why does the pointer array "equivalence" not work in the following case?
void foo(int** x) {
cout << x[0][1];
}
int main( ) {
int a[2][2] = {{1,2},{2,3}};
foo(a);
}
thank you
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6What doesn't work? what error message are you getting?Daniel– Daniel2011-05-12 20:47:23 +00:00Commented May 12, 2011 at 20:47
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possible duplicate of In C, are arrays pointers or used as pointers?ildjarn– ildjarn2011-05-12 20:48:55 +00:00Commented May 12, 2011 at 20:48
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The pointer array "equivalence"user695652– user6956522011-05-12 20:49:06 +00:00Commented May 12, 2011 at 20:49
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@ildjam, no its not duplicate, there the 1D case is discussed for which the pointer array "equivalence" worksuser695652– user6956522011-05-12 20:51:00 +00:00Commented May 12, 2011 at 20:51
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4@user695652: ...Because there's no such thing as "pointer array equivalence" and there never was. In fact, your example is a direct proof of that.AnT stands with Russia– AnT stands with Russia2011-05-12 20:53:06 +00:00Commented May 12, 2011 at 20:53
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3 Answers
The memory model of int** and int[2][2] is different.
int a[2][2] is stored in memory as:
&a : a[0][0]
&a + 4 : a[0][1]
&a + 8 : a[1][0]
&a + 12: a[1][1]
int** x:
&x : addr1
&x + 4 : addr2
addr1 : x[0][0]
addr1 + 4: x[0][1]
addr2 : x[1][0]
addr2 + 4: x[1][1]
while addr1 and addr2 are just addresses in memory.
You just can't convert one to the other.
It doesn't work because only the first level of the multidimensional array decays to a pointer. Try this:
#include <iostream>
using std::cout;
void foo(int (*x)[2]) {
cout << x[0][1];
}
int main( ) {
int a[2][2] = {{1,2},{2,3}};
foo(a);
}
Comments
because the type is not int **. this right for foo function
foo(int *[2]);
type of pointer a is not int ** , exactly int* [2]..
