CODEWITHSUNDEEP
c++integer
tej-kweku
tej-kweku

Reputation: 433

Is there an actual 8-bit integer data type in C++

In c++, specifically the cstdint header file, there are types for 8-bit integers which turn out to be of the char data type with a typedef. Could anyone suggest an actual 8-bit integer type?

Upvotes: 32

Views: 95103

Answers (2)

winapiadmin
winapiadmin

Reputation: 149

There are __int8, __int16, __int32, __int64 default data type (this also applied for C programming language), without using the <cstdint> header and there aren't __int128 for Microsoft-specific (lot of answers is at here, and even intrinsic function __shiftleft128, __shiftright128 use two __int64 for example).

int8_t is char are also applied for unsigned types, for example, uint8_t are unsigned char.

Upvotes: 1

dlmeetei
dlmeetei

Reputation: 10391

Yes, you are right. int8_t and uint8_t are typedef to char on platforms where 1 byte is 8 bits. On platforms where it is not, appropriate definition will be given.

Following answer is based on assumption that char is 8 bits

char holds 1 byte, which may be signed or unsigned based on implementation.

So int8_t is signed char and uint8_t is unsigned char, but this will be safe to use int8_t/uint8_t as actual 8-bit integer without relying too much on the implementation.

For a implementer's point of view, typedeffing where char is 8 bits makes sense.

Having seen all this, It is safe to use int8_t or uint8_t as real 8 bit integer.

Upvotes: 39

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