std::chrono::time_point compiler error when converting from a variable

Question

I have a variable of type long long that represents a time point in nanoseconds.

I'm trying to wrap it using std::chrono::time_point yet the compiler (VS 2017) is giving me troubles.

here is a code that compiles:

std::chrono::time_point<std::chrono::steady_clock> tpStart(std::chrono::nanoseconds(10ll)); std::chrono::time_point<std::chrono::steady_clock> tpEnd = std::chrono::steady_clock::now(); double d = std::chrono::duration<double>(tpEnd - tpStart).count();

Now if i switch the value 10ll with a variable, the line that computes the duration fails to compile:

constexpr long long t = 10ll; std::chrono::time_point<std::chrono::steady_clock> tpStart(std::chrono::nanoseconds(t)); std::chrono::time_point<std::chrono::steady_clock> tpEnd = std::chrono::steady_clock::now(); double d = std::chrono::duration<double>(tpEnd - tpStart).count();

Here is the error code:

error C2679: binary '-': no operator found which takes a right-hand operand of type 'overloaded-function' (or there is no acceptable conversion)

Any Idea why is this happening? how can I convert a variable of type long long to a std::chrono::time_point?

Martin Morterol · Accepted Answer · 2019-06-16 13:22:57Z

TLDR : It's a most vexing parse case

prog.cc:8:59: warning: parentheses were disambiguated as a function declaration [-Wvexing-parse] std::chrono::time_point<std::chrono::steady_clock> tpStart(std::chrono::nanoseconds(t));

Fix with {} instead of ()

constexpr long long t = 10ll; std::chrono::time_point<std::chrono::steady_clock> tpStart{std::chrono::nanoseconds{t}}; std::chrono::time_point<std::chrono::steady_clock> tpEnd = std::chrono::steady_clock::now(); double d = std::chrono::duration<double>(tpEnd - tpStart).count();

Why it's a most vexing parse?

std::chrono::time_point<std::chrono::steady_clock> pStart (std::chrono::nanoseconds(t)); // ^^^^ Type 1 ^^^^^ ^name^ ^^^^ Type 2 ^^^^

So we can reproduce with :

constexpr long long t = 10ll; int fail (double (t) ); fail = 6; // compilation error here

Why ? Let's remove some noise :

int fail (double (t) ); // <=> int fail (double t); // <=> int fail (double) ; // Most vexing parse.

We can fix it by switching to {}

int Ok {double {t} };

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