# compound literals

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Constructs an unnamed object of specified type in-place, used when a variable of array, struct, or union type would be needed only once.

### Syntax

`(` type `)` `{` initializer-list `}` |  | (since C99)

where

type | - | a type name specifying any complete object type or an array of unknown size, but not a VLA
initializer-list | - | list of initializers suitable for initialization of an object of type

### Explanation

The compound literal expression constructs an unnamed object of the type specified by type and initializes it as specified by initializer-list.

The type of the compound literal is type (except when type is an array of unknown size; its size is deduced from the initializer-list as in array initialization).

The value category of a compound literal is lvalue (its address can be taken).

The unnamed object to which the compound literal evaluates has static storage duration if the compound literal occurs at file scope and automatic storage duration if the compound literal occurs at block scope (in which case the object's lifetime ends at the end of the enclosing block).

### Notes

Compound literals of const-qualified character or wide character array types may share storage with string literals.

    (const char []){"abc"} == "abc" // might be 1 or 0, implementation-defined

Each compound literal creates only a single object in its scope:

    int f (void)
    {
        struct s {int i;} *p = 0, *q;
        int j = 0;
    again:
        q = p, p = &((struct s){ j++ });
        if (j < 2) goto again; // note; if a loop were used, it would end scope here,
                               // which would terminate the lifetime of the compound literal
                               // leaving p as a dangling pointer
        return p == q && q->i == 1; // always returns 1
    }

Because compound literals are unnamed, a compound literal cannot reference itself (a named struct can include a pointer to itself)

Although the syntax of a compound literal is similar to a cast, the important distinction is that a cast is a non-lvalue expression while a compound literal is an lvalue.

### Example

Run this code

    int *p = (int[]){2, 4}; // creates an unnamed static array of type int[2]
                            // initializes the array to the values {2, 4}
                            // creates pointer p to point at the first element of the array
    const float *pc = (const float []){1e0, 1e1, 1e2}; // read-only compound literal

    int main(void)
    {
        int n = 2, *p = &n;
        p = (int [2]){*p}; // creates an unnamed automatic array of type int[2]
                           // initializes the first element to the value formerly held in *p
                           // initializes the second element to zero
                           // stores the address of the first element in p

        struct point {double x,y;};
        void drawline1(struct point from, struct point to);
        void drawline2(struct point *from, struct point *to);
        drawline1((struct point){.x=1, .y=1},  // creates two structs with block scope
                  (struct point){.x=3, .y=4}); // and calls drawline1, passing them by value
        drawline2(&(struct point){.x=1, .y=1},  // creates two structs with block scope
                  &(struct point){.x=3, .y=4}); // and calls drawline2, passing their addresses
    }

### References

- C11 standard (ISO/IEC 9899:2011):

  - 6.5.2.5 Compound literals (p: 85-87)

- C99 standard (ISO/IEC 9899:1999):

  - 6.5.2.5 Compound literals (p: 75-77)

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