C Programming - Expressions - Discussion

@ALL.

int x=3;
i=i++; which means i=i+1; => i=4;
printf("%d",i); prints 4.

Isn't it depend on the compiler. I mean in c++,
int x = 10;
x = ++x;
cout<<x;
it will still give 10 as output.

It's a post increment and the value of i will be added before.

According to me, the answer is 3. Run the code in the compiler.

Hi All, Answer is wrong and the correct answer is 3, how means;

int i=3;
i=i++;
in above line i post incremented but result (i++ value is 3 in current line)3 is assigned to i.
So final output is 3.

(if you want output as 4 try this,
int i=3;
i=++i;
)

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.