Constexpr Variable Appears Uninitialized Inside Lambda

Question

In the code sample below, I would expect the output to be as follows, since x is a static constexpr variable.

However, g++ warns me that x is uninitialized inside the lambda function when I compile with -Wall, and the last three lines of the output vary, presumably because the value of uninitialized memory on the stack is being printed. Here is one possible output produced by the program when it is compiled with the options -Wall -std=c++0x. Why is the output not what I expect it to be?

If it helps, the expected output is produced when constexpr is deleted in the declaration constexpr T x.

Example Program

#include <algorithm>
#include <iostream>

struct _foo
{
        template <class T>
        struct traits
        {
                static constexpr T val = 5;
        };

        template <class T>
        constexpr T getval() const { return traits<T>::val; }
} foo;

struct _test
{
        template <class T>
        void bar(const T& t)
        {
                int arr[] = { 1, 2, 3 };
                constexpr T x = foo.getval<T>();
                std::cout << x << std::endl;
                std::for_each(arr, arr + 3, [&](int i) {
                        std::cout << x << std::endl;
                });
        }
} test;

int main()
{
        test.bar(5u);
        return 0;
}

Note: Leading underscores are reserved in global scope, so technically this code has undefined behavior. — Adam H. Peterson
– Adam H. Peterson, Commented Apr 20, 2016 at 23:27

void-pointer · Accepted Answer · 2012-02-15 15:02:36Z

2

This was indeed a compiler bug. I reported it here, and it was confirmed a while ago. It doesn't crash on older versions of g++, because the constexpr keyword is just parsed and ignored.

answered Feb 15, 2012 at 15:02

void-pointer

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Comments

Seth Carnegie · Accepted Answer · 2012-01-28 03:03:07Z

0

You have to capture x in the lambda's capture list:

std::for_each(arr, arr + 3, [&](int i) { // notice the & (capture the outside env
                                         // by reference
    std::cout << x << std::endl;
});

answered Jan 28, 2012 at 3:03

Seth Carnegie

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5 Comments

void-pointer Over a year ago

I just added that as you posted this answer. Same problem.

void-pointer Over a year ago

I'm using g++ 4.6.1 on Linux Mint 64-bit. Also, the fact that it's not reported as an error without the ref-capture may be another compiler bug.

void-pointer Over a year ago

Apparently g++ 4.5 doesn't support constexpr; it merely parses and ignores it. gcc.gnu.org/bugzilla/show_bug.cgi?id=51604

void-pointer Over a year ago

I submitted a bug report here for g++ 4.6.1: gcc.gnu.org/bugzilla/show_bug.cgi?id=52026.

Richard Smith Over a year ago

You don't need to capture x, because it is not ODR-used by the lambda, because it satisfies the requirements for appearing in a constant expression and the lvalue-to-rvalue conversion is immediately applied. See [basic.def.odr]p2.

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