c如何使用memcpy将十六进制值复制到数组中

Question

I am working with C , and I have a uint8_t array of hex values and I want to add another value to it via memcpy dynamically eg

I have an array: uint8_t sample['0x23', '0x34',...] etc ...and I want to copy another Hex character into the array later on. How is this done? If this question has been answered before, sorry, I couldn't find it. I am new to C , and these pointers are giving me hell. Thanks in advance.

Answer 1

You could write simply

sample[4] = '\x0A';

If you want to use memcpy then the valid code will look like

memcpy( &sample[4], "\x0A", 1 );

or

memcpy( sample + 4, "\x0A", 1 );

That is you need to use string literal instead of character constant.

Answer 2

I wonder if what is really being asked is how to add a new value to the end of the array.

In C, there is no "array" class, so you must either manage the memory and associated pointers yourself, or use a pre-defined function library that does it all for you. Let's assume the former ie the programmer is doing the management themselves.

Before you can assign or copy a new value into your array, you'll need to ensure there is enough space. One way to manage this follows:

int     array_size = 10;
uint8_t sample[array_size];
uint8_t i = 4;    /* assuming array index cannot exceed 255 */

if (i < array_size) {
    sample[i] = 0x0a;
}

The key here is that your sample array occupies just 10 bytes of memory and you need to make sure that your accesses to the sample array do not fall outside that bound.

If indeed the desire is to append a value to the end of the array, then another variable will have to be added to keep track of the last insertion point/index and use that for i in the example.