_Aion_
Nothing Else
Hello.
Someone can explain me what difference between
(int64_t)0 to (int64_t)1?
Someone can explain me what difference between
(int64_t)0 to (int64_t)1?
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C++ One doubt
- Thread starter _Aion_
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short explanation: just the value changes, and the cast is redundant.
In every strongly typed language, (A) value generally is called an explicit cast.
It means fitting the bits value (which has a given type, lets say it has the type B for the sake of the explanation.) into the type A representation of the same bits value.
For example, uint32 and int32 have the same number of bits available to allocate values, but they are different representation of those bits. In other words, the number of different values they can allocate is the same, but they represent different ranges.
Since int32 uses one bit for the sign (+ -) it may only hold values from –2,147,483,648 to 2,147,483,647
Uint has a range of 0 to 4,294,967,295.
EDIT: I just saw this is a C++ thread. I'm most familiar with C#, but the same explanation holds for strongly typed language so...
It means fitting the bits value (which has a given type, lets say it has the type B for the sake of the explanation.) into the type A representation of the same bits value.
For example, uint32 and int32 have the same number of bits available to allocate values, but they are different representation of those bits. In other words, the number of different values they can allocate is the same, but they represent different ranges.
Since int32 uses one bit for the sign (+ -) it may only hold values from –2,147,483,648 to 2,147,483,647
Uint has a range of 0 to 4,294,967,295.
EDIT: I just saw this is a C++ thread. I'm most familiar with C#, but the same explanation holds for strongly typed language so...
Last edited:
Shyzoul
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c/c++ is not 'strongly' typed language because pointers can bypass typing system, i think they call it 'weakly' typed language
it's not redundant, if you're dealing with bitwise operations then you could need a 8byte long variable otherwise it will be overflowed, since by default integers have 4 bytes long.
nothing both are just numbers like any other, if you meant a type conversion '(int64_t)' would be the same as '0LL' or '(long long)0' which mean a variable capable to represent numbers up to 8 bytes long depending on your system, it can be 4 bytes long if your machine doesn't support 64bit-long instructions extensions.