error: no match for 'operator='

[ycai77]

I tried to compile the following code using g++. Could anyone help point out what is wrong here. Thanks a lot. Happy holidays.

==== code ====
#include <iostream>

using namespace std;

#define NUM 10

struct TypeA {
int aa;
};

int main() {

TypeA *c = NULL;
c = new TypeA;
c = NULL;
cout << c << endl;

TypeA *d[NUM];

for (int i = 0; i < NUM; i++) {
d[i] = new TypeA[NUM];
}

for (int i = 0; i < NUM; i++) {
for (int j = 0; j < NUM; j++) {
d[i][j] = NULL;
}
}


return 0;
}


==== compiling result

-bash-2.05b$ g++ main1.cc
main1.cc: In function `int main()':
main1.cc:26: error: no match for 'operator=' in '*(d[i] + (+(j * 4))) = 0'
main1.cc:7: error: candidates are: TypeA& TypeA::operator=(const TypeA&)

[helios]

Did you mean Packet **memory_pool[NUM_MBS];?

[ycai77]

No.

[helios]

(You just said yes...)

No? Well, then why are you trying to assign NULL to an object?

[ycai77]

Hi, first of all, thanks for your reply.

My purpose is as follows.
I want to create a 2-D pointer array in which each element is a pointer pointing to .... To make the program safer, I would like to make all pointers in my 2-D pointer array point to NULL. And then I will check if an element points to NULL before trying to use it.

Here I come up with a simpler program to reflect the problem. I update my first post. Thanks,

[helios]

So then I was right. You did mean to make memory_pool a pointer to a pointer.

T **twoDarray=new T *[width*height];
for (int a=0;a<width*height;a++)
    twoDarray[a]=0;
//...
twoDarray[y*width+x]=new T;

[ycai77]

Thanks for your help. It works. However, it implements a 2-D array as a 1-D one.

But I really want to use something like towDarray[i][j], which is a pointer. How should I do it? thanks,

[helios]

This way is cheaper memory-wise, and probably computationally, since there's only one level of indirection. If you really want to do it like that, it's just a matter of adding more indirection.
I'm going to obviate the initialization to zero to save me some typing:

T ***a=new T **[w];
for (int b=0;b<w;b++)
    a[b]=new T *[h];
//...
a[x][y]=new T;

[ycai77]

Hi, you are so helpful. THANKS A LOT. I agree! I implemented your code as follows. I thought they did the same thing. Please point out if you find something incorrect.

======

    TypeA **e[NUM];
    
    for (int i = 0; i < NUM; i++) {
        e[i] = new TypeA *[NUM];
    }
    
    for (int i = 0; i < NUM; i++) {
        for (int j = 0; j < NUM; j++) {
            e[i][j] = NULL;
            cout << e[i][j] << endl;
        }
    }

=======

BTW, the reason for e[i][j] is to ease the operations.

[helios]

None of the asterisks in my snippet are optional. You have to be very careful about that.

Is NUM a small (let's say <20) constant? If so, you could do T *array[NUM][NUM];
If it's not a small constant, you can still do it like that, but if the object is not created with new, you could overflow the stack.

[ycai77]

Is T *array[NUM][NUM] the simplest way?

[helios]

Depends, like I said, on how you're going to allocate the object, and whether NUM is a small constant. If you're absolutely sure you're always going to allocate TypeA dynamically (e.g. TypeA *obj=new TypeA;), then yes, and the value of NUM doesn't matter for the most part.