Why is 0x00000100 = 256?

Question

Shouldn't 0x00000100 = 4.

I understand that 0x00000001 = 1 since 2^0 and 0x00000010 = 2 since 2^1. What is wrong with my thinking?

initVariable(&variable1, "variable1", "1, 2, 3", 0x00000100);

assertIntegerEquals(variable1.address, 4); // 0x00000100 = 4?

My assertion fails because it says that 256 != 4

Answer 1

Numbers that begin with 0x are interpreted as hexadecimal (base 16) in C.

So 0x10 == 16, and 0x100 == 256, 0x10000 == 65536, etc.

Powers of two are:

• 2⁰ = 0x1
• 2¹ = 0x2
• 2² = 0x4
• 2³ = 0x8
• 2⁴ = 0x10
• 2⁵ = 0x20
• 2⁶ = 0x40
• 2⁷ = 0x80
• 2⁸ = 0x100

Answer 2

No, the 0x means hexadecimal (i.e. base-16) and not binary, which is what you seem to be confusing it with.

If you want to use binary literals in your code, then see this SO question, which mentions the gcc extension that allows 0b00000100 to be used to represent 4 in binary.

Answer 3

0x00000100 is in base-16 because 0x prefix means you are using hexadecimal notation.

So 0x00000001 = 1*16⁰ = 1, 0x00000010 = 1*16¹, 0x00000100 = 1*16² = 256 and 0x00000123 = 1*16² + 2*16¹ + 3*16⁰ = 256 + 32 + 3 = 291

To play with base-2, base-10 and base-16 notation, you can try this site: http://www.mathsisfun.com/binary-decimal-hexadecimal-converter.html