How the post-increment operator works exaclty. Are there some exceptions?

Question

Consider this piece of code

int j = 0;
int x = 0;

for(int i=0;i<5;i++){
    j = x++;
    cout<<x<<" ";
}

Output:

1 2 3 4 5

Now consider this

int j=0;

for(int i=0;i<5;i++){
    
    j = j++;
    cout<<j<<" "; 
}

Output:

0 0 0 0 0

My doubt is why is j not being incremented after it is assigned the value 0. Isn't j=j++; equivalent to j = j; j++; and if it isn't then what's going on with the first case. I know it's a silly doubt but I couldn't figure this out by myself.

`j++` evaluates to the previous value of `j`, zero on every occasion. By assigning it back to `j` you basically reset the latter. — bipll, Aug 03 '20 at 09:40

score 5 · Accepted Answer · answered Aug 03 '20 at 10:14

Isn't j=j++; equivalent to j = j; j++;

No absolutely not, and this is one of the most common misunderstandings about the pre and post increment operators.

The true equivalent of j = j++; is

int temp = j;
++j;
j = temp;

Beginners think that the post increment happens after everything else. But it's not true. The increment happens after the value is taken, not after the rest of the statement is executed. You can see that in the code above, the value of j is taken temp = j;, then the increment happens ++j;, then the value is used j = temp;.

And as has been said several times, this code only has a defined meaning from C++17, before that it was undefined.

score 2 · Answer 2 · answered Aug 03 '20 at 09:40

2

My doubt is why is j not being incremented after it is assigned the value 0. Isn't j=j++; equivalent to j = j; j++; and if it isn't then what's going on with the first case.

No, because j = j++; is undefined. See this and other examples in https://en.wikipedia.org/wiki/Undefined_behavior#Examples_in_C_and_C++

answered Aug 03 '20 at 09:40

artm

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4

This is no longer undefined behavior in C++17: https://stackoverflow.com/questions/47702220/what-made-i-i-1-legal-in-c17 – VLL Aug 03 '20 at 09:50
@vll thanks, didn't know about that change in C++17 – artm Aug 03 '20 at 09:51

Sai Sreenivas · Answer 3 · 2020-08-03T10:28:23.110

2

j = j++;

First, the right-hand side expression evaluates i.e j++.
As we know j++ returns the previous value of j (i.e value before incrementing j). So, here j++ returns 0.
Finally, this turns out to be j = 0;

The same thing happens every iteration and that's the reason only 0's are printed.

Note:

As others mentioned, this is defined behavior from C++17

edited Aug 03 '20 at 10:28

answered Aug 03 '20 at 09:40

Sai Sreenivas

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Martins3 · Answer 4 · 2020-08-03T09:55:16.347

1

It's a undefined behavior under c++11, and both clang++ and g++ complain about it.

test.cpp:7:10: warning: multiple unsequenced modifications to 'j' [-Wunsequenced]
    j = j++;
      ~  ^
1 warning generated.

But it's no longer ub in c++17.

edited Aug 03 '20 at 09:55

answered Aug 03 '20 at 09:44

Martins3

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score 0 · Answer 5 · answered Aug 03 '20 at 09:47

0

j = j++;

lead to undefined behavior，there is a concept named sequence point in C++, see: https://en.wikipedia.org/wiki/Sequence_point

answered Aug 03 '20 at 09:47

AAAAAXXXXXXX

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AFAIK, the concept of sequence points no longer exists in C++ since C++ 11. Instead, operations are defined as being sequenced before or after other operations. – Violet Giraffe Aug 03 '20 at 09:50
3

This is no longer undefined behavior in C++17: https://stackoverflow.com/questions/47702220/what-made-i-i-1-legal-in-c17 – VLL Aug 03 '20 at 09:50
@vll: Great point! You should post an answer. – Violet Giraffe Aug 03 '20 at 09:53

How the post-increment operator works exaclty. Are there some exceptions?

5 Answers5