C++ array with command line variable - c++

Question

My array looks something like this;
unsigned char send_bytes[] = { 0x0B, 0x11, 0xA6, 0x05, 0x00, 0x00, 0x70 };
One of the values is a variable that can change all the time.. so I tried something like this;
const char* input = "0x05";
unsigned char send_bytes[] = { 0x0B, 0x11, 0xA6, input, 0x00, 0x00, 0x70 };
When I compile I get a warning;
warning: initialization makes integer from pointer without a cast
I am a little confused on the conversion I need to do.. since the array has hex strings in it.. and the input string is a char..

Answer 1

in the first line you are declaring a pointer to const char, and initializing to the beginning of string "0x05", that's fine, but it is not the thing you are trying to do.
in the second line, you try to initialize the fourth array element (an unsigned char) with the value of the pointer you assigned to the input variable in the first line. The compiler says you are pretending to embed a pointer value (the address of "0x05" string) into a char variable, so that's why it complained. And also it is not what you intend.
also, take into account that if you are using binary data (from the fact you are initializing arrays with hex numbers) you had better to use unsigned char for binaries, as signed char is valid only for -128 to +127 values, you can expect some more unpredictable behaviour. Perhaps, a declaration typedef unsigned char byte; can do things easier.
typedef unsigned char byte;
byte send_bytes[] = { 0x0b, 0x11, 0xa6, 0x00, 0x00, 0x00, 0x70 };
byte &input = send_bytes[3]; /* input is an alias of send_bytes[3] */
BR,
Luis

Answer 2

Maybe explaining exactly what const char* input = "0x05"; does will clear things up for you.
First the compiler computes the string data and creates it as a static object:
const char data[5] = { 0x30, 0x78, 0x30, 0x35, 0x0 };
Then your variable is initialized:
const char *input = &data[0];
Note that input is a pointer with a value that depends entirely upon the location the compiler chooses to store the string data at, and has nothing to do with the contents of the string. So if you say char c = input; then c basically gets assigned a random number.
So you should be asking yourself "Where is the value 0x05 that I want to store in the send_bytes array?" In your code it's encoded as text, rather than as a number that your program can use directly. You need to figure out how to convert from a string of symbols following the hexadecimal scheme of representing numbers into C++'s native representation of numbers.
Here are a couple hints. Part of the operation involves associating values with each digit symbol. The symbol '0' is associated with the value zero, '1' with the value one, and so on, according to the usual hexadecimal system. Second, once you can get the associated value of a symbol, then you can use those values in some basic arithmetic operations to figure out the value of the number represented by the whole string of symbols.
For example, if you have the symbols '1' '2' and 'a', in that order from left to right then the arithmetic to compute what number is represented is 1 * 16 * 16 + 2 * 16 + 10.

Answer 3

The error string is pretty much telling you exactly what's wrong.
input is of type const char* (a pointer to a const char), whereas your array send_bytes is of type unsigned char[] (an array of unsigned chars).
First, signed and unsigned values are still different types, though your error message isn't referring to that specifically.
In reality, your input value isn't a string (as there is no true string type in C++), but a pointer to a character. This means that the input string doesn't hold the byte x05, but rather the bytes {x30, x78, x30, x35, x00}.
The compiler is saying Hey, I've no idea what you're trying to do, so I'm just converting the address that string I don't understand (input) to an unsigned char and adding it to the array.
That means if the string "0x05" starts at location 0xAB, your array will ultimately contain { 0x0B, 0x11, 0xA6, 0xAB, 0x00, 0x00, 0x70 }.
You're going to either have to convert from a string to an integer using a radix of 16, or just not use a string at all.
I'd also recommend reading up on pointers.

Answer 4

The array doesn't have "hex strings" in it - if they were, they would be enclosed in quotation marks, like all strings.
The literals are integers written in hexadecimal notation, and equivalent to
unsigned char send_bytes[] = { 11, 17, 166, input, 0, 0, 112 };
Since it's an array of unsigned char you should put an unsigned char there:
unsigned char input = 0x05;
unsigned char send_bytes[] = { 0x0B, 0x11, 0xA6, input, 0x00, 0x00, 0x70 };

Answer 5

You had better to put in your code:
unsigned char send_bytes[] = { 0x0b, 0x11, 0xa6, 0x00, 0x00, 0x00, 0x70 };
unsigned char &input = send_bytes[3]; /* input is an alias of send_bytes[3] */
this way you can do things like:
input = 0x26;
send_packet(send_bytes);