why same code in two technology behaving different

Question

Below is my code snippet in C.

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

output : 8

public static void main(String[] args){

        int x = 7;

        x =  x++;
        System.out.println(x);
    }

output : 7

i am not getting why both language giving different output. I've referred below link What is x after "x = x++"?

First code snippet is [Undefined Behaviour](http://en.wikipedia.org/wiki/Undefined_behavior) in C. — Rohit Jain, Aug 01 '13 at 11:46
the exact same post as this famous one!!! [link here][1] [1]: http://stackoverflow.com/questions/7911776/what-is-x-after-x-x?lq=1 — user2497624, Aug 01 '13 at 11:47

score 2 · Answer 1 · edited May 23 '17 at 10:25

2

In java after x++ there is no change in the value of x

x = x++; equal to

int i= x;
x = x + 1;
x = i;

so x remains same as i

You can read more from here :Why are these constructs (using ++) undefined behavior?

edited May 23 '17 at 10:25

Community

1
1

answered Aug 01 '13 at 11:44

Suresh Atta

120,458
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Why this is equivalent to int i= x; x = x + 1; x = i; in java not in C – Rohit Kumar Aug 01 '13 at 11:45
@RohitKumar Please go through the link attached. – Suresh Atta Aug 01 '13 at 11:48

score 0 · Answer 2 · answered Aug 01 '13 at 11:46

0

In the second example the assignment first saves the value of x, then sets x to its value plus 1, and, finally, resets x back to its original value. Kind of:

int temp=x;
x=x+1;
x=temp;

answered Aug 01 '13 at 11:46

Lavinia

11
2

score 0 · Answer 3 · answered Aug 01 '13 at 12:08

0

x=x++;

This gives arbitrary results in C, mainly depending on compiler. Read about sequential points in C. You may refer to C Programming by Dennis ritchie.

answered Aug 01 '13 at 12:08

user2550754

884
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why same code in two technology behaving different

3 Answers3

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