Need help with error C2664

I have a question as to why the two shown func args are different? See comments (yes I am novice)

struct MyFileClass
{
.....other code

void MyFileOut( TCHAR* pS, DWORD pVar)
{
TCHAR PrntStrg1[] = _T("The value of passed ptr->");
.... other code

MyFilePrint(&PrntStrg1, Scratch_PP); //<-this pulls error C2664:
//cannot convert parameter 1 from 'TCHAR (*__w64 )[26]' to 'TCHAR *'

MyFilePrint(PrntStrg1, Scratch_PP); //<-this does not
// doesn't PrntStrg1 and &PrntStrg1 evaluate to the same thing ?

......other code
} // end MyFileOut

int MyFilePrint(TCHAR* pS, DWORD pVar)
{
....other code
} // end MyFilePrint

} // end MyFileClass

The type of PrntStrg1 is TCHAR[] which is (almost) the same as TCHAR*.
The type of PrntStrg1* is TCHAR*[] which is (almost) the same as TCHAR**.

Do you see the difference?

Quote:

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Originally Posted by VictorN

The type of PrntStrg1* is TCHAR*[] which is (almost) the same as TCHAR**.

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I wouldn't say they are almost the same, because they are not convertible to each other.

Code:

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void foo(char** test);
void bar()
{
  char array[] = "test";
  foo(&array); // error C2664: 'foo' : cannot convert parameter 1 from 'char (*)[5]' to 'char **'
}

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Quote:

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Originally Posted by D_Drmmr

I wouldn't say they are almost the same, because they are not convertible to each other.

Code:

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void foo(char** test);
void bar()
{
  char array[] = "test";
  foo(&array); // error C2664: 'foo' : cannot convert parameter 1 from 'char (*)[5]' to 'char **'
}

---

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Yes, you are right.
I just tried to explain the OP problem as short as possible... So sorry for non-correct explanation! :blush:

Quote:

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Originally Posted by VictorN

The type of PrntStrg1 is TCHAR[] which is (almost) the same as TCHAR*.
The type of PrntStrg1* is TCHAR*[] which is (almost) the same as TCHAR**.
Do you see the difference?

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Quote:

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Originally Posted by D_Drmmr

I wouldn't say they are almost the same, because they are not convertible to each other.

Code:

---


void foo(char** test);
void bar()
{
  char array[] = "test";
  foo(&array); // error C2664: 'foo' : cannot convert parameter 1 from 'char (*)[5]' to 'char **'
}

---

---

Quote:

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Originally Posted by VictorN

Yes, you are right.
I just tried to explain the OP problem as short as possible... So sorry for non-correct explanation! :blush:

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Thanks for the explanations. I kinda see what your are taking about. I guess I need to do some experimental evaluations to make it more clear to me. I had always thought an Char array was the same as the address of the first character, and that in essense it was the same as Char ptr. So I guess you are saying that I'm passing the address of the ptr instead of the ptr. Anyhow at least you guys have shown me where I have some learning to do. I thank you much.

Quote:

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Originally Posted by J_W

... So I guess you are saying that I'm passing the address of the ptr instead of the ptr.

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Exactly! :thumb:

Quote:

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Originally Posted by J_W

I had always thought an Char array was the same as the address of the first character, and that in essense it was the same as Char ptr.

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An array is an array (a contiguous buffer of a certain type), and a pointer is a pointer. They are not the same thing.

What you're confused about is that

1) a pointer can be used syntactically to traverse, retrieve, or set an element within an array

2) When you "pass an array" to a function, the receiving function receives a pointer to the first element of the array. Then item 1) takes effect within the function.

But do not make the mistake that an array is a pointer and a pointer is an array. If you want further proof:

Code:

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#include <iostream>

using namespace std;
int main()
{
  char x[100];
  char* y;
  cout << sizeof(x) << endl;
  cout << sizeof(y) << endl;
}

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What values are printed when I attempt to retrieve the size of each of those types? I can bet they are vastly different.

Regards,

Paul McKenzie

Quote:

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Originally Posted by Paul McKenzie

. . . . .
What you're confused about is that

1) a pointer can be used syntactically to traverse, retrieve, or set an element within an array

2) When you "pass an array" to a function, the receiving function receives a pointer to the first element of the array. Then item 1) takes effect within the function.

But do not make the mistake that an array is a pointer and a pointer is an array. If you want further proof:

Code:

---

#include <iostream>

using namespace std;
int main()
{
  char x[100];
  char* y;
  cout << sizeof(x) << endl;
  cout << sizeof(y) << endl;
}

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Regards, Paul McKenzie

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Point understood, thank you. 100 bytes compared to 4.