The difference between size of datatype and sizeof(data type)

Question

I was learning C++ and come across the following question. I'm just a beginner and I got confused. Isn't sizeof() function supposed to return the size of the datatype? Why could a data object has different size from its sizeof()? I don't understand the explanation of the answer.

Suppose in a hypothetical machine, the size of char is 32 bits. What would sizeof(char) return?

a) 4

b) 1

c) Implementation dependent

d) Machine dependent

Answer:b

Explanation: The standard does NOT require a char to be 8-bits, but does require that sizeof(char) return 1.

Answer 1

The sizeof operator yields the size of a type in bytes, where a byte is defined to be the size of a char. So sizeof(char) is always 1 by definition, regardless of how many bits char has on a given platform.

This applies to both C and C++.

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From the C11 standard, 6.5.3.4

2. The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand....
   

Then,

4. When sizeof is applied to an operand that has type char, unsigned char, or signed char, (or a qualified version thereof) the result is 1.

From the C++11 standard, 5.3.3

1. The sizeof operator yields the number of bytes in the object representation of its operand. The operand is either an expression, which is an unevaluated operand (Clause 5), or a parenthesized type-id.... ... sizeof(char), sizeof(signed char) and sizeof(unsigned char) are 1.

(emphasis mine)

Answer 2

You're just confused with the difference between bytes and octets.

A byte is the size of one character. This yields to the always true sizeof(char) == 1, because sizeof return the size in bytes

While an octet consists out of 8 bits.

On almost all modern platforms, the size of a byte is coincidentally the same as of an octet. That's the reason why it's a common error to mix them up, even book authors and professors are doing this.