You are facing issue in
s1.name=”Paolo”;
because, in the LHS, you’re using an array type, which is not assignable.
To elaborate, from C11, chapter §6.5.16
assignment operator shall have a modifiable lvalue as its left operand.
and, regarding the modifiable lvalue, from chapter §6.3.2.1
A modifiable lvalue is an lvalue that
does not have array type, […]
You need to use strcpy() to copy into the array.
That said, data s1 = {“Paolo”, “Rossi”, 19}; works fine, because this is not a direct assignment involving assignment operator. There we’re using a brace-enclosed initializer list to provide the initial values of the object. That follows the law of initialization, as mentioned in chapter §6.7.9
Each brace-enclosed initializer list has an associated current object. When no
designations are present, subobjects of the current object are initialized in order according
to the type of the current object: array elements in increasing subscript order, structure
members in declaration order, and the first named member of a union.[….]
typedef struct{
char name[30];
char surname[30];
int age;
} data;
defines that data should be a block of memory that fits 60 chars plus 4 for the int (see note)
[—————————-,——————————,—-]
^ this is name ^ this is surname ^ this is age
This allocates the memory on the stack.
data s1;
Assignments just copies numbers, sometimes pointers.
This fails
s1.name = “Paulo”;
because the compiler knows that s1.name is the start of a struct 64 bytes long, and “Paulo” is a char[] 6 bytes long (6 because of the trailing in C strings)
Thus, trying to assign a pointer to a string into a string.
To copy “Paulo” into the struct at the point name and “Rossi” into the struct at point surname.
memcpy(s1.name, “Paulo”, 6);
memcpy(s1.surname, “Rossi”, 6);
s1.age = 1;
You end up with
[Paulo0———————-,Rossi0————————-,0001]
strcpy does the same thing but it knows about termination so does not need the length hardcoded.
Alternatively you can define a struct which points to char arrays of any length.
typedef struct {
char *name;
char *surname;
int age;
} data;
This will create
[—-,—-,—-]
This will now work because you are filling the struct with pointers.
s1.name = “Paulo”;
s1.surname = “Rossi”;
s1.age = 1;
Something like this
[—4,–10,—1]
Where 4 and 10 are pointers.
Note: the ints and pointers can be different sizes, the sizes 4 above are 32bit as an example.