Why is not ampersand needed in the following code?

Why is not ampersand needed in the following code?

By : Tijo Thomas
Date : September 23 2020, 02:00 PM
should help you out Because you have declared a as an array, an expression using that variable's name, on its own, will usually 'decay' to a pointer to the first element of the array (What is array decaying? - but see also the excellent comment added by Eric Postpischil for exceptions). This is similar to using a char[] string where, as you correctly noted, you don't need to use the & operator when passing it as an argument to scanf.
Adding i to this 'base address' of the a array will give the address of (i.e. a pointer to) the 'i'th element of the array.
code :

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EF CTP4: For Code Only, no database generation needed, how much DB info is needed?

EF CTP4: For Code Only, no database generation needed, how much DB info is needed?

By : user2265886
Date : March 29 2020, 07:55 AM
To fix the issue you can do Yes, the EF does need information on string field lengths, foreign keys, etc., in the model. For example, if a DB FK has a cascade, the EF needs to know that so that it doesn't force you to manually delete detail records; if the EF is aware of the cascade it will let the DB handle that. Similarly, if the EF is aware that a key is store-generated (e.g., auto-incremented), it won't complain when you don't set it on a new record, because it will presume that the DB will do that.
However, the code-only approach takes a "convention over configuration" approach. You don't have to specify values which the EF can guess. You can read about those here.
Ruby: replace in string ampersand to backslash with ampersand

Ruby: replace in string ampersand to backslash with ampersand

By : Christian Gahlert
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Simple question, but I can't find simple answer
code :
2.0.0-p0 :018 > "a&b".gsub('&', '\\\&')
 => "a\\&b"
2.0.0-p0 :019 > _.chars.to_a
 => ["a", "\\", "&", "b"]
2.0.0-p0 :023 > puts "a&b".gsub('&', '\\\&')
ampersand in code, what does it do

ampersand in code, what does it do

By : mini cpp
Date : March 29 2020, 07:55 AM
seems to work fine The ampersand is a bitwise AND. It means that you're comparing on the bit level. For each bit, the resulting bit is 1 if and only if the 2 incoming bits are 1.
code :
1 & 2 = 0
2 & 3 = 2
7fffffff = (0111) (1111) (1111) etc...
Batch script (bat or cmd) - extra spaces before ampersand (or automatically escape ampersand)

Batch script (bat or cmd) - extra spaces before ampersand (or automatically escape ampersand)

By : dro
Date : March 29 2020, 07:55 AM
hope this fix your issue Note: I understand the ^ symbol can be used to escape & symbol in batch scripts. But I need the escaping to be done automatically, as the paths are constructed at run-time. The sample script below is merely to show the problem and it is not the actual script. Please read my explanation after script output. , Enable delayed expansion and try this:
code :
Set "test1=c:\file&folder\"
Set "test2=!test1!currentPath"
echo !test2!
set PATH=C:\Another\Folder;!PATH!
Ampersand or no ampersand before string variable name in scanf() and printf()?

Ampersand or no ampersand before string variable name in scanf() and printf()?

By : Pankaj Vaidhya
Date : March 29 2020, 07:55 AM
this one helps. str1 gives a char pointer char* to the first element, which will be allocated at the first address in the array. &str1 gives an array pointer to the array as whole, of type char(*)[100]. The address of the array is always the same as the address of the first element. By using the %s format specifier you tell printf to treat the passed pointer as a char*. So no matter which pointer type you pass, it will get converted to this type. As it happens, this particular pointer conversion from char(*)[100] to char* has to yield the very same address, so the code works no matter which pointer type you use.
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