How do you copy one array in C to another?

Hi all, this problem doesn't involve strings.

I have written the following:

Code:

---

#include <stdio.h>
int main()
{
int age[4];
int same_age[4];
age[0]=23;
age[1]=34;
age[2]=65;
age[3]=74;

same_age=age;

printf("%d\n", same_age[0]);
printf("%d\n", same_age[1]);
printf("%d\n", same_age[2]);
printf("%d\n", same_age[3]);
return 0;
}

---

However this returns an error stating: 'incompatible types in assignment'?
Thanks

You cannot write to an array, only to an individual cell in an array. So you must copy into same_age[0], into same_age[1], ..., ..., ....

thanks for reply, but how, i don't want to do it individually.

Use memcpy() or a loop.

Use a for loop.

http://www.cprogramming.com/tutorial/c/lesson3.html

i thought memcpy() was a function for strings?

Quote:

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Originally Posted by cus

i thought memcpy() was a function for strings?

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An array of ints can be viewed as an array of bytes when you are not interested in the individual ints at that point.

i have no idea how to use memcpy()

Quote:

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Originally Posted by cus

i have no idea how to use memcpy()

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So look it up.

you can use "reference"
when you write

Code:

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same_age=age;

---

or

Code:

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same_age=&age[0];

---

that way same age will point to the same starting place of address as "age"
and same_age will be the same as age.

Quote:

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Originally Posted by transgalactic2

you can use "reference"
when you write

Code:

---

same_age=age;

---

or

Code:

---

same_age=&age[0];

---

that way same age will point to the same starting place of address as "age"
and same_age will be the same as age.

---

I feel obligated to point out that you cannot, in fact, do this. It is quite simply not possible to assign anything at all to same_age. Not a value, not a "reference" (whatever that is supposed to mean).

same_age is a pointer (points to the first cell of the array)
i put in it the address of the first cell of age (&age[0])

Code:

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same_age=&age[0];

---

so theoretically i cant see why it cant happen?

same_age, by itself, does indeed decay to a const pointer to the first cell of the array. And I admit that I (as well as others, alas) often elide the word "const" in the description. But nevertheless, it is a const pointer -- which means its value cannot be changed.

Edit: Not actually quite true: it decays to a pointer that is not an lvalue. In this case a distinction without a difference (but it allows arguments to be passed without losing const, so that's important).

Quote:

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Originally Posted by transgalactic2

same_age is a pointer (points to the first cell of the array)

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No, same_age is an array, not a pointer. It is true that an array is converted to a pointer to its first element when passed as an argument, but in the scope of the main function same_age is an array, and one cannot assign to an array.

EDIT:

Quote:

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Originally Posted by tabstop

And I admit that I (as well as others, alas) often elide the word "const" in the description. But nevertheless, it is a const pointer -- which means its value cannot be changed.

---

The 1999 edition of the C standard omits the word "const", hence there is no elision in the first place.

so its a const pointer
it cannot be changed in value.

Quote:

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Originally Posted by tabstop

Not actually quite true: it decays to a pointer that is not an lvalue. In this case a distinction without a difference (but it allows arguments to be passed without losing const, so that's important).

---

Yes, that is true.

Quote:

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Originally Posted by transgalactic2

so its a const pointer
it cannot be changed in value.

---

However, this is still not true. It is an array, not a pointer of any sort. Granted, the difference only shows up in the use of sizeof, but there is a semantic difference nonetheless.

No, it's an array, and not a pointer, therefore you cannot assign an address to it.
If you pass an array to a function, a pointer to its first element is passed instead.

wow... I started an arguement lol, I have looked-up memcpy(), but only found syntax and not examples of how it should be used. However it is late now and I shall post back in the morning. And yes by the way 'same-age' is an array.

Quote:

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Originally Posted by cus

wow... I started an arguement lol, I have looked-up memcpy(), but only found syntax and not examples of how it should be used.

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I am worried that you believe there is a difference.

Quote:

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Originally Posted by tabstop

I am worried that you believe there is a difference.

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a difference to what exactly?

Very well, I shall make an example for you.

You've looked up memcpy, so you know that it will copy n bytes from the source pointed to by the second argument to the destination pointed to from the first argument. Simply calculate your bytes.

Code:

---

#include <string.h>
int main()
{
    int age[4] = {59, 89, 23, 20};
    int sameAge[4];

    memcpy(sameAge, age, sizeof age[0] * 4);

    return 0;
}

---

Like that. But a loop would work just as well. memcpy is essentially that in a function.

Quote:

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Originally Posted by cus

a difference to what exactly?

---

A difference between the syntax and how it should be used.

thank you very much for your useful post, the information that memcpy(), is essentially a functions explains fully. And also answers the question to my previous post to 'tabstop'

Incidentally, in this case you could just write:

Code:

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memcpy(sameAge, age, sizeof sameAge);

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but the multiplication would be more like what you would do if this were in another function.

thanks for the help, i have done it this way:

Code:

---

#include <string.h>
#include <stdio.h>
int main()
{
    int age[4] = {169, 8, 17, 20};
    int sameAge[4];
    int i;

memcpy(sameAge, age, sizeof age);
for( i=0; i<4; i++)
{       
printf("%d\n", sameAge[i]);
}
    return 0;
}

---

Consequently, it can also be written as:

Code:

---

memcpy( sameAge, age, sizeof(age) );

---

Which is my preferred method. It also saves you headaches since sizeof something does not work for everything, but sizeof(something) always works.

yes thanks, i think i just read that on wikipedia, due to the type name always need to be enclosed in parenthesis i.e.

printf("%u,%u", (unsigned) sizeof c, (unsigned) sizeof(int) );