casting pointer to pointer to an array

Question

Below is the code snippet. I want to know if line no. 17 typecasting is valid and proper in c?

#include <stdio.h>

typedef int twoInts[2];

void print(twoInts *twoIntsPtr);
void intermediate (twoInts twoIntsAppearsByValue);

int main () {
    twoInts a;
    a[0] = 0;
    a[1] = 1;
    print(&a);
    intermediate(a);
    return 0;
}
void intermediate(twoInts b) {
    print((int(*)[])b); // <<< line no. 17 <<<
}

void print(twoInts *c){
    printf("%d\n%d\n", (*c)[0], (*c)[1]);
}

Also, when i change the definition intermediate to

void intermediate(twoInts b) {
    print(&b);
}

I am getting below warnings while compiling and o/p is not proper.

1.c:17:11: warning: passing argument 1 of print from incompatible pointer type
     print(&b);
           ^
1.c:5:6: note: expected int (*)[2] but argument is of type int **
 void print(twoInts *twoIntsPtr);

As per my understanding, array is decaying to pointer to int in funtion argument. What is the exact reason?

Let me just add that: void print(twoInts *twoIntsPtr); is identical to: void print(int (*twoIntsPtr)[2]); Or in other words, don't hide arrays and pointers behind typedefs. — this
– this, Commented Oct 14, 2015 at 10:51
The best and proper solution to the problem, is indeed to never hide an array behind a typedef. Then you could use plain int*pointers without the need to obfuscate the program to oblivion. — Lundin
– Lundin, Commented Oct 14, 2015 at 11:19

P.P · Accepted Answer · 2015-10-14 10:14:41Z

3

a is an array. So when you pass it to intermediate(), it gets converted into a pointer to its first element (aka "array decaying").

So, the b in:

void intermediate(twoInts b) {
    print(&b);
}

has type int*, not int[2] as you might seem to expect. Hence, &b is of type int** rather than int (*)[2] that's expected by print() function. Hence, the types don't match.

If you change intermediate() to:

void intermediate(twoInts *b) {
    print(b);
}

then you wouldn't need to pass the address of the &b and it will as expected and the types would match correctly.

answered Oct 14, 2015 at 10:14

P.P

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