大小小于变量的指针类型

Question

Working with embedded systems, in order to have more resolution in a incremental sequence, I have two variables, one always following the other.

Specifically, I set a goal value using a 8 bits variable, but to go from one point (current value) to another I do it using 32 bits steps.

For example (that is a stupid example, but it just to show how I want to use it, in my code there are some temporizations which require the 32 bits varaibles to allow a slow change):

/* The variables */
char goal8bits;         // 8 bits
long int current32bits; // 32 bits    
char current8bits;      // 8 bits    
long int step32bits;    // 32 bits

/* The main function (in the real code that is done periodically with a specific period) */
current32bits = CONVERT_8BITS_TO_32BITS(current8bits);  // E.g: 0xAB -> 0xABABABAB
if (goal8bits < current8bits) {
   current32bits += step32bits;
}
current8bits = CONVERT_32BITS_TO_8BITS(current32bits);  // E.g: 0x01234567 -> 0x01

/* Other parts of the code */
I use current8bits to know the current value in the middle of a transition.

My question is if I can use a char pointer and make it point to the 32 bits variable one, so I do not need to update it each time I change it. The previous example will look like this:

/* The variables */
char goal8bits;         // 8 bits
long int current32bits; // 32 bits
char *current8bits = (char *)&current32bits;      // Pointer to 8 bits
long int step32bits;    // 32 bits

/* The main function (in the real code that is done periodically with a specific period) */
if (goal8bits < *current8bits) {
   current32bits += step32bits;
}

/* Other parts of the code */
I will use *current8bits to know the current value in the middle of a transition.

Do you see any problem in doing that? Can it lead to a problem wih endianism?

Thank you!

Answer 1

Yes, it is endian dependent code, to make it portable you can use a mask and the left shift operator:

uint8_t goal8bits = 0x01;               // 8 bits
uint32_t current32bits = 0x01234567;    // 32 bits
uint32_t step32bits = 1;                // 32 bits

if (goal8bits < ((current32bits & 0xFF000000) >> 24)) {
    current32bits += step32bits;
}

Answer 2

If you know the endianless of your system, and it is static you have to select from

char *current8bits = (char *)&current32bits;

or

char *current8bits = (((char *)&current32bits)+3);

If you have to test it, and your system cannot give you such of info you can derive it at application startup

uint32_t temp = 0x01020304;
uint8_t *temp2 = (uint8_t *)(&temp);
if (*temp2 == 0x01)
{
   char *current8bits = (char *)&current32bits;
}
else
{
   char *current8bits = (((char *)&current32bits)+3);
}

Another good solution is the top-voted and checked-as-answered answer HERE .