float m[4][3];

[xontrn]

For

float a[9];

we know just using the array name a gives a float* to the first
element in the array.

For

float m[4][3];

what is the type of the array name m?

[Victor Bazarov]

xo****@yahoo.com wrote:

For

float a[9];

we know just using the array name a gives a float* to the first
element in the array.

For

float m[4][3];

what is the type of the array name m?

Since 'm' is an array of arrays, the array-to-pointer conversion of
'm' would yield a pointer to an array.

V
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[xontrn]

On Jul 30, 8:01 am, "Victor Bazarov" <[e-mail]:

[e-mail] wrote:

For

float a[9];

we know just using the array name a gives a float* to the first
element in the array.

For

float m[4][3];

what is the type of the array name m?

Since 'm' is an array of arrays, the array-to-pointer conversion of
'm' would yield a pointer to an array.

Okay, then why doesn't this work? Isn't T* d[3] a pointer to an
array?

#include <iostream>
using namespace std;

template<typename T>
class C
{
public:
C(T* d[3], int m, int n)
{
mData = new T*[m];
for(int i = 0; i < m; ++i)
mData[i] = new T[n];

for(int i = 0; i < r; ++i)
for(int j = 0; j < c; ++j)
mData[i][j] = d[i][j];

r = m;
c = n;
}

~C()
{
for(int i = 0; i < r; ++i)
delete[] mData[i];

delete[] mData;
}

T& operator()(int i, int j)
{
return mData[i][j];
}

void print()
{
for(int i = 0; i < r; ++i)
{
for(int j = 0; j < c; ++j)
{
cout << data[i][j] << "\t";
}
cout << endl;
}
}

private:
T** data;
int r;
int c;
};

int main()
{
float m[2][3];
m[0][0] = 1.0f;
m[0][1] = 2.0f;
m[0][2] = 3.0f;
m[1][0] = 4.0f;
m[1][1] = 5.0f;
m[1][2] = 6.0f;
m[2][0] = 7.0f;
m[2][1] = 8.0f;
m[2][2] = 9.0f;

C<floatT(m, 2, 3);

T.print();
}

V
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[Victor Bazarov]

xo****@yahoo.com wrote:

On Jul 30, 8:01 am, "Victor Bazarov" <[e-mail]:

>[e-mail] wrote:

>>For

>>float a[9];

>>we know just using the array name a gives a float* to the first
element in the array.

>>For

>>float m[4][3];

>>what is the type of the array name m?

Since 'm' is an array of arrays, the array-to-pointer conversion of
'm' would yield a pointer to an array.

Okay, then why doesn't this work? Isn't T* d[3] a pointer to an
array?

No,

T *d[3]

is an array of three pointers. You need to use parentheses.

[..]

V
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[xontrn]

On Jul 30, 11:08 am, "Victor Bazarov" <[e-mail]:

[e-mail] wrote:

On Jul 30, 8:01 am, "Victor Bazarov" <[e-mail]:

[e-mail] wrote:
For

>float a[9];

>we know just using the array name a gives a float* to the first
element in the array.

>For

>float m[4][3];

>what is the type of the array name m?

Since 'm' is an array of arrays, the array-to-pointer conversion of
'm' would yield a pointer to an array.

Okay, then why doesn't this work? Isn't T* d[3] a pointer to an
array?

No,

T *d[3]

is an array of three pointers. You need to use parentheses.

It still didn't work. How did you do the parentheses?

>
V
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[Ivan Vecerina]

<xo****@yahoo.comwrote in message
news:11**********************@m37g2000prh.googlegr oups.com...
: >float m[4][3];
: >
: >what is the type of the array name m?
: >
: Since 'm' is an array of arrays, the array-to-pointer conversion
of
: 'm' would yield a pointer to an array.
: >
: Okay, then why doesn't this work? Isn't T* d[3] a pointer to an
: array?
: >
: No,
: >
: T *d[3]
: >
: is an array of three pointers. You need to use parentheses.
:
: It still didn't work. How did you do the parentheses?

Did you try:
T (*d)[3]
hth -Ivan
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[xontrn]

Did you try:
T (*d)[3]

Thank you, I was doing it wrong.

So just in how I should read this:

T* d[3] = "d is an array of 3 T*"

T (*d)[3] = "d is a pointer to a T[3]"

hth -Ivan
--http://ivan.vecerina.com/contact/?subject=NG_POST<- email contact form

[Victor Bazarov]

xo****@yahoo.com wrote:

>Did you try:
T (*d)[3]

Thank you, I was doing it wrong.

So just in how I should read this:

T* d[3] = "d is an array of 3 T*"

T (*d)[3] = "d is a pointer to a T[3]"

Right. By extension,

T (*d[5])[3] = "d is an array of 5 pointers to T[3]"

V
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