...then why is the code below give same value (meatballs) for address and the actual content? And how to make sense of %d value of meatballs, is it random?
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
int main()
{
int i;
int meatballs[5]= {1,2,3,4,5};
printf("\nmeatsballs %p %p %d \t \n",meatballs, &meatballs,
meatballs);
return 0;
}
meatballs is an array, and it has type int[5]. However C doesn't allow passing arrays by value, so array decays into pointer for function call. This is equivalent of pointer to first element of array &meatballs[0], and it has type int*.
&meatballs is a pointer to an array, and it has type int(*)[5]. Since it's a pointer it can be passed to a function.
As you can see, both 1. and 2. return same address, but they have different types: pointer to integer array vs pointer to single integer.
Note: Types void* and int* don't necessarily have the same representation(1), and for %p only valid type is void*(2) or you will get undefined behaviour(3). Always convert pointer to void* when printing addresses:
printf("%p %p", (void*)meatballs, (void*)&meatballs);
%d. Resulting type is again int* as in case 1, but it is interpreted as int. This is clear undefined behaviour, and output is garbage.To print integer array element, use any of the following methods:
printf("%d %d %d", meatballs[i], *(meatballs + i), *meatballs);
First two will print array element at index i, and last will print first element. I recommend using the meatballs[i] in most cases, as it's most clear.
References to C standard (draft):
A pointer to void shall have the same representation and alignment requirements as a pointer to a character type.48)... ...Pointers to other types need not have the same representation or alignment requirements.
p The argument shall be a pointer to void. ...
... If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.
meatballs or &meatballs to void* as both are simply memory addresses which are void * to begin with. However, for the purpose of discussion it does no harm.
void* is only required to have same representation as char* (N1570, 6.2.5 p28), and %p strictly requires void* (7.21.6 p8).char * and void * pointers for any purpose to the extent %p cares. Are you saying something else that I'm missing? They are compatible types. Specifically, note:48 addresses void * and char * compatibility: 48) The same representation and alignment requirements are meant to imply interchangeability as arguments to functions, return values from functions, and members of unionsvoid* and int* are not interchangeable, and that's why cast is needed on printf. But as you say, according to note 48, cast to char* should also suffice.You have to use the unary dereference operator *, so that the actual value is shown as %d expects an integer.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
int main()
{
int i;
int meatballs[5]= {1,2,3,4,5};
printf("\nmeatsballs %p %p %d \t \n",meatballs, &meatballs,
*meatballs);
return 0;
}
Look at the last parameter: *meatballs. This code actually prints 1 as value and is the first element in your array. So your guess is right and it is only a bug.
