Identifier with the same name in both expression and declaration of range-based for

Question

Is it legal to declare a loop variable in a ranged-based for loop with the same name I use in the expression statement of the loop? I hope the example makes it clear.

#include <iostream>
#include <vector>

struct bar {
    std::vector<int> nums;
};

int main()
{
    bar b;
    b.nums = {1, 2, 3};

    for(int b : b.nums)
        std::cout << b << std::endl;   
}

gcc 4.8 gives an error while clang 3.2 allows it.

@JesseGood Thanks for digging that up. Turns out I didn't find anything on bugzilla because I was searching for "ranged based for" instead of "range based for". Don't know how that term manifested in my head, same mistake happened in the title, too. — Stephan Dollberg
– Stephan Dollberg, Commented May 6, 2013 at 21:26

Bill Lynch · Accepted Answer · 2013-05-06 21:15:16Z

11

From my reading of C++2011 6.5.4, your code of:

bar b;

for(int b : b.nums)
    std::cout << b << std::endl;

Should be converted to:

bar b;

{
   auto && __range = b.nums;
   for (auto __begin = __range.begin(), __end = __range.end(); __begin != __end; ++__begin ) {
       int b = *__begin;
       std::cout << b << std::endl;
   }
}

This to me means that clang is correct.

answered May 6, 2013 at 21:15

Bill Lynch

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6 Comments

rodrigo Over a year ago

IMHO, I'd say that this is a defect in the standard. The visibility of int b should be from the : until the end of the for statement.

Bill Lynch Over a year ago

@rodrigo: I agreed until I thought about it some more. What would this mean: for (auto & x : x)?

rodrigo Over a year ago

That would be as meaningless as the simple declaration auto &x = x; and for the same reasons.

GManNickG Over a year ago

@rodrigo: Why? Why make that meaningless when what we have now currently works and makes sense?

rodrigo Over a year ago

@GManNickG: Why? For consistence with other parts of the language: everywhere when a name is introduced its scope starts just after the name itself. For example void *p = &p; is perfectly legal. The range-based-for is an exception to the rule and for no real reason other than a likely unwanted side-effect of the standard wording.

|

Andy Prowl · Accepted Answer · 2013-05-06 21:10:44Z

9

Clang is right.

Paragraph 6.5.4/1 of the C++11 Standard defines the range-based for statement as follows:

For a range-based for statement of the form
for ( for-range-declaration : expression ) statement
let range-init be equivalent to the expression surrounded by parentheses
( expression )
and for a range-based for statement of the form
for ( for-range-declaration : braced-init-list ) statement
let range-init be equivalent to the braced-init-list. In each case, a range-based for statement is equivalent to
{
    auto && __range = range-init;
    for ( auto __begin = begin-expr,
          __end = end-expr;
          __begin != __end;
          ++__begin ) {
        for-range-declaration = *__begin;
        statement
    }
}

From the above, it is visible that variable b, which corresponds to the for-range-declaration, is declared inside a nested block statement, while the initializer range-init (which corresponds to b.nums) appears in the parent scope, where b should resolve to the object of type bar.

edited May 6, 2013 at 21:10

answered May 6, 2013 at 21:04

Andy Prowl

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4 Comments

Andy Prowl Over a year ago

@sftrabbit: You're right, I overlooked that initially. I edited the answer, thank you

rodrigo Over a year ago

It is not so easy... from a literal interpretation of the standard, in auto && __range = range-init; the range-init is actually b.num, but the int b does not enter in the definition until 5 lines later... But again, seeing the original code, it would be expected that the int b hides the bar b.

Joseph Mansfield Over a year ago

Apologies for leading you astray. My mistake! I agree with this answer.

Andy Prowl Over a year ago

@sftrabbit: That's all right, your observation was OK and I should have triple-checked rather than rushing into answering ;) Lesson learnt

Ville Voutilainen · Accepted Answer · 2015-10-01 19:25:35Z

2

For what it's worth, this bug has now been fixed on gcc trunk. :)

answered Oct 1, 2015 at 19:25

Ville Voutilainen

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