Where are default parameters evaluated in c++? - c++

Question

If there is a function like the following,
void logData(std::string data, int line=__LINE__);
is the default parameter evaluated where the function is called, or where it is declared? Is the behavior standard, or does it depend on the compiler?

Answer 1

You asked:
is the default parameter evaluated where the function is called, or where it is declared? Is the behavior standard, or does it depend on the compiler?
The value of the default parameter will be the line number of the file where the function is declared. That is as per the standard.
From the C++11 Standard:
16.8 Predefined macro names
...
_ _ LINE _ _
The presumed line number (within the current source file) of the current source line (an integer constant).
Since __LINE__ is a pre-processor macro, the code being compiled by the compiler does not have __LINE__ in it. In its place, it will see an integer constant that represents the line number of the line in the file, which is the file where the function is declared.

Answer 2

This sounds like a question a Python programmer might ask but you've picked a variant that has two answers.
The first part is __LINE__ which is a pre-processor macro. Pre-processing is, as the name suggests, done before some "processing" step. In this case, it happens before the compiler tries to compile the code. pre-processing involves replacing "#include" statements with the pre-processed contents of the file, evaluating and replacing macros, #define and #if statements etc.
So when the compiler sees
void logData(... __LINE__)
on line 123 of a file, the __LINE__ is replaced with the constants 123. You can avoid this by using macros, e.g.
enum Level { DEBUG, NOTE, WARNING, ERROR };
#define FLD __FILE__, __LINE__, DEBUG
#define FLN __FILE__, __LINE__, NOTE
#define FLW __FILE__, __LINE__, WARNING
#define FLE __FILE__, __LINE__, ERROR
void log(const char* file_, size_t line_, Level level_, const char* what);
...
logData(FLD, "debug message");
As to where the arguments are "evaluated", call site; so you will not run into Python's
from __future__ import print_function
def f(inlist, l=[]):
if l:
l.append("unexpected surprise")
l.extend(inlist)
return l
f([1])
print(f([2]))
http://ideone.com/GvmsL6
[1, 'unexpected surprise', 2]
in c++
#include <iostream>
#include <vector>
#include <string>
using strvec = std::vector<std::string>;
strvec f(strvec in, strvec out={})
{
if (!out.empty())
out.push_back("unexpected surprise");
out.insert(out.end(), in.cbegin(), in.cend());
return out;
}
int main()
{
f({"a"});
auto vec = f({"b"});
for (auto& str: vec) {
std::cout << str << "\n";
}
}
See http://ideone.com/17ilNL