3 ms·

Never quite understood why compound literals are lvalues, but fine, whatever, I guess, it's so that you can write "&(struct Foo){};" instead of "struct Foo tmp;

Never quite understood why compound literals are lvalues, but fine, whatever, I guess, it's so that you can write "&(struct Foo){};" instead of "struct Foo tmp; &tmp;"... which, on a tangential note, reminds me about Go: the proposals to make things like &5 and &true legal in Go were rejected because "the implied semantics would be unclear" even though &structFoo{} is legal and apparently has obvious semantics.

leni536 4y ago
In C compound literals have a relatively long lifetime compared to C++ temporaries. With these lifetime rules it makes sense that they are lvalues, although I like C++ rvalues (especially prvalues) more. https://cigix.me/c17#6.5.2.5.p5 https://cigix.me/c17#6.5.2.5.p5 > If the compound literal occurs outside the body of a function, the object has static storage duration; otherwise, it has automatic storage duration associated with the enclosing block.
cataphract 4y ago
It's useful when a function has a out or in/out struct parameter whose value at the end you're not interested in. Or in functions where the struct is an input parameter, but they return it as a return value too, which you can then assign to a pointer variable or immediately pass to another function. Note that the struct values thus created have longer lifetimes than temporary C++ objects created directly inside the argument list of a function call.
flohofwoe 4y ago
That's extemely useful for invoking a function that takes a pointer to a struct, but you don't want 'taint' the code with a temporary variable that's only needed for the function call: func(&(bla_t){ .x = 0, .y = 1 });
gallier2 4y ago
Compound literals are anonymous variables, i.e. like variables except that there is no label visible in the C program. If you look at the assembly you'll see that they are declared exactly like a variable except with a compiler generated internal label.

leni536 4y ago

In C compound literals have a relatively long lifetime compared to C++ temporaries. With these lifetime rules it makes sense that they are lvalues, although I like C++ rvalues (especially prvalues) more. https://cigix.me/c17#6.5.2.5.p5 https://cigix.me/c17#6.5.2.5.p5 > If the compound literal occurs outside the body of a function, the object has static storage duration; otherwise, it has automatic storage duration associated with the enclosing block.

cataphract 4y ago

It's useful when a function has a out or in/out struct parameter whose value at the end you're not interested in. Or in functions where the struct is an input parameter, but they return it as a return value too, which you can then assign to a pointer variable or immediately pass to another function. Note that the struct values thus created have longer lifetimes than temporary C++ objects created directly inside the argument list of a function call.

flohofwoe 4y ago

That's extemely useful for invoking a function that takes a pointer to a struct, but you don't want 'taint' the code with a temporary variable that's only needed for the function call: func(&(bla_t){ .x = 0, .y = 1 });

gallier2 4y ago

Compound literals are anonymous variables, i.e. like variables except that there is no label visible in the C program. If you look at the assembly you'll see that they are declared exactly like a variable except with a compiler generated internal label.