Alex Rivera | Logout

volatile variables as argument to function

Asked 2010-07-21T20:44:14.503
12

Having this code:

typedef volatile int COUNT;       

COUNT functionOne( COUNT *number );

int  functionTwo( int *number );

I can't get rid of some warnings..

I get this warning 1 at functionOne prototype

[Warning] type qualifiers ignored on function return type

and I get this warning 2, wherever I call functionTwo with a COUNT pointer argument instead of an int pointer

[Warning] cast discards qualifiers from pointer target type

obviously variables/pointers can't be "cast" to volatile/un-volatile.. but every arguments must be specified as volatile too? so how can I use any library function if it's already defined for non-volatile variable?

EDIT: Using gcc -std=c99 -pedantic -Wall -Wshadow -Wpointer-arith -Wcast-qual -Wextra -Wstrict-prototypes -Wmissing-prototypes …

EDIT: After Jukka Suomela advice this is a code sample for warning two

typedef volatile int COUNT;       

static int functionTwo(int *number) {
    return *number + 1;
}

int main(void) {
    COUNT count= 10;
    count = functionTwo(&count);
    return 0;
}
c pointers arguments warnings volatile
Edit
Report

1 Answer

2

I don't understand why you'd want to have the volatile qualifier on a function return type. The variable that you assign the function's return value to should be typed as a volatile instead.

Try making these changes:

typedef int COUNT_TYPE;
typedef volatile COUNT_TYPE COUNT;       

COUNT_TYPE functionOne( COUNT number );

COUNT_TYPE functionTwo( COUNT_TYPE number );

And when calling functionTwo(), explicitly cast the argument:

functionTwo( (COUNT_TYPE)arg );

HTH, Ashish.

answered 2010-07-21T21:52:05.807

Your Answer