Storing Volatile Chars in an Array

Question

Disclaimer: I'm quite new to C and to programming in general so this question may show a lack of basic knowledge.

I'm trying to store a set of temperature readings of type 'volatile char' in an array. The temperature reading and array are declared respectively as follows:

volatile char TempCelsiusDisplay[] = "+abc.d C";
volatile char temps[60];

The rest of the code appears to work fine, until I try to store a temperature value in the array

temps[59] = TempCelsiusDisplay;

Which throws up the error:

Error: A value of type "volatile char *" cannot be assigned to an entity of 
type "char" in "main.cpp", Line: 70, Col: 20

Is anybody able to explain why this is happening? It seems to me the way I have declared the array is not the correct way to declare a list of volatile chars, however I don't really understand what is happening or how to fix it.

Thanks in Advance :)

As the compiler is telling you, `temps[59]` is a single `char`, `TempCelsiusDisplay` is an array of `volatile char`s. So, I am presuming you want to **copy** the characters in the first array into the second array. Also, keep in mind that `temps[59]` is the last character in the `temps` array; you probably don't want to start copying to that place because it will overflow. — vgru, Mar 21 '18 at 08:38
Possible duplicate of [How to copy a char array in C?](https://stackoverflow.com/questions/16645583/how-to-copy-a-char-array-in-c) — vgru, Mar 21 '18 at 08:41
Alternatively, if you want to store multiple character arrays into `temps`, then you want `temps` to be an array of character arrays, i.e. `char * volatile temps[60]` (I also believe the pointer should be volatile in that case, not the data), but it's hard to recommend exactly how to do the assignment since it's not specified how you allocate and free the arrays for storing individual temperatures. — vgru, Mar 21 '18 at 08:46

Matthias Grün · Answer 1 · 2018-03-21T09:20:24.037

Temps is an array of single chars, so temps[59] is of type char. So by saying

temps[59] = TempCelsiusDisplay;

you are trying to assign a value of type char[] (char array) to a value of type char (single character), which of course is not possible.

If you want to copy the string in TempCelsiusDisplay into temps you could use strcpy_s:

strcpy_s((char*)temps, sizeof(temps), (const char*)TempCelsiusDisplay);

If, on the other hand, you want each of the entries in array temps to be able to hold a temperature string like the one contained in TempCelsiusDisplay, you would need to declare an array of strings (char pointers) like so:

volatile char* temps[60];

Then you could copy the string TempCelsiusDisplay into one of the strings in the array:

temps[0] = (char*) malloc(10);    // allocate memory (example: 10 bytes)
strcpy_s((char*)temps[0], 10, (const char*)TempCelsiusDisplay);
// ... use string
free((void*) temps[0]);    // free memory

Optionally, you could save yourself the trouble of dynamically allocating memory by declaring a multidimensional array, like so:

volatile char temps[60][10];

This declares 60 10-character arrays.

Then you can copy the string like before:

strcpy_s((char*)temps[0], sizeof(temps[0]), (const char*)TempCelsiusDisplay);

Storing Volatile Chars in an Array

1 Answers1