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Modifying non-const char array that is referred to by const char array


Modifying non-const char array that is referred to by const char array

By : jrembold
Date : October 23 2020, 06:10 PM
around this issue The key here is "an object defined with a const-qualified type". What matters is how you define (IOW, "allocate memory for") the object being pointed at. If the "object" was not "created" as const, then it does not matter how many const or non-const pointers and references you use - you can still modify it.
So,
code :
const int i = 0;
int *p = (int*)&i;
*p = 1;
int i = 0;
const int *cp = &i;
int *p = (int*) cp;
*p = 1;
const int *cp = new const int(0);
int *p = (int*) cp;
*p = 1;


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I got this warning: sizeof on array function parameter will return size of 'const char *' instead of 'const char []'

I got this warning: sizeof on array function parameter will return size of 'const char *' instead of 'const char []'


By : Julian Carlos Tapia
Date : March 29 2020, 07:55 AM
wish helps you That's because const char source[] in argument position is just syntactic sugar for const char *source. See, e.g., Steven Summit's C notes.
In this particular case, you'll want to call strlen. When not dealing with strings, pass the size of the array as a separate argument.
g++ treats returned string literal as const char pointer not const char array

g++ treats returned string literal as const char pointer not const char array


By : Pradeep Thiyagarajan
Date : March 29 2020, 07:55 AM
Does that help It appears that this is a bug affecting several versions of gcc which has been reported over and over again, most recently about a month ago against the most recent version, 4.8.2. See http://gcc.gnu.org/bugzilla/show_bug.cgi?id=24666
Initializing to a dynamically allocated const char array from a const char array

Initializing to a dynamically allocated const char array from a const char array


By : Alain Cheng
Date : March 29 2020, 07:55 AM
this will help You cannot copy-initialize an array from another array directly, hence the error. You either need to manually copy the elements (or use std::copy), or, better, if you want a "copy"-able array use std::array instead.
But, as mentioned in the comments, the best probably is to use an std::string here, and use its std::string::c_str() member function in case you need to pass around const char* pointers to old C-like interfaces.
In C++, why overload a function on a `const char array` and a private struct wrapping a `const char*`?

In C++, why overload a function on a `const char array` and a private struct wrapping a `const char*`?


By : David Croffutt
Date : March 29 2020, 07:55 AM
hop of those help? The author is trying to help you avoid accidental performance problems that happen when you repeatedly hash strings. Since hashing strings is expensive, you probably want to do it once and cache it somewhere. If they have an implicit constructor, you could hash the same string repeatedly without knowing or intending to do so.
So the library provides implicit construction for string literals, which can be computed at compile-time via constexpr but explicit construction for const char* in general since those can't generally be done at compile-time and you want to avoid doing it repeatedly or accidentally.
code :
void consume( hashed_string );

int main()
{
    const char* const s = "abc";
    const auto hs1 = hashed_string{"my.png"}; // Ok - explicit, compile-time hashing
    const auto hs2 = hashed_string{s};        // Ok - explicit, runtime hashing

    consume( hs1 ); // Ok - cached value - no hashing required
    consume( hs2 ); // Ok - cached value - no hashing required

    consume( "my.png" ); // Ok - implicit, compile-time hashing
    consume( s );        // Error! Implicit, runtime hashing disallowed!
                         // Potential hidden inefficiency, so library disallows it.
}
consume(hashed_string(hs1));
consume(hashed_string(hs2));
consume(hashed_string("my.png"));
const char s[100] = "abc";
consume( s );  // Compiles BUT it's doing implicit, runtime hashing. Doh.

// Decay 's' back to a pointer, and the library's guardrails return
const auto consume_decayed = []( const char* str ) { consume( str ); }
consume_decayed( s ); // Error! Implicit, runtime hashing disallowed!
const char array template vs char const* function overload

const char array template vs char const* function overload


By : Engr.Shah
Date : March 29 2020, 07:55 AM
With these it helps During the overload resolution usual functions are always preferred over function templates.
It means, that if there is an overload of foo which might accept appropriate arguments (even if the function template seems to be a better match), then the compiler will choose this overload.
code :
foo( 1, c_strg_1, "bla" );
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