Why const is ignored in auto keyword on constant reference
Date : March 29 2020, 07:55 AM
I hope this helps . Because the type deduction rules for auto deduce to an object type. You are copy initializing a new int from the one referenced by b. That's by design. We want to create new objects without explicitly specifying their type. And more often than not, it's a new object that is a copy some other object. Had it deduced to a reference, it would defeat the intended purpose. decltype(auto) d = b; // d and b now refer to the same object
d = 15; // And this will error because the cv-qualifiers are the same as b's
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How to set a variable to constant defined using "const" keyword?
Tag : php , By : Phil Austin
Date : March 29 2020, 07:55 AM
seems to work fine I need to store an array to a constant. I am using PHP 5.6, so cant use define. I am getting that data from database. , You can save array like this define('VAR_DATA', implode(',',array(1, 2, 3)));
$array = explode(',',VAR_DATA);
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Defining constant in c with const keyword
Tag : c , By : CSCI GOIN KILL ME
Date : March 29 2020, 07:55 AM
I wish this helpful for you Using the const type qualifier doesn't make something a constant. A constant in C has its own definition. See § 6.7.3 ¶ 6 of the C11 standard for a description of the const keyword: float user_array[] = {5.1, 7.2, 5.1, 8.45, 23.0, 67.123, 5.1};
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Why declare a constant pointer using the const keyword when the reference (const pointer) is available?
Tag : cpp , By : user177837
Date : March 29 2020, 07:55 AM
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const keyword and constant expressions in C99 and C11
Date : March 29 2020, 07:55 AM
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