Binary representation of a char array

Question

I have a union, that is 8 fields of uint8, 2 fields of int32, or one int64.

I am trying to save an array of 8 values there: 0, 0, 0, 1, 0, 0, 0, 1

There is a bug in the software that I am trying to solve unfortunately, I don't understand the binary representation of this array.

My union:

union {
    uint8_t u08[8];
    uint32_t u32[2];
    uint64_t u64;
} data;

My debugger output:

Name : u08
Details:"\0\0\0\001\0\0\0\001"
Default:0x4001e6a0 <ucHeap+66948>
Decimal:1073866400
Hex:0x4001e6a0
Binary:1000000000000011110011010100000
Octal:010000363240

Name : u32
Details:{1, 1}
Default:0x4001e6a0 <ucHeap+66948>
Decimal:1073866400
Hex:0x4001e6a0
Binary:1000000000000011110011010100000
Octal:010000363240

Name : u64
Details:4294967297
Default:4294967297
Decimal:4294967297
Hex:0x100000001
Binary:100000000000000000000000000000001
Octal:040000000001

My questions are:

why does a u64 have a different value than the others? as far as I understand, c saves space only for one variable of a union to store the value.
what does this binary representation of the uint8 array mean? Even if it was ASCII (which it isn't, as I am sending direct values, not chars), then '0' is 110000 and '1' is 110001

I'm using MPC5748G from NXP with S32DS (eclipse based IDE) for the debugging.

Something else is wrong in your program and/or your debugger. Is it an in-process debugger, so it could have been affected by corruption in the program? Edit the question to provide a minimal reproducible example. — Eric Postpischil
– Eric Postpischil, Commented Dec 29, 2021 at 12:56
For the stated inputs to u08 it looks as though the output for u64 is the only correct one. — Weather Vane
– Weather Vane, Commented Dec 29, 2021 at 13:45

0___________ · Accepted Answer · 2021-12-29 14:24:08Z

What you see is not you think you see. Everything is correct there. u08 & u32 are arrays. u64 is an integer value and the debugger displays different information

Name : u08
Details:"\0\0\0\001\0\0\0\001"    -- actual values stored in the array
Default:0x4001e6a0 <ucHeap+66948> -- address of the array u08
Decimal:1073866400                -- same address in different bases
Hex:0x4001e6a0
Binary:1000000000000011110011010100000
Octal:010000363240

Name : u32
Details:{1, 1}                     -- actual values stored in the array
Default:0x4001e6a0 <ucHeap+66948>  -- address of the array u32
Decimal:1073866400                 -- same address in different bases
Hex:0x4001e6a0
Binary:1000000000000011110011010100000
Octal:010000363240

Name : u64
Details:4294967297                 -- value stored un the u64 integer
Default:4294967297                 -- same value in different bases
Decimal:4294967297
Hex:0x100000001
Binary:100000000000000000000000000000001
Octal:040000000001

{0, 0, 0, 1, 0, 0, 0, 1} is in big endian representation:

0x00000001 (1) in u32
0x0000000100000001 (0x100000001 without zeroes) in u64

if you want to see the same information for all the members you need to make the last member an array:

union {
    uint8_t u08[8];
    uint32_t u32[2];
    uint64_t u64[1];
} data;

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