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I'm working with a Zedboard and I'm printing to the screen memory addresses of consecutive 32-bit float numbers.

So the print generates this:

Result 0: 5374.557617
Memory is A000000
Result 1: 5474.651233
Memory is A000004
Result 2: 5574.557129
Memory is A000008
 ....

So memory is incremented 4 each time.I can't understand why a memory with 4 more bits can keep a 32 bit number. The only way this makes sense is is hexadecimal is convertible to bytes and so +4 bytes=+32 bits.

However this isn't the way I learned, everyone says that hexadecimal shall me converted to bits

Surprisingly I haven't found anything on the Internet about this question in particular.

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    \$\begingroup\$ RAM is very, very, very rarely addressed in such a way as to specify bit addresses. Byte addresses are the norm. \$\endgroup\$
    – brhans
    Apr 20, 2015 at 12:23

1 Answer 1

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So memory is incremented 4 each time.

Nope. The address is incremented by 4. RAM Memory is typically addressed byte-wise, that means there is one byte of data at each address.

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  • \$\begingroup\$ Ok, thank you for the answer. Just to clarify, when I increment the address like this: 0x40400000 + 0x58 am I incrementing 58 bytes? \$\endgroup\$
    – João Pereira
    Apr 20, 2015 at 13:20
  • \$\begingroup\$ Nope, you are incrementing by 88 Bytes. Remember that 0x is hex prefix. \$\endgroup\$
    – Turbo J
    Apr 20, 2015 at 17:29
  • \$\begingroup\$ Oh ok. So if i do 0x40400000 + 58 i'll be incrementing 58 bytes but if I do 0x40400000 + 0x58 i'll be incrementing (58)_16=(88)_10 bytes. Correct? \$\endgroup\$
    – João Pereira
    Apr 20, 2015 at 21:33

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