problem with double array assignment with a single type

i have two structures declared as such:

Code:

---

struct rtpkt {
        int src;          /* source router                                  */
        int dst;          /* destination router (immediate neighbor)        */
        int mincost[N];  /* minimun cost to nodes 0, 1, 2, and 3            */
};

struct dt
{
  int dist[N][N];    /* dist[i][j] is the distance to node i via node j */
};

---

And i have a line in the .c file:

Code:

---

            struct rtpkt send_neighbor;
            send_neighbor.mincost = dt[node].dist[node];

---

and the error that im getting is (cant assign two different types) not a quote.
What am i doing wrong here?

-thank you

The left side is an array of ints, the right side is one int. Perhaps you want to put the number into a particular slot in your array?

Quote:

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Originally Posted by tabstop

The left side is an array of ints, the right side is one int. Perhaps you want to put the number into a particular slot in your array?

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how is the right side one int? in my understanding, (apparently a wrong one)
doublearray[10][10] is an array of double pointers, so doublearray[1] is pointing to a location in memory that is the start of a 10*sizeof(int) block.
so is onearray[10], which onearray is pointing to the same sort of structure.

what is wrong from my understanding?

We don't have any doublearray or onearray anywhere here. I didn't quote your original, but now what you have is an array and an array. However, arrays are not assignable (in C or C++). If you want each element of the first array to have the corresponding value in the second array, you'll need to use a for-loop to make it happen.

memcpy works in C...

Quote:

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Originally Posted by Elysia

memcpy works in C...

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That would require him to have the same data types on both sides, which I'm not sure he does, what with this "double" stuff appearing out of nowhere.

Hrm, well, let's leave that subject untouched until more information appears.

Quote:

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Originally Posted by tabstop

We don't have any doublearray or onearray anywhere here. I didn't quote your original, but now what you have is an array and an array. However, arrays are not assignable (in C or C++). If you want each element of the first array to have the corresponding value in the second array, you'll need to use a for-loop to make it happen.

---

I think i confused you with 2D array when i said double.
so considering the variable
int dist[N][N]
and
int mincost[N]
and can't have mincost to point to dist[N], by setting mincost = dist[N] ?

No, you cannot. Use memcpy.

Or use a more obvious for loop to copy each member of the array.

Why?
1. It's the more natural thing to do.
2. Chances are, you'll screw up the 'size' parameter of memcpy at some point, and either copy too little, or too much data.
3. It might actually be quicker at run-time, depending on your implementation of memcpy.
4. When you decide to use C++, then the assignment approach will seem natural, and applying the technique to complex objects (with an overloaded assignment operator) will continue to "do the right thing". memcpy in this instance would be a disaster.

Quote:

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Originally Posted by Salem

3. It might actually be quicker at run-time, depending on your implementation of memcpy.

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Normally I would agree, but I don't think I can do that with this one...
memcpy is an optimized and fast function for copying memory, so it will probably be faster, if anything. Or should be.

Quote:

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Originally Posted by armen

I think i confused you with 2D array when i said double.
so considering the variable
int dist[N][N]
and
int mincost[N]
and can't have mincost to point to dist[N], by setting mincost = dist[N] ?

---

No you cannot do that because the name of an array mincost is not a variable and as Salem suggested use a loop after an assignment like

Code:

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int *ip = mincost

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Your code seems misleading, since you are actually creating a new pointer and assigning to it instead of assigning to the destination array.
Although it does support assigning a structure to a structure, which pretty much means it is possible to assign a range of values to another range of values, just in a different form. You may think it is strange, but it's just how it is.

I see how it can be misconstrued; ofcourse the OP can simply use a pointer and an offset in a loop for array assignment as in

Code:

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*(mincost + i) = *(dist[0] + i);

---

Quote:

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Originally Posted by Elysia

Normally I would agree, but I don't think I can do that with this one...
memcpy is an optimized and fast function for copying memory, so it will probably be faster, if anything. Or should be.

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But it's probably better at copying 400-40000 bytes than 40 bytes. After all, the compiler may well come up with roughly the same loop as a short memcpy. And there is the overhead of the call or stuffing the right values in the right registers for a rep movs sequence (assuming x86). So I'm with Salem here. Use a loop.

Obviously, if we have 1000 indices at each size, memcpy() will probably be a tad faster.

--
Mats

Quote:

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Originally Posted by itCbitC

I see how it can be misconstrued; ofcourse the OP can simply use a pointer and an offset in a loop for array assignment as in

Code:

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*(mincost + i) = *(dist[0] + i);

---

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Translated code is:

Code:

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mincost[i] = dist[0][i];

---

This is a late response but thanks to all your responses.
I used a loop, since there are only 15 items.
But I wanted to code correctly,(considering the case of having 999999999999 items).
-cheers

You need to use dynamic memory management - ie malloc and free. And keep track of the size so you can resize it and know how many times to iterate.