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How do i handle this situation C++ sizeof problem


By : TalZ
Date : September 23 2020, 04:00 AM
seems to work fine If I have understood correctly you are trying to get the size of the passed array in the function.
Do it the following way
code :
template <size_t N>
void f( int* ( &b )[N] )
{
    cout << sizeof( b ) << endl;
}
void f(int* b[], size_t n )
{
    printf( "%zu\n", n * sizeof( *b ) );
}
f(a, sizeof( a ) / sizeof( *a ) );


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Any situation where long is preferred over int when sizeof(int)=sizeof(long)


By : Venkata Jagannath Ta
Date : March 29 2020, 07:55 AM
wish of those help Don't use either of them. In modern C (starting with C89) or C++ there are typedef that have a semantic that helps you to write portable code. int is almost always wrong, the only use case that I still have for that is the return value of library functions. Otherwise use
bool or _Bool for Booleans (if you have C++ or C99, otherwise use a typedef) enum for applicative case distinction size_t for counting and indexing unsigned types when you use integers for bit patterns ptrdiff_t (if you must) for differences of addresses

C/C++: sizeof(short), sizeof(int), sizeof(long), sizeof(long long), etc... on a 32-bit machine versus on a 64-bit machin


By : Sapigokil Sama
Date : March 29 2020, 07:55 AM
This might help you Looks right to me. In c/c++ int isn't specifically defined in terms of bit-size. When creating a project you can select a "console application". VS2012 still supports C, but they mostly lump projects into C/C++. There is a compiler option (/TC I think) which will force the compiler into C compliance. By default it will imply the language by the file extension. MS C support isn't ideal, it doesn't include stdbool.h for instance.
If you want to control the bit size of your data you can use stdint.h which contains exact width int datatypes.

Is there a situation when an array is not converted to pointer except of sizeof?


By : facecream07
Date : March 29 2020, 07:55 AM
help you fix your problem
Is there a situation when an array is not converted to pointer except of sizeof?
code :
void foo(int **a) { }

int b[5][10];

foo(b);  // Compilation error; b becomes &b[0], which is pointer type,
         // and thus doesn't then become &&b[0][0]
int a[5];
int b[5];

a = b;   // Compilation error; a becomes &a[0], which isn't an lvalue

Where in the C++ Standard does it say that sizeof(wchar_t) <= sizeof(long) and sizeof(bool) <= sizeof(long)?


By : Koco
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further It's hard to imagine a sizeof(wchar_t)>N && sizeof(long)<=N implementation for any value of N, let alone one on which anyone would consider using wchar_t. That said:
In fact, I think it's wrong. 4.5 "Integral promotions" p2 explicitly allows for the possibility:

Does C correctly handle sizeof(...) and sizeof ... in this case?


By : pradip
Date : March 29 2020, 07:55 AM
I hope this helps you . In both cases, the last rofl is the variable name. A variable name is in scope as soon as it appears; and for the remainder of the current scope, that identifier in an ordinary context(*) always means the variable name.
The sizeof operator does not introduce any special cases for name lookup. In fact, there are no language constructs that will use the hidden meaning of an identifier.
code :
typedef int A;      // "A" declared as ordinary identifier, meaning a type name

struct A { A A; };  // "A" declared as struct tag and member name -- OK as these are three different name spaces. Member type is "int"

A main()            // int main() - ordinary context
{
    struct A A();   // "A" declared as ordinary identifier, meaning a function name; hides line 1's A
    // A C;         // Would be error: ordinary A is a function now, not a typedef for int
    struct A B;     // OK, struct tags have separate name space
    A:+A().A;       // OK, labels and struct members have separate name space, calls function
    goto A;         // OK, label name space
}
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