x=10;

y=x++ + x++;

Logically the output should be 21, but its 20 only.

You can try more examples like y=x++ + x++ + ++x;

P.S.: Correct me if i am wrong.

very very thanks dear friends

Can we really increment the value of all the "i's" at one go? We need to follow the rules. We should first increment the value of 'i' which is inside the parenthesis and then the others. We need to follow the sequence. Following the sequence we get the result as 6 instead of 9. I guess 6 is the correct answer.

Correct me if i am wrong!

Correct me if i am wrong!

@musicfreak: U r right....logically the answer should be 6, but u try to run the code, u will get the answer 9. The reason given is that when the value of i is incremented first time thats not the end of the statement, so the compiler goes till the end of statement, doingg all the increment operation. After that the values are put in our expression. A bit awkward, but this was the best explanation i could find.....😛

@All: Nicepeeyush is right... Confirmed... 😀

Yes very logical explanation indeed!

@musicfreak: try reading "Exploring C" by yashwant kanitkar......you will forget what logic is.....😉

ok thanx peeyush😀

exactly logically it would be 6

logical reason behind this can be explained with an example of another c program.Becasue when unary operators come into action it uses a concept called stack and reverse order execution.

According to your approach answer is 0224.

But c will execute in reverse order from right to left ++num,num++,++num, num++ and now you think answer is 1133.But it is not since it will put it in stack and take it in reverse order so the

correct answer is 3311

Hope this might answer your queries

Correct me if i am wrong.

#includeyou make think that output of the program might be num++,++num,num++,++num executed in this way.you will execute from left to right.void main() { int num=0; printf("%d%d%d%d",num++,++num,num++,++num); getch(); }

According to your approach answer is 0224.

But c will execute in reverse order from right to left ++num,num++,++num, num++ and now you think answer is 1133.But it is not since it will put it in stack and take it in reverse order so the

correct answer is 3311

Hope this might answer your queries

Correct me if i am wrong.

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