Can anyone explain the output..?

Question

include<stdio.h>
int main()
{
int x=1;
x=( ++x)+(++x)+x;
printf("%d", x) ;

}
//output =9

And this

#include <stdio.h>

int main() {
int x=10;
x+=x++ + ++x +x;
printf("%d",x);
//Output =46

return 0;
}

Answer 1

This is getting annoying

https://stackoverflow.com/questions/949433/why-are-these-constructs-using-pre-and-post-increment-undefined-behavior

Answer 2

Sheldon Duncan You really went out of your way to write an answer, but the truth is that everything containing multiple assignments/increments is undefined in C. The result is not predictable and may depend on the compiler

Answer 3

Sheldon Duncan While undefined behavior and sequence points apply to C/C++, this is not the case with other languages like Java and C#.

That might help explain the disconnect.

Answer 4

(2) In the second main method, x is instantiated to 10.

You then have x += x++ + ++x + x, which can be changed to x = x + x++ + ++x + x, since x += x is another way of saying x = x + x.

However, we'll keep += the same for reasons I'll explain later.

To make things clearer, let's add parentheses around the increments:

x += (x++) + (++x) + x

The first increment is a post increment, not a pre increment, so it occurs after the rest of the expression is evaluated.

The second increment is a pre increment, however, so that x and every x after it has increased by 1.

As such, we have:

x += (10++) + 11 + 11

However, += actually has lower precedence then the remainder (at least, in Java it does - it's probably also like this in Python). So, the x before the += becomes 11 before the rest is evaluated.

Thus, it's now:

x = 11 + (10++) + 11 + 11

Which is:

x = 43++

Which can be simplified to:

x = 44

Answer 5

Anna that's really interesting. In programming contests for Java, the result and evaluation is expected to be exact. Either I don't get something, or stuff is just different in Java.

Answer 6

(1) I don't know C, but the fundamental logic is the same in a lot of languages, so I think I can answer this.

For the first main method, you have x instantiated to 1.

You then increment it by 1 twice in the expression x = (++x) + (++x) + x.

However, pre-increment (before-increment) is evaluated on the first x, making it 2, then is evaluated on the second x, making it 3. The last x is now 3, too, making it x = 2 + 3 + 3 = 8.

Answer 7

1.
++x ; increments x first and assigns x the value of 2,

++x ; increments x first and assigns x the value of 3,

now equation looks like this;

x = 3 + 3 + 3

2.
x++ ; assigns the value of 10 to x, than increments to 11,

++x ; increments x first and assigns x the value of 12

now equation looks like this;

x += 10 + 12 + 12 ; (value of x is 12)
x = 12 + 34

Answer 8

Good examples for understanding posfix, prefix increments and precedence. Evaluation for right side of statments and assigns for left side of =.

I thinks this question, but the second is more dificulty to see. But Esref Efe Ozcivelek explain well. I think he is explained your doubt.

I think that you can suggest the examples for quizz factory. Excuse my english.

Thanks.

Below a code that i use for study this question. Show the 34 in calculation of second:
https://code.sololearn.com/cxl7W0WI4H69/?ref=app