Find the output for the following code

rengaraj · 2009-12-11T06:47:05+00:00

Sir / madam, My question is Find the output for the following code void main() { int i=7; printf("%d",i++*i++); } Advance Thanks, R.Rengaraj {Java Certification Quiz} {On this forum my 20th Question.}

Find the output for the following code

rengaraj
@rengaraj-89Rrct

Updated: Oct 7, 2024

Views: 1.1K

Sir / madam,
My question is

Find the output for the following code

void main()
{
int i=7;
printf("%d",i++*i++);
}

Advance Thanks,
R.Rengaraj
{Java Certification Quiz}
{On this forum my 20th Question.}

0

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sarveshgupta

Member • Dec 11, 2009

Is the answer 56?

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rengaraj

Member • Dec 11, 2009

hi SarveshGupta,
Your answer is right, but i am confused with the answer. Can you explain me the process ?
R.Rengaraj

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sarveshgupta

Member • Dec 11, 2009
i++ means using post increment that is when we use it in any expression it uses the current value of that variable that is being carried forward from previous statement and then increment after compiler reads it so that any next time compiler reads the variable again it will read the incremented value

now we are doing i++ * i++ this implies

first time the value read is 7 which was the initial value of i and it gets incremented so for next operand i++ it takes the value 8 and after using 8 it increments it to 9

since we are using them only till here we get the value as 7*8 =56
if we try to print the value of i at the end of the program it will be 9
```
void main()
{
int i=7;
printf("%d \n",i++*i++);
printf("%d",i);
}
```
will give the output as:

56
9
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gaurav.bhorkar

Member • Dec 11, 2009

@sarvesh
You explained it perfectly, buddy.

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rengaraj

Member • Dec 11, 2009
sarveshgupta
i++ means using post increment that is when we use it in any expression it uses the current value of that variable that is being carried forward from previous statement and then increment after compiler reads it so that any next time compiler reads the variable again it will read the incremented value

now we are doing i++ * i++ this implies

first time the value read is 7 which was the initial value of i and it gets incremented so for next operand i++ it takes the value 8 and after using 8 it increments it to 9

since we are using them only till here we get the value as 7*8 =56
if we try to print the value of i at the end of the program it will be 9
```
void main()
{
int i=7;
printf("%d \n",i++*i++);
printf("%d",i);
}
```
will give the output as:
Dear sarveshgupta,
Your answer is right, but you stated as 7*8 why i can be 8*7,
bcoz, if execution is being handled from right to left.
R.Rengaraj
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gaurav.bhorkar

Member • Dec 12, 2009

rengaraj
Dear sarveshgupta,
Your answer is right, but you stated as 7*8 why i can be 8*7,
bcoz, if execution is being handled from right to left.
R.Rengaraj
Good Point. In fact printf statement handles execution from right to left (in C).
In this case it won't affect the answer in either case.

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Sahithi Pallavi

Member • Dec 12, 2009

@sarvesh : Very good explanation.!

@rengaraj : Nice point..!

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rengaraj

Member • Dec 12, 2009

Sir,
Thanks to all of you. For your excellent co-operation.
R.Rengaraj

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sarveshgupta

Member • Dec 12, 2009

Ya I am sorry for writing it that way but i actually meant 8 * 7 as I know printf reads from right to left

I wrote it simply missing the order of writing

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rengaraj

Member • Dec 12, 2009

No, Problem Sarveshgupta

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