Is there a way to define a temporary variable for use as a function argument when calling a function? There is a function:
int hello(const int *p);
call it with
int a = 10;
hello(&a);
but the the var a won't be used after that, so I want them in one line, something like this:
hello(&(int a = 10)); // can't compile
hello(&int(10)); // can't compile
What is the right form to express this?
Answer
The proper syntax for a compound literal is as follows:
hello(&(const int){10});
Compound Literal
According to IBM:
The syntax for a compound literal resembles that of a cast expression. However, a compound literal is an lvalue, while the result of a cast expression is not. Furthermore, a cast can only convert to scalar types or void, whereas a compound literal results in an object of the specified type.
const one could be placed in the same area as static const data. They could all be conflated too, in the same way as string literals can, e.g. &(const int){10} might be the same address for every time that expression occurs in your program.I think you have to define a first. But you can do it inside curly braces this effectively limiting scope of a:
{
int a = 10;
f (&a);
}
You can use the scope concept to do this.
Since "a" is an automatic integer, its scope is limited to the block in which it is declared.
Thus you can simply enclose the declaration and the function call inside a block, like this:
{
int a = 10;
hello(&a);
}
