Suppose left and right are two pair<T1, T2objects.
Then
'left < right'
returns
left.first < right.first || !(right.first < left.first) && left.second
< right.second.
Suppose left.first is NOT less than right.first.
Then isn't the subexpression '!(right.first < left.first)' equivalent
to 'left.first == right.first' ?
Is there any reason for mentioning like '!(right.first < left.first)'
instead of
'left.first == right.first' ?
Kindly clarify.
Thanks
V.Subramanian 2 1391 su************* *@yahoo.com, India wrote:
Suppose left and right are two pair<T1, T2objects.
Then
'left < right'
returns
left.first < right.first || !(right.first < left.first) && left.second
< right.second.
Suppose left.first is NOT less than right.first.
Then isn't the subexpression '!(right.first < left.first)' equivalent
to 'left.first == right.first' ?
Is there any reason for mentioning like '!(right.first < left.first)'
instead of
'left.first == right.first' ?
operator< is generally the first operator defined for a given type,
since the standard containers and algorithms default to std::less.
operator==(a, b) is then defined as !(a < b) && !(b < a). Using
operator< in the first place will therefore tend to be a little more
efficient.
James Kanze wrote:
On Feb 4, 5:45 am, Jeff Schwab <j...@schwabcen ter.comwrote:
>subramanian10. ..@yahoo.com, India wrote:
>>Suppose left and right are two pair<T1, T2objects.
>>Then 'left < right' returns left.first < right.first || !(right.first < left.first) && left.second < right.second.
>>Suppose left.first is NOT less than right.first.
>>Then isn't the subexpression '!(right.first < left.first)' equivalent to 'left.first == right.first' ?
Not if one of the types involved doesn't define an operator==.
>>Is there any reason for mentioning like '!(right.first < left.first)' instead of 'left.first == right.first' ?
>operator< is generally the first operator defined for a given type, since the standard containers and algorithms default to std::less. operator==(a, b) is then defined as !(a < b) && !(b < a). Using operator< in the first place will therefore tend to be a little more efficient.
For many types, an operator== can be implemented at far less
runtime cost than an operator<.
I didn't say it couldn't be, just that it often isn't. operator== can
be defined in terms of operator<, but the reverse is not true; so, if
you need both, you either have to define operator== in terms of
operator<, or you have to implement them both manually and be careful to
keep them in sync. While the (typically) constant-factor speed-up
*might* be worth the increased risk of bugs, I would certainly wait for
a profiler to tell me so. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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