C Programming - Expressions - Discussion

I am very appreciate you to give me such like chance to see myself how I understand the course!!!!
Thanks to you!

@Karthikeyan You are absolutely correct.

Let me give my points too.

i = i++; // You may have little confusion here.

The above statement will be executed as given below:

Step-1: i = i; // Here i = 3
Step-2: i = i+1; // Here i = 3 + 1 = 4.

Therefore, 4 is the correct answer.

Step 1: int i=3
Step 2: i=i++ post increment so still i=3 in this statement
Step 3: after completion of the above statement i will be increment by 1 so i becomes 4
step 4: in printf statement i value is printed i.e:4

int i = 3;
i = 3+3;
i = 6;
i++;
i = 7;
i++;
i = 8;

Finally i = 8 will be print;

According to me the answer is 3.

I tried this code in DEV C++ blog and I got the output as 3. But here, the compiler says 4. Why?

Well, careful guys,

First of all let me make this very clear that,

----> priority of "Post Increment" is less than that of "Assignment Operator".

So the first assignment will be done then increment in value of 'i' will take place.

That means, "i" is initialized with 3.

Hence,

3 will be assigned to 'i'. // i ==> i = 3.

But after assignment value of "i" will get increase by 1. // i++ ==> i = 4.

Therefore,

New updated value of "i" will be 4. And we printing value of 'i' in printf () statement.

If the program had been written like this then the output will be '3'.

#include

int main ()
{
int i=3;
int m;
m= i++;
printf ("%d\n", m) ;
Return 0;
}
Here, we are the printing value of m . the output will be '3'.
But,
In the given question. We are the printing value of 'i'.

Hence, Option (B) i.e. 4 is correct.

All of you are wrong. The post-increment effect is damaged by the assignment operator. Thus the value of i=3.

Always go with the last operation here i++ is the last operation so the answer is 4.

I tried this code in code block and I got the output as 3. But here, the compiler says 4. Why?