If by "shallow copy", you mean that after assignment of a struct containing an array, the array would point to the original struct's data, then: it can't. Each element of the array has to be copied over to the new struct. "Shallow copy" comes into the picture if your struct has pointers. If it doesn't, you can't do a shallow copy.
When you assign a struct containing an array to some value, it cannot do a shallow copy, since that would mean assigning to an array, which is illegal. So the only copy you get is a deep copy.
Consider:
#include <stdio.h>
struct data {
char message[6];
};
int main(void)
{
struct data d1 = { "Hello" };
struct data d2 = d1; /* struct assignment, (almost) equivalent to
memcpy(&d2, &d1, sizeof d2) */
/* Note that it's illegal to say d2.message = d1.message */
d2.message[0] = 'h';
printf("%s
", d1.message);
printf("%s
", d2.message);
return 0;
}
The above will print:
Hello
hello
If, on the other hand, your struct had a pointer, struct assignment will only copy pointers, which is "shallow copy":
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct data {
char *message;
};
int main(void)
{
struct data d1, d2;
char *str = malloc(6);
if (str == NULL) {
return 1;
}
strcpy(str, "Hello");
d1.message = str;
d2 = d1;
d2.message[0] = 'h';
printf("%s
", d1.message);
printf("%s
", d2.message);
free(str);
return 0;
}
The above will print:
hello
hello
In general, given struct T d1, d2;, d2 = d1; is equivalent to memcpy(&d2, &d1, sizeof d2);, but if the struct has padding, that may or may not be copied.
Edit: In C, you can't assign to arrays. Given:
int data[10] = { 0 };
int data_copy[10];
data_copy = data;
is illegal. So, as I said above, if you have an array in a struct, assigning to the struct has to copy the data element-wise in the array. You don't get shallow copy in this case: it doesn't make any sense to apply the term "shallow copy" to a case like this.
