Passing an Array to a structure in C - c

Question

In order to fill a structure with integers (to then be passed on further in the program) I thought the following would work:
main() {
struct songs {int pitch[5], length[5];} songs[4];
int i[5]={1,22,23,14,52};
int k=0;
songs[0].pitch=i;
for (k=0; k<5; k++) printf("%d\n",songs[0].pitch[k]);
}
however this comes up with an error "incompatible types in assignment"
if I dont however pass this array to the structure, using the following:
main() {
int i[5]={1,22,23,14,52};
int k=0;
for (k=0; k<5; k++) printf("%d\n",i[k]);
}
it compiles and will display the content of the array.
I realise there is probably a simple fix to this, but any help would be brilliant!
Thanks in advance

Answer 1

C89 does not allow you to assign an array to another array. Here's the relevant bit from C99, but the text is much the same in C89 with the exception of the mention of the C99 only type _Bool. (I only have paper copies of C89)
Arrays don't fit any of these conditions -- they aren't arithmetic types, they aren't of structure or union type, and they're not pointers1. Therefore you can't use the assignment operator on them.
You can, however, use memcpy. If you replace this line:
songs[0].pitch=i;
with this line:
memcpy(songs[0].pitch, i, sizeof(i));
you'll get the behavior you expect. (After including <string.h> first of course)
1 Technically speaking 6.3.2.1/3 says that the array is converted into an rvalue pointer before it is seen by operator=, but such an assignment is still prohibited because 6.5.16/2 requires that the left side of an assignment be an lvalue.

Answer 2

Due to the way C handles arrays you can't assign to one like that. You need to copy over the values, either individually or via memcpy (or similar functions). Example:
for (k=0; k<5; k++){
songs[0].i[k] = i[k];
}
or:
memcpy(songs[0].i, i, sizeof i);
Note that memcpy requires you to include <string.h>

Answer 3

You will need to copy the array elements individually. Also the main function really should return an int.
int main( void ) {
struct songs {int pitch[5], length[5];} songs[4];
int i[5]={1,22,23,14,52};
int qq, k=0;
for( qq=0; qq<5; qq++) {
songs[0].pitch[qq]=i[qq];
}
for (k=0; k<5; k++) printf("%d\n",songs[0].pitch[k]);
return 0;
}

Answer 4

C does not let you copy arrays into arrays. You have to do an element by element copy.
struct songs {int pitch[5], length[5];} songs[4];
int i[5]={1,22,23,14,52};
int k=0;
for (k=0; k<5; k++) songs[0].pitch[k] = i[k];
for (k=0; k<5; k++) printf("%d\n",songs[0].pitch[k]);

Answer 5

when you declare an array, you have 2 options:
int * i; // #1; you do this when you don`t know the size
int i[size]; // #2
you can not assign an array to another array.
what you need to do, as others suggested, is to copy all elements of the array one by one:
for (int j = 0; j < 5; j++) {
songs[0].pitch[j] = i[j];
}
remember that using the index operator, [], is the same as dereferencing the pointer.
even more, when you say pitch[j] you actually move the pointer j positions forward and dereference it; like you would be saying *(pitch+j)

Answer 6

the following, compiles, runs, works.
notice the correction to the declaration of main
notice the addition of #include string.h to support memcpy function
notice the proper return statement at the end
and most importantly,
notice the correct method of copying an array
as a side note:
making the struct tag name and the instance name the same
is very bad programming practice
#include <stdio.h>
#include <string.h>
int main()
{
struct songs
{
int pitch[5];
int length[5];
} songs[4];
int i[5]={1,22,23,14,52};
int k=0;
memcpy( songs[0].pitch, i, sizeof( i ) );
for (k=0; k<5; k++)
{
printf("%d\n",songs[0].pitch[k]);
} // end for
return(0);
}

Answer 7

Problem is here,
songs[0].pitch=i;
Use int* in the structure, You can't assign an array to another array and name of the structure and the structure object.
This would work.
struct Songs {int* pitch, length[5];} songs[4];