Why this output?

Question

I have been trying to understand of this following C-program:

#include <stdio.h>

int arr[] = {1,2,3,4};
int count;

int incr(){
 return ++count;
}

int main(){  
   arr[count++]=incr();
   printf("%d %d",count,arr[count]);    
   return 0;
}

The program gives 1 2 as output,what I am not gtting is why the value of count is 1 here and not 2 (since there are two increments)?

You should initialize the count variable; `count = 0` otherwise the value will be undefined. — Drahakar, Apr 29 '11 at 04:08
@Drahakar : Global variables are initialized to their default values. In OP's code count would be initialized to 0 — Prasoon Saurav, Apr 29 '11 at 04:10

score 5 · Accepted Answer · edited May 23 '17 at 10:32

5

The order of evaluation of operands of = operator is unspecified in arr[count++]=incr(); and since both the operands are trying to modify the same global variable count the result would be different on different compilers depending upon the order of evaluation.

EDIT

Actually the behavior is undefined (which means anything can happen) because "the prior value of the variable count is not accessed (only) to determine the value to be stored."

edited May 23 '17 at 10:32

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answered Apr 29 '11 at 04:05

Prasoon Saurav

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either way, by the `printf` line, the value of `count` should be 2, not 1, though, no? – nicolaskruchten Apr 29 '11 at 04:08
No, because it's undefined behaviour it can actually be _anything_. Your compiler decides to make it 1, but there's no guarantee on what it could be – Ben Stott Apr 29 '11 at 04:14
1

Yep the value would be undefined since it's I think it's same as `a[i++]= ++i;` – Quixotic Apr 29 '11 at 04:34
2

I don't believe that this is undefined. I sent a PR to GCC about this. See http://gcc.gnu.org/bugzilla/show_bug.cgi?id=48814 – Johannes Schaub - litb Apr 29 '11 at 06:28

Johannes Schaub - litb · Answer 2 · 2011-04-29T06:45:11.217

incr() will return either 1 or 2. It depends on whether the implementation first increments count and then calls incr(), or whether it first calls incr() and then increments count.

Please note that this choice does not mean that behavior is undefined. Since before a function is entered, and after a function is left, there is a sequence point at each point, both increments we have here are separated by a sequence point, so that the increment in main, if it started to happen before the call, will be finished once entering incr(), and if it happens after incr() was called, will not have yet started until incr() has left.

We have multiple scenarios here:

First do the increment for count++, then call incr(). This will write 2 into arr[0].
First call incr(), then do the increment for count++. This will write 1 into arr[1].

So, count is always 2, and arr[count] is always 3 (it wasn't overwritten). So it should output 2 3, not 1 2.

I think that if you do the following, you have more options

int main(){  
   arr[++count]=incr();
   printf("%d %d",count,arr[count]);    
   return 0;
}

Now, the value read from ++count can be different than count+1, because there is nothing that stops incr() to be called after incrementing count but before reading it. In this case we have

First do the increment for ++count, then call incr(), then read from count. This will write 2 into arr[2].
First do the increment for ++count, then read from count, and then call incr(). This will write 2 into arr[1].
First call incr(), then do the increment for ++count and read from it. This will write 1 into arr[2].

In this case, you can either have output 2 2 or 2 1 or 2 3.

Why this output?

2 Answers2