Thread: 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 a for loop.

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

i thought memcpy() was a function for strings?

i have no idea how to use memcpy()

Originally Posted by cus

i have no idea how to use memcpy()

So look it up.

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.

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:

same_age=&age[0];

so theoretically i cant see why it cant happen?

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