How to copy one struct in C to another?

hi all, with the help from people in this forum i achieved this process when concerning arrays: as seen here: http://cboard.cprogramming.com/showthread.php?t=110716
I would like to achieve the same effect with structure's now but am receiving some errors.

Code:

---

#include <stdio.h>
#include <string.h>

typedef struct one
{
int x;
int y;
} point;

point top = {2,15};

struct two
{
int a;
int b;
}blunt;

int main ()
{
memcpy(two, one, sizeof one);
return 0;
}

---

The errors are as follows:
error: ‘two’ undeclared (first use in this function)
error: (Each undeclared identifier is reported only once
error: for each function it appears in.)
error: ‘one’ undeclared (first use in this function.

I'm sure I am not referencing the source structure and destination structure, for memcpy () correctly, also is sizeof 'one' correct? Does the second structure have to be a typedef?

So struct one and struct two are types, just like int and float and char. You need to copy into actual variables. You have one variable of type struct one ("struct one" is the same type as "point", due to the typedef), called top. You have a variable blunt of type struct two.

You also need to pass the address of those variables to memcpy, because it expects pointers.
Arrays automatically decays to pointers to their first element when passed as an argument, so you got away with not doing that on those.

does that satisfy the question though? how to copy one structure into another in C? wouldn't that be copying variables into another?

i'm not saying your wrong btw, I dont know the answer to that myself to be honest.

Quote:

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Originally Posted by cus

does that satisfy the question though? how to copy one structure into another in C? wouldn't that be copying variables into another?

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Yes it would. In computer memory, it's all bits and data anyway, so memcpy takes that data and puts it into the other struct.
The type of the variables determine how the compiler - and ultimately, your code - treats that data.

Quote:

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Originally Posted by cus

does that satisfy the question though? how to copy one structure into another in C? wouldn't that be copying variables into another?

i'm not saying your wrong btw, I dont know the answer to that myself to be honest.

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What do you think the difference is? The variables are structures.

yes haha, offcourse they are, thanks for that :) I'l have a try in a moment

ok i tried these two in memcpy(), and wasn't sure what 'sizeof' should be and the errors i'm now getting seem to re-enforce that:

Code:

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memcpy(&blunt, &point, sizeof one);

---

Quote:

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memcpy(&blunt, &point, sizeof point);

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The errors are:
error: expected expression before ‘point’
error: too few arguments to function ‘memcpy’

I guess point and one are structure types. If so you have to wrap them with ()'s, i.e. sizeof(one), sizeof(point), sizeof(int), etc. However, int num; sizeof num is fine.

Quote:

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Originally Posted by cus

ok i tried these two in memcpy(), and wasn't sure what 'sizeof' should be and the errors i'm now getting seem to re-enforce that:

Code:

---

memcpy(&blunt, &point, sizeof one);

---

The errors are:
error: expected expression before ‘point’
error: too few arguments to function ‘memcpy’

---

Once again, point is a type, like int or float or char. You must memcpy into the address of a variable. The only variable you have of that type is "top".

That is why you should always do sizeof(something), not sizeof something as some like to do!

thanks tabstop, twomers and elysia.

I'm presuming this works now, i got your hint tabstop.
I changed the second structure to a typedef, referenced 'top' in memcpy() correctly and created another variable for the second structure so i could read the values into it. Thanks again. btw is this ok?

Code:

---

#include <stdio.h>
#include <string.h>

typedef struct one
{
int x;
int y;
} point;

point top = {2,15};

typedef struct two
{
int a;
int b;
} blunt;

blunt bottom;

int main ()
{
memcpy(&bottom, &top, sizeof top);
printf("The second structure values are %d and %d\n", bottom.a, bottom.b);

return 0;

}

---

Also is there a way to show my problem has been solved i.e. a forum feature - or even a thank system?

Looks OK. Does it work? I recommend adopting Elysia's approach and always use sizeof().

Quote:

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Originally Posted by twomers

Looks OK. Does it work? I recommend adopting Elysia's approach and always use sizeof().

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yes it works :) eliysia recommended this method in a previous post to avoid errors, i just forgot to put it in, i shall use it always from now on. By the way how long do post's on this forum stay active for? Will they still be here for example if I searched in a couple of months time?

There are posts going back to 2001 so a long time. A Brief History of Cprogramming.com contains the archive. The posts in the other forums are the more active ones.

Why do you have two structures with the same members?

If you have the same type, then
bottom = top;
is a completely natural thing to do.

Purely to carry out the exercise of copying one structure to another, the values are irrelevant. Unless you have another way to copy one struct to another?

Well the memcpy() only works because both structures are memberwise alike, which essentially makes one of them redundant. struct assignment which I showed is nothing more than a disguised memcpy anyway.

If you had this say

Code:

---

typedef struct one
{
    int x;
    int y;
} point;

typedef struct two
{
    float a;
    float b;
} blunt;

---

Then you would have to do memberwise assignments yourself, as in
bottom.a = top.x;
kinda thing.

yes I take your point - so how would you copy one structure to another

That's it, you've seen both ways.

Generally in the real world you wouldn't have two structs definitions with the same members. You'd have one struct definition and multiple struct variables that are instances of that struct type. E.g. one called "a" and another called "b".
Then the real answer to the question posed becomes:

Code:

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a = b;

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Type safety is a good thing, go with it instead of getting into a habbit of unnecessarily subverting it using memcpy. Otherwise you'll shoot yourself in the foot.