Why doesn't compound literals assignment work without a typecast

Question

I have a question about literals in C.

int a;
//a is an integer that is assigned an integer literal 414
a = 414;

float b;
//b is a float that is assigned a float literal of 3.14
b = 3.14;

struct point {
    int x,y;
};

struct point b;
//{5,6} is a compound literal that is assigned to a struture.
b = {5,6}; //doesn't work.

b = (struct point){5,6}; //works.

That doesn't seem to work without a typecast? What is the reason for this?

score 5 · Accepted Answer · edited Jun 20 '20 at 09:12

5

(struct point){5,6} as a whole is a compound literal.

C11 §6.5.2.5 Compound literals

A postfix expression that consists of a parenthesized type name followed by a brace enclosed list of initializers is a compound literal.

edited Jun 20 '20 at 09:12

Community

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answered Sep 01 '14 at 10:22

Yu Hao

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Since the type can be inferrd from the decleration of point b before the initialization,why doesn't it work? – liv2hak Sep 01 '14 at 10:25
3

@liv2hak `(struct point)` is not a cast, it's part of the compound literal. – Yu Hao Sep 01 '14 at 10:27
Can't that be infered from the Symbol Table? – liv2hak Sep 01 '14 at 10:32
3

@liv2hak: It cannot be inferred. `{5,6}` could be an `int[2]`, or a `struct { char a, b; }`, ... – mafso Sep 01 '14 at 10:53

Why doesn't compound literals assignment work without a typecast

1 Answers1

C11 §6.5.2.5 Compound literals

Linked

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