CODEWITHSUNDEEP
cstruct
user9993
user9993

Reputation: 6170

Convert struct field to unsigned char string

I'm on an embedded system so do not have access to a most of the standard library. I have a struct which contains char values.

I have a simple print function which simply outputs an unsigned char string to an attached screen. It does not support format specifiers like printf does.

This is the struct:

typedef struct my_options{
    char test;
} my_options;

And this is where I'm trying to output the value:

 struct my_options options;
 print(options.test);  //Here I get "implicit conversion of int to ptr"

How do I achieve this?

Upvotes: 2

Views: 701

Answers (4)

Elias Van Ootegem
Elias Van Ootegem

Reputation: 76415

your member test if of the type char, where the print function expects an argument of the type const char * (assuming the const bit here, but that's what I'd expect, this as an asside). Passing the address of test then would seem like the appropriate solution, but is it?

No, of course it isn't. There is no absolute guarantee that the next byte after test will be '\0' (a string terminating char). What you, then, ought to do is create a wrapper string:

char char_wrapper[2] = {};//initializes according to standard
//but as Lundin pointed out, self-documenting code is important:
char_wrapper[0] = options.test;?
char_wrapper[1] = '\0';//explicit, so it's clear what this code does
print(char_wrapper);

That should work just fine.
You can, of course, write this as a one-liner:

char char_wrapper[2] = {options.test, '\0'};//same as before, only in 1 statement
print(char_wrapper);//print "string"

That should do it, really. You don't even have to explicitly write the terminating char, since the standard specifically states:

An array of character type may be initialized by a character string literal, optionally enclosed in braces. Successive characters of the character string literal (including the terminating null character if there is room or if the array is of unknown size) initialize the elements of the array.

6.7.8 Initialization cf p. 138, semantics, point 14

Be that as it may, I'd still prefer to browse the web, or just set about writing your own minor implementation of printf so you can use format specifiers. Heck, it's one of the first exercises in the K&R book, and there's tons of solutions floating about on the net. check those out, and adapt them to your specific needs.
Or, perhaps define print to accept a size_t argument, to specify how many chars you want to pass to the output stream. and call it like so print(options.test, 1);

Upvotes: 1

user2591612
user2591612

Reputation:

print is looking for a char*. You are passing an char which can also be represented as an int. Thus, the function is trying to implicitly convert an int to a pointer as it is telling you. 

char ptr[2];
ptr[0] = options.test;
ptr[1] = '\0';

Would wrap the char into a char array, which in C will decay into a pointer when you pass it to a function.

Upvotes: 0

unwind
unwind

Reputation: 399919

Create a temporary 2-character long string that has the character to print, and then the terminator. Then pass that string to your print() function:

void print_options(const struct my_options *opt)
{
  char tmp[] = { opt->test, '\0' };
  print(tmp);
}

Upvotes: 2

Chris Middleton
Chris Middleton

Reputation: 5934

Create a char array to hold your char and then print it:

char wrapper[2];
wrapper[0] = options.test;
wrapper[1] = '\0';
print(wrapper);

Upvotes: 3

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