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Why is the size of the data type different when the value is directly passed to the sizeof operator?


Why is the size of the data type different when the value is directly passed to the sizeof operator?

By : ayambakar
Date : September 23 2020, 01:00 PM
around this issue Constants, like variables, have a type of their own:
6.5 : A floating point constant of type double 90000 : An integer constant of type int (if int is 32 bits) or long (if int is 16 bits) 'A' : A character constant of type int in C and char in C++
code :


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Opposite behavior of Marshal.SizeOf and sizeof operator for boolean and char data types in C#

Opposite behavior of Marshal.SizeOf and sizeof operator for boolean and char data types in C#


By : zhengzy
Date : March 29 2020, 07:55 AM
I hope this helps . Sorry, you really do have to consider the technicalities to make sense of these choices. The target language for pinvoke is the C language, a very old language by modern standards with a lot of history and used in a lot of different machine architectures. It makes very few assumptions about the size of a type, the notion of a byte does not exist. Which made the language very easy to port to the kind of machines that were common back when C was invented and the unusual architectures used in super-computers and digital signal processors.
C did not originally have a bool type. Logical expressions instead use int where a value of 0 represents false and any other value represents true. Also carried forward into the winapi, it does use a BOOL type which is an alias for int. So 4 was the logical choice. But not a universal choice and you have to watch out, many C++ implementations use a single byte, COM Automation chose two bytes.
float data type(5.0) being assigned 8 bytes rather that 4 using sizeof operator in C

float data type(5.0) being assigned 8 bytes rather that 4 using sizeof operator in C


By : Anto
Date : March 29 2020, 07:55 AM
may help you . Because all floating point literals default to double. If you want it to be a float then use the f suffix as in 5.0f.
Read e.g. this floating point constant (literal) reference for more information.
Output data type of sizeof() operator

Output data type of sizeof() operator


By : user2360623
Date : March 29 2020, 07:55 AM
I wish this helpful for you The sizeof operator evaluates to a value of type size_t. This type is unsigned and typically larger than an int, which is why you get the warning.
Using the wrong format specifier to printf invokes undefined behavior. You can get away with this however for types smaller than int due to the rules of integer promotions in section 6.3.1.1p2 of the C standard:
code :
printf("size of long int is %zu\n", sizeof(mylint));
How can a sizeof operator calculate size of return type of a function?

How can a sizeof operator calculate size of return type of a function?


By : anup thakur
Date : March 29 2020, 07:55 AM
Hope that helps No, sizeof never evaluates its operands (except for C variable length arrays but, since this is a C++ question, that's irrelevant). From C++17 [expr.sizeof]:
The data type produced by the sizeof operator

The data type produced by the sizeof operator


By : MosheKri
Date : March 29 2020, 07:55 AM
wish of those help std::size_t is a typedef-name or type alias, that is a synonim of some other existing type; it's not a distinct type and exists only for the purpose of name lookup.
so, for example, you cannot overload/specialize/etc... against size_t VS its underlying type ...
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