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c++unsignedbit-representation
Luiz Fernando
Luiz Fernando

Reputation: 195

C++ and unsigned types

I'm reading the C++ Primer 5th Edition, and I don't understand the following part:

In an unsigned type, all the bits represent the value. For example, an 8-bit unsigned char can hold the values from 0 through 255 inclusive.

What does it mean with "all the bits represent the value"?

Upvotes: 2

Views: 211

Answers (5)

xskxzr
xskxzr

Reputation: 13040

The value of an object of trivially copyable type is determined by some bits in it, while other bits do not affect its value. In the C++ standard, the bits that do not affect the value are called padding bits.

For example, consider a type with 8 bits where the last 4 bits are padding bits, then the objects represented by 00000000 and 00001111 have the same value, and compare equal.

In reality, padding bits are often used for alignment and/or error detection.

Knowing the knowledge above, you can understand what the book is saying. It says there are no padding bits for an unsigned type. However, the statement is wrong. In fact, the standard only guarantees unsigned char (and signed char, char) has no padding bits. The following is a quote of related part of the standard [basic.fundamental]/1:

For narrow character types, all bits of the object representation participate in the value representation.

Also, the C11 standard 6.2.6.2/1 says

For unsigned integer types other than unsigned char, the bits of the object representation shall be divided into two groups: value bits and padding bits (there need not be any of the latter).

Upvotes: 1

MSalters
MSalters

Reputation: 179779

This is mostly a theoretical thing. On real hardware, the same holds for signed integers as well. Obviously, with signed integers, some of those values are negative.

Back to unsigned - what the text says is basically that the value of an unsigned number is simply 1<<0 + 1<<1 + 1<<2 + ... up to the total number of bits. Importantly, not only are all bits contributing, but all combinations of bits form a valid number. This is NOT the case for signed integers. Therefore, if you need a bitmask, it has to be an unsigned type of sufficient width, or you could run into invalid bit patterns.

Upvotes: 0

Aganju
Aganju

Reputation: 6395

For example, one byte contains 8 bits, and all 8 bits are used to counting up from 0.

For unsigned, all bits zero = 00000000 means 0, 00000001 = 1, 00000010 = 2, 00000011 = 3, ... up to 11111111 = 255.
For a signed byte (or signed char), the leftmost bit means the sign, and therefore cannot be used to count. (I am optically separating the leftmost bit!) 0 0000001 = 1, but 1 0000001 = -1, 0 0000010 = 2, and 1 0000010 = -2, etc, up to 0 1111111 = 127, and 1 1111111 = -127. In this example, 1 0000000 would mean -0, which is useless/wasted, so it can mean for example 128.

There are other ways to code the bits into numbers, and some computers start from the left instead from the right. These details are hardware specific, and not relevant to understand 'unsigned', you only need to care about that when you want to mess in the code with the single bits (not recommended).

Upvotes: 0

John Burger
John Burger

Reputation: 3672

You should compare this to a signed type. In a signed value, one bit (the top bit) is used to indicate whether the value is positive or negative, while the rest of the bits are used to hold the value.

Upvotes: 1

Jovibor
Jovibor

Reputation: 789

It means that all 8 bits represent an actual value, while in signed char only 7 bits represent actual value and 8-th bit (the most significant) represent sign of that value - positive or negative (+/-).

Upvotes: 0

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