what is the value of a++ + a if value of a is 5?

Question

i am a beginner in c, and i am finding it difficult to understand the post and pre increment i have given my code below,i already compiled it in a turbo c++ compiler and i got output as a = 6 and b = 10 but since the post increment operator is used the output should be a = 6 and b = 11 ,why is it not happening?could someone explain it..

#include<stdio.h>    
int main()    
{
    int a=5,b;
    b = a++ + a;    
    printf("\na = %d and b = %d",a,b);    
    return 0;    
}

I'm voting to close this question as off-topic because of it. — Martin James, Dec 11 '15 at 15:54
If this is the work of a professor then it's as depressing as it is concerning. — Bathsheba, Dec 11 '15 at 15:59
If you turn up the warning level in your compiler, does it not emit a diagnostic about the line in question? I don't know about Turbo C++, but gcc or g++ certainly would do. — John Bollinger, Dec 11 '15 at 16:08
@Bathsheba it's even more depressing to imagine that anyone outside academia would write it:( — Martin James, Dec 11 '15 at 16:11

score 7 · Accepted Answer · answered Dec 11 '15 at 15:47

7

The behaviour of a++ + a; is undefined in C. This is because the + is not a sequencing point and you're essentially attempting to increment and read a in the same expression.

So you can't guarantee a particular answer.

In order to understand prefix and postfix increments, use statements like b = a++; and b = ++a;

answered Dec 11 '15 at 15:47

Bathsheba

231,907
34
361
483

what is a sequencing point? – prashanth Dec 11 '15 at 17:46
when i ran it in ubuntu linux compiler i got output as a=6 and b=11 – prashanth Dec 11 '15 at 17:46
@prashanth You should Google that: and make your teacher aware of them; constructively of course. – Bathsheba Dec 11 '15 at 18:28
@prashanth basically a *sequence point* is a semicolon `;`. There are a few more rules you may want to read more about in the [Annex C of the Standard](http://port70.net/~nsz/c/c11/n1570.html#C). – pmg Dec 11 '15 at 20:49

score 3 · Answer 2 · answered Dec 11 '15 at 16:14

What happens in the following?

b = a++ + a;

1) Is a incremented and its original value is then added to the original value of a?

2) Is a incremented and its original value is then added to the new value of a?

3) Is a on the right side fetched first and then added to the original value of an incremented a?

C allows any of theses approaches (and likely others) as this line of code lacks a sequence point which would define evaluation order. This lack of restriction allows compilers often to make optimized code. It comes at a cost as the approaches do not generate the same result when accessing a in the various ways above.

Therefore it is implementation defined behavior. Instead:

b = a++; 
b = b + a;

or

b = a; 
b = b + a++;

score 1 · Answer 3 · answered Dec 11 '15 at 15:57

After int a = 5; the value of a is 5

b = a; // b is 5;

After int a = 5; the value of a++ is 5

b = a++; // b is 5

but the side effect of a++ is to increase the value of a. That increase can happen anytime between the last and next sequence points (basically the last and next semicolon).

So

/* ... */;
b = a++ + a;
#if 0
    /* side-effect */ 5 + 6
    5 /* side-effect */ + 6
    5 + /* side effect mixed with reading the value originating a strange value */ BOOM
    5 + 5 /* side effect */
#endif

what is the value of a++ + a if value of a is 5?

3 Answers3