Pointer to a function parameter

[econy]

I write a function like:

void func1(char chx[], int y)
{
int wsn = 0;

 wsn = \*(int \*) (&chx);
 if (wsn == 0) {
    ...
 }

}

Compiler works well, no warning, no error. But when the code is running, seems it get a wild pointer. the code crashed.

[Lost User]

Should be:

void func1(char chx[], int y)
{
int wsn = 0;

wsn = \*(int \*)chx;
if (wsn == 0) {
...
}

}

The variable name chx is already a pointer to the array, adding the addressof operator creates a pointer to that pointer.

[econy]

Thanks, but I tested, &chx = chx.

[Lost User]

No it doesn't. I have no idea how you tested this but it is not correct. Build the following, step through it with your debugger and you will see that they are different.

 int\* pAddressOf = (int \*) (&chx);
 int\* pNormal = (int \*)(chx);

[econy]

I tested in Visual studio IDE.

char s1[] = {'H', 'e', 'l', 'l', 'o'};

printf("%lX,%lX",s1,&s1);

And I get same value for s1 and &s1.

[econy]

And I tested your code in Visual studio, use Printf("%lX"), It shows same value too

[Lost User]

That may be true, but sending those values to printf is not the same as casting to pointers. You can confirm it by printing out the values of the two pointers created in my two lines of code.

[Lost User]

See http://publications.gbdirect.co.uk/c_book/chapter5/arrays_and_address_of.html[^] for a good explanation of this issue.