Does the equal sign make a difference in brace ini

Question

Here are two ways to initialize a variable in C++11:

T a {something};
T a = {something};

I tested these two in all scenarios I could think of and I failed to notice a difference. This answer suggests that there is a subtle difference between the two:

For variables I don't pay much attention between the T t = { init }; or T t { init }; styles, I find the difference to be minor and will at worst only result in a helpful compiler message about misusing an explicit constructor.

So, is there any difference between the two?

Answer 1

The only significant difference I know is in the treatment of explicit constructors:

struct foo
{
    explicit foo(int);
};

foo f0 {42};    // OK
foo f1 = {42};  // not allowed

This is similar to the "traditional" initialization:

foo f0 (42);  // OK
foo f1 = 42;  // not allowed

See [over.match.list]/1.

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Apart from that, there's a defect (see CWG 1270) in C++11 that allows brace-elision only for the form T a = {something}

struct aggr
{
    int arr[5];
};

aggr a0 = {1,2,3,4,5};  // OK
aggr a1 {1,2,3,4,5};    // not allowed