C pointers and arrays: [Warning] assignment makes pointer from integer without a cast
In this case a[4]
is the 5th
integer in the array a
, ap
is a pointer to integer, so you are assigning an integer to a pointer and that's the warning.
So ap
now holds 45
and when you try to de-reference it (by doing *ap
) you are trying to access a memory at address 45, which is an invalid address, so your program crashes.
You should do ap = &(a[4]);
or ap = a + 4;
In c
array names decays to pointer, so a
points to the 1st element of the array.
In this way, a
is equivalent to &(a[0])
.
What are you doing: (I am using bytes instead of in for better reading)
You start with int *ap
and so on, so your (your computers) memory looks like this:
-------------- memory used by some one else --------
000: ?
001: ?
...
098: ?
099: ?
-------------- your memory --------
100: something <- here is *ap
101: 41 <- here starts a[]
102: 42
103: 43
104: 44
105: 45
106: something <- here waits x
lets take a look waht happens when (print short cut for ...print("$d", ...)
print a[0] -> 41 //no surprise
print a -> 101 // because a points to the start of the array
print *a -> 41 // again the first element of array
print a+1 -> guess? 102
print *(a+1) -> whats behind 102? 42 (we all love this number)
and so on, so a[0] is the same as *a, a[1] = *(a+1), ....
a[n] just reads easier.
now, what happens at line 9?
ap=a[4] // we know a[4]=*(a+4) somehow *105 ==> 45
// warning! converting int to pointer!
-------------- your memory --------
100: 45 <- here is *ap now 45
x = *ap; // wow ap is 45 -> where is 45 pointing to?
-------------- memory used by some one else --------
bang! // dont touch neighbours garden
So the "warning" is not just a warning it's a severe error.