指针作为C ++中的参数

Question

I'm new to C++, coming from mostly working with Java and I'm having a problem with a function I'm trying to write. I'm sure it's something simple, but nonetheless, it's giving me fits, so prepare for a painfully newbie question.

I'm trying to write a function as follows:

void foo(u_char *ct){

/* ct is a counter variable, 
it must be written this way due to the library 
I need to use keeping it as an arbitrary user argument*/

/*here I would just like to convert the u_char to an int, 
print and increment it for each call to foo, 
the code sample I'm working from attempts to do it more or less as follows:*/

int *counter = (int *) ct;
printf("Count: %d\n", *counter);
*counter++;

return;

}

When I try to run this in XCode (something I'm also new to using), I get a EXE_BAD_ACCESS exception on the printf() portion of foo. I'm really not sure what is going on here but I suspect that it has something to do with conflating values, pointers and references, something I don't yet have a strong gasp of how C++ understands them coming from Java. Does anyone see where I'm slipping up here?

Thanks.

Answer 1

An u_char would be 1 byte in memory (the name suggests it's just an unsigned char), an int is typically 4 bytes. In printf , you tell the runtime to read an int (4 bytes) from the address where counter resides. But you only own 1 byte there.

Answer 2

EDIT (based on comments down here where poster says it's called actually with the address of an int: foo((u_char*)&count) ):

void foo(u_char *ct)
{
   int *pcounter = (int *)ct;  // change pointer back to an int *
   printf("Count: %d\n", *pcounter);
   (*pcounter)++;  // <<-- brackets here because of operator precedence.
}

Or even shorter (the wild c-style for which everbody but newbies loves this language):

void foo(u_char *ct)
{
   printf("Count: %d\n", (*(int *)ct)++);
}