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'this' cannot be used in a constant expression C++


By : Akshay Ravi
Date : September 15 2020, 11:00 PM
With these it helps I tried to run the bwlow code, but I am getting an error , The error is probably not on the line you mentioned but here
code :
int result[V];
std::vector<int> result(V);


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Is a glvalue integral constant expression a constant expression?


By : Armando Moran
Date : March 29 2020, 07:55 AM
I hope this helps . N4527 5.20 [expr.const]p3 ,
Is a an integral constant expression?
code :
int b[a]{};
static_assert(a,"");
switch(1){
  case a:
    ;
}

'Constant Expression Required' Error while keeping formal argument as a constant


By : sakurabadesu
Date : March 29 2020, 07:55 AM
should help you out In C++, const is not always constexpr. Back in the days, constexpr didn't exist, so the only way of having a compile time constant was to either use const with a literal, or to use enum, because both of these are easy for the compiler to check the value.
However, in C++11, we added constexpr, which guaranties that a constexpr variable has a value available at compile-time, and state that constexpr function can be evaluated aat compile time if all arguments are constexpr too.
code :
void display(const int const1=5)
{
    constexpr int const2 = 5;

    // ...
}
template<int const1 = 5>
void display()
{
    constexpr int const2 = 5;

    int array1[const1];
    int array2[const2];
}
// const1 is 5
display();

// const1 is 10
display<10>();

Get string length function (strlen) of a constant char array is not a constant expression


By : Yuusuke Takeuchi
Date : March 29 2020, 07:55 AM
wish help you to fix your issue Standard strlen is not constepxr, therefore, it cannot be used in constexpr context. However, GCC knows about strlen, therefore it is able to compute the length of the string, in some circumstances - even if it is not mandated/permitted by the standard.
If you are concerned only by arrays, you can use std::size to get their size:
code :
template <class T, std::size_t N>
constexpr std::size_t size(const T (&array)[N]) noexcept
{
    return N;
}

ASP.NET MVC4: An attribute argument must be a constant expression , typeof expression or array creation expression of an


By : Scott Cooper
Date : March 29 2020, 07:55 AM
hope this fix your issue First you create a Resource .resx file this will contain your localised strings.
When you declare the attribute you set the ResourceType argument. This causes the Name, ShortName and Description arguments to be used as a resource key instead of a value.

Why do we need the two definitions: integral constant expression and converted constant expression?


By : Gabriel Silva
Date : March 29 2020, 07:55 AM
this one helps. Both definitions are needed because there are things you can do with one but not the other. And no, not every integral constant expression is really a converted constant expression. For the obvious example, a converted constant expression prohibits narrowing conversions, but an integral constant expression doesn't.
Therefore I can't do this:
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