Ill formed code snippets

Question

can somebody please tell me the difference between the following two code snippets:

//Code snippet A: Compiles fine
int main()
{
    if(int i = 2)
    {
        i = 2 + 3;
    }
    else
    {
        i = 0;
    }
}

//Code Snippet B: Doesn't compile :(
int main()
{
    if((int i = 2))
    {
        i = 2 + 3;
    }
    else
    {
        i = 0;
    }
}

If you notice the diff is just an extra parenthesis at if statement. I am using g++ compiler.Error is "expected primary-expression before âintâ"

Why would you want to do this? Neither code makes much sense. — graham.reeds, Nov 11 '09 at 13:36
duplicate: http://stackoverflow.com/questions/1516919/declaring-and-initializing-a-variable-in-a-conditional-or-control-statement-in-c — Khaled Alshaya, Nov 11 '09 at 13:40

paxdiablo · Accepted Answer · 2009-11-11T13:55:28.377

Section 6.4 of the C++ standard (draft n2914 of c++0x) has this to say about the format of if statements:

Selection statements choose one of several flows of control.
   selection-statement:
      if ( condition ) statement
      if ( condition ) statement else statement
      switch ( condition ) statement
   condition:
      expression
      type-specifier-seq attribute-specifieropt declarator = initializer-clause
      type-specifier-seq attribute-specifieropt declarator braced-init-list

That bit at the end means a condition can be either an expression or a decalarator-type construct.

And the minute the parser hits that second parenthesis, it becomes an expression, so no declarations allowed, I'm afraid.

The snippet:

if (int i = 2) { ... } else { ... }

is perfectly valid C++ in which the if section defines an integer i for the duration of the if/else and sets it to 2. It then uses that 2 as the input to the if (2 is always true, being non-zero).

The snippet if((int i = 2)) is no different syntactically to int x = (int i = 2;); if (x) which is not valid C++.

score 10 · Answer 2 · answered Nov 11 '09 at 13:50

10

Snippet A is fine - the if condition delares and initialises a variable that can be interpreted as a boolean. The variable is defined within the if and any else blocks.

Snippet B is wrong, because you can't put parentheses around a declaration - you can only put them round expressions. The following is also wrong for the same reason:

int main()
{
    int i;       // OK
    (i = 2);     // OK

    (int x = 2); // compile error
}

answered Nov 11 '09 at 13:50

Mike Seymour

249,747
28
448
644

Thanks Mike! simple and the best answer :) ..wish i could vote..not registered yet. – White_Pawn Nov 11 '09 at 14:01
Well, register so you can. Although I can't agree it's the *best* answer, it's certainly a good succinct one, especially for those here not willing to sit through my usual essays to find a gem of good information:-) I've already upvoted it. – paxdiablo Nov 11 '09 at 14:10
upvoting is to encourage..right? ..and am philanthropist anyways :) – White_Pawn Nov 11 '09 at 14:50

score 5 · Answer 3 · answered Nov 11 '09 at 13:41

Here is another variant of second snippet error:

int main()
{
    int i = (int j = 0);
    return 0;
}

You can't declare variables inside ANY expression. Second () inside for is the same case - you can declare variable i inside for() but not inside nested expression placed into ().

Hope I explained it in proper words, maybe more correct explanation exists. Actually both code fragments are odd.

John Carter · Answer 4 · 2009-11-11T16:30:08.117

1

Both look wrong.

Just because code compiles it doesn't mean it's valid. I'm surprised that the first case is compiling, since it seems like i isn't defined in the else case.

edit: I'm wrong here, definitions in the if condition are valid for that whole if / else statement.

edited Nov 11 '09 at 16:30

answered Nov 11 '09 at 13:38

John Carter

53,924
26
111
144

4

The scope of the int declarator is to the end of the *entire* if statement. – paxdiablo Nov 11 '09 at 13:41
1

Not only that, but the expression inside the if is performing an assignment not a comparison, and as such the else case will never be reached. – Alex Deem Nov 11 '09 at 13:44
The first case is perfectly valid C++. It doesn't have to be useful. 'if (0)' is also perfectly valid C++ despite it's uselessnes. – paxdiablo Nov 11 '09 at 13:49
yeah..compilation and coding these is not an issue ..I was just trying to understand "the ill formed cpp statements". – White_Pawn Nov 11 '09 at 13:59
..I mean, the snippets are useless and we will never write such codes..but there are few not so known things about c++. – White_Pawn Nov 11 '09 at 14:05
My mistake, I didn't realise that `if (int i = 0) {} else { i = 1; } ` was valid code. – John Carter Nov 11 '09 at 14:33
else is not that useful with literals, but you'd generally :) see things like `if (X* p = foo())` – UncleBens Nov 11 '09 at 14:40
Makes sense, maybe I knew it, I'd just never thought that through. :) – John Carter Nov 11 '09 at 16:28

Ill formed code snippets

4 Answers4