char** function parameters

[confusedcoder]

Suppose I have the following function:

void doSomething(cha r** array, int size);

Normally, I would call the function with an array of, say, 3
elements...

char* strs[] = { "str1","str2"," str3" };
doSomething(str s, 3);

That works fine. But if I want to pass it only one char*, I would
think this would work....

char* str = "str1";
doSomething(&st r, 1);

It compiles with a warning about incompatible pointer types, and
needless to say it doesn't behave as I expected. "str1" does not get
passed into the function. What am I doing wrong? How can I pass a
single char* into the function?

[Rajan]

Primarily what you need to do is to avoid the warning you need to
typecast
by doing doSomething((ch ar **)&str,1);
So let's say your function is

void doSomething(cha r** s, int size)
{
printf("%s\n",* s);
}
This needs to be done because you are not printing s[0], instead are
trying to print address of s[0];
Remember this is a double pointer.

[Stephane Zuckerman]

> Primarily what you need to do is to avoid the warning you need to

typecast
by doing doSomething((ch ar **)&str,1);

I'm not sure I'm understanding why you would have to absolutely try to
hush the compiler. Warnings are there for a good reason, most of the time.
The only thing we need to know is "is this warning announcing something
that was unexpected ?"... well, IMO at least :)

--
"Je deteste les ordinateurs : ils font toujours ce que je dis, jamais ce
que je veux !"
"The obvious mathematical breakthrough would be development of an easy
way to factor large prime numbers." (Bill Gates, The Road Ahead)

[confusedcoder]

Thanks for all the feedback!