The empty default constructor like Widget() {}; is seen as a user defined default constructor, while Widget() = default; is not. This leads to default initialization in the former case, while value initialization in the latter, in definitions involving the form Widget w = new Widget(), Widget w{} etc.
this.value = default!; as I saw in a different question here, then it compiles just fine. But I don't understand what the ! is doing here, and it's pretty hard to google, since google seems to ignore punctuation in most cases. What does default! do?
- name: Create default user: name: "default_name" when: my_variable is not defined - name: Create custom user: name: "{{my_variable}}" when: my_variable is defined But as I mentioned, there's a lot of optional variables and this creates a lot of possibilities. Is there something like the code above?
Is it possible to set default values for some struct member? I tried the following but, it'd cause syntax error: typedef struct { int flag = 3; } MyStruct; Errors: $ gcc -o testIt test.c test....
As far as i see it the answer is 'default' is optional, saying a switch must always contain a default is like saying every 'if-elseif' must contain a 'else'. If there is a logic to be done by default, then the 'default' statement should be there, but otherwise the code could continue executing without doing anything.
INNER JOIN is the default if you don't specify the type when you use the word JOIN. You can also use LEFT OUTER JOIN or RIGHT OUTER JOIN, in which case the word OUTER is optional, or you can specify CROSS JOIN.
On your computer, open Chrome. At the top right, select More Settings. Select Reset settings Restore settings to their original defaults Reset settings.
