CODEWITHSUNDEEP
c
Barioth
Barioth

Reputation: 3

Affecting a "static" struct in another struct in heap

So I solved my problem, wasn't that hard, but I'm wandering why the first version wasn't working. So here, an example of my problem :

typedef struct a {
    int i;
}   A;

typedef struct b {
    int i;
    A a;
}   B;

typedef B * PB;

PB create_B(int ia, int ib) {
    PB b = malloc(sizeof(B));
    b->i = ib;
    b->a = {ia};
}

I get an error from the compiler saying :

"error: expected expression before ‘{’ token"

at line: b->a = {ia};

But i don't really get what is the problem.

I solved it casting the structure :

PB create_B(int ia, int ib) {
    PB b = malloc(sizeof(B));
    b->i = ib;
    b->a = (A){ia};
}

But the types are well defined no ? I mean it's kind of obvious to me that {ia} is of type A since b->a is of type A aswell.

I am probably wrong on this last saying (compiler is probably right). So if you have an example where this situation isn't obvious and really need a cast, it would be really appreciated, or jut an explanation at least.

Thank you for your time.

Barioth

PS : I could have done that too i guess

PB create_B(int ia, int ib) {
    PB b = malloc(sizeof(B));
    *b = (B){ib, {ia}};
}

But i still need a cast...

Upvotes: 0

Views: 40

Answers (1)

Eric Postpischil
Eric Postpischil

Reputation: 223712

{ia} is not an expression in C and cannot be used in an expression statement.

{ia} is part of a syntax for definitions. In a definition, a list of initial values may be given in braces after an =. (And the braces may be omitted when initializing a scalar object.) This is a special syntax for initializing in a definition and is not an assignment like using = in an expression.

(A){ia} is an expression; it is a construct called a compound literal. After a type name in parentheses, initial values are given inside braces. This creates an object of the stated type. That object may then be used in an expression, such as as the right operand of an assignment.

Upvotes: 1

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