Suppose I have the following code:
main ()
{
char string[20];
printf(" The String is %s \n " , &str);
}
What would printf(" The String is %s \n " ,&str); give?
Suppose str points to location 200, what would &str give??
views:
102answers:
3
A:
You should get a warning saying that
%s expects char* but argument 2 has char (*)[20] type.
The address is not printed, in fact nothing is printed.
sheki
2010-09-28 11:07:18
SORRY str is mean to be string
Timmy
2010-09-28 11:09:36
There is no way that printf can know the type of the variables being passed in...it's a var_args function. It just ASSUMES that it will be `char *` and will treat it as such. It will work fine.
banister
2010-09-28 11:13:56
@banister: Clever compilers are allowed to treat the functions defined by the language, like `printf`, specially - and in fact gcc and glibc together do this, and can present such a warning.
caf
2010-09-28 11:19:02
A:
Assuming you actually initialized the array with a string (which you didn't, but let's assume you did) then:
It is of type char (*)[20]. It'll give the same output as
printf("The String is %s\n", str)
&str points to the same memory location as str but is of different type; namely pointer-to-array. The type of str is of type pointer-to-char.
banister
2010-09-28 11:10:48