In thsi example:
int a[2][2]={{1,2},{3,4}};
int *p=a[0];
cout<<p;
cout<<&a[0][0];
Both gives the same output. Then why am I not able to call function (say fun)like this and loop through the array:
fun(a[0]);
fun(int *p)
{
cout<<p[1][1];
}
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JoeG
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Gautam Kumar
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This is not a exact duplicate but You should have a look at this excellent faq entry: http://stackoverflow.com/questions/4810664/how-do-i-use-arrays-in-c – Alok Save Jul 11 '11 at 10:51
3 Answers
3
fun(a[0]); //this looks OK
void fun(int *p) // this is OK if you add return type'
^^^^
{
cout<<p[1][1]; //NOT OK! You can't have 2 indices on an `int*`
cout << p[1]; // OK, will print a[0][1]
}
Armen Tsirunyan
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Very nice. To be clear, this function `fun` accesses the 1-dimensional slices of the original 2-dim array. I don't really understand if that's what the OP wanted, so I thought best to spell that out. – Kerrek SB Jul 11 '11 at 10:56
2
To answer your question: when you write:
p = a[0];
a[0] (now 0th element to 1D array) actually decays to a pointer p. So both are not exactly the same type, though they appears to be. When you write:
fun(a[0]);
You are actually passing the 0the element of the array which is now a 1D array. So you can receive in either of below ways:
fun(int *p); // decay to pointer to 1D array
fun(int (&a)[2]); // receive array by reference
In both the case fun() has now a 1D array.
To make the things simpler, Pass reference to array:
void fun(int (&p)[2][2])
{
cout<<p[1][1]; // ok !
}
Usage:
fun(a); // not a[0]
iammilind
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No, in this case he can't pass a[0] to fun, which is apparently what he wants – Armen Tsirunyan Jul 11 '11 at 10:42
