Question

volatile char* sevensegment_char_value;

void ss_load_char(volatile char *digits) {
    ...
    int l=strlen(digits);
    ...
 }

 ss_load_char(sevensegment_char_value);

In the above example I’ve got warning from avr-gcc compiler

Warning 6   passing argument 1 of 'strlen' discards 'volatile' qualifier from pointer target type [enabled by default]

So I have to somehow copy the value from volatile to non-volatile var? What is the safe workaround?

Answer 1

There is no such thing like a “built in” Workaround in C. Volatile tells the compiler, that the contents of a variable (or in your case the memory the variable is pointing at) can change without the compiler noticing it and forces the compiler to read the data direct from the data bus rather than using a possibly existing copy in the registers.
Therefore the volatile keyword is used to avoid odd behaviour induced through compiler optimizations. (I can explain this further if you like)

In your case, you have a character buffer declared as volatile. If your program changes the contents of this buffer in a different context like an ISR for example, you have to implement sort of a synchronisation mechanism (like disabling the particular interrupt or so) to avoid inconsistency of data. After aquiring the “lock” (disabling the interrupt) you can copy the data byte by byte to a local (non-volatile) buffer and work on this buffer for the rest of the routine.

If the buffer will not change “outside” of the context of your read accesses I suggest to omit the volatile keyword as there is no use for it.

To judge the correct solution, a little bit more information about your exact use case would be needed.