CODEWITHSUNDEEP
cstm32f4stm32cubeide8-bit
T1P0Z
T1P0Z

Reputation: 13

How to read a whole uint_8 variable?

I have a 8 bit variable received_control. This variable should store some 8 bit data. In my use case, I need be able to read all the bits, even those that are 0. Problem is that if my most significant bit value is 0, it won't get stored in my variable.

Variable in question:

  uint8_t received_control = (0 << 0) |
                             (1 << 1) |
                             (0 << 2) |
                             (1 << 3) |
                             (0 << 4) |
                             (1 << 5) |
                             (0 << 6) |
                             (0 << 7);

Output from debugger (via watcch expression)

Name : received_control
Details:42 '*'
Default:42 '*'
Decimal:42
Hex:0x2a
Binary:101010
Octal:052

I'm using a STM32F407 and I program it in STM32CubeIDE. I was curious if there is integrated solution within C, as the solutions I saw online were not suitable.

Upvotes: -4

Views: 112

Answers (2)

user30482
user30482

Reputation: 333

The problem appears to be that you're not using leading-zero formatting. See the documentation for printf() or whichever formatting function you're using. Most likely, you would need something along the lines of of printf("Binary: %08b\n", value).

Note that the binary format was non-standard until late 2024 ("C17" and below) when it was brought into the standard ("C23" and newer). This means that some compilers or implementations of the C standard libraries may not support the binary format specifier.

Upvotes: 2

Gerhardh
Gerhardh

Reputation: 12404

it won't get stored in my variable

This is not true. You have some misconception about how variables work. An 8 bit integer variable always holds 8 bits. No matter what value you assign to it, there will be 8 bits.

If your statement was correct, what value would you assume, these bits should hold after your assignment? As they are present in the variable, they must have some value. And for integer variables, the more significant bits are simply set to 0 or 1 depending on signedness of the variable and the assigned value.

What you observed is not related to storing data in a variable but it is about a visual representation of that value. You told printf to print an integer variable in binary representation. Normally, printf takes as few characters for printing as possible. If you print as hexadecimal, it will not print 000000002A but only 2A. Unless for some formatting purposes, these leading 0 don't serve any purpose. Therefore you must explicitely ask for it if you want to get more than minimum digits.

Upvotes: 2

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