C++ unknown errors

Question

I have the following function in C++:

std::wstring decToDeg(double input) { int deg, min; double sec, sec_all, min_all; std::wstring output; sec_all = input * 3600; sec = Math::Round(static_cast<int>(sec_all) % 60, 3); //code from @spin_eight answer min_all = (sec_all - sec) / 60; min = static_cast<int>(min_all) % 60; deg = static_cast<int>(min_all - min) / 60; output = deg + L"º " + min + L"' " + sec + L"\""; return output; }

When I try to compile I get this error:

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

What can I do to correct these two errors in my function?

EDIT: solved

std::wstring decToDeg(double input) { int deg, min; double sec, sec_all, min_all; sec_all = input * 3600; sec = Math::Round(static_cast<int>(sec_all) % 60, 3); min_all = (sec_all - sec) / 60; min = static_cast<int>(min_all) % 60; deg = static_cast<int>(min_all - min) / 60; std::wostringstream output; output << deg << L"º " << min << L"' " << sec << L"\""; return output.str(); }

Also, you're trying to sum numbers with strings, which can't be done like this in C++ — Kiril Kirov
– Kiril Kirov, Commented Nov 22, 2012 at 7:37
Could you tell me what to do to solve these errors? I am totally new to C++... — Victor
– Victor, Commented Nov 22, 2012 at 7:40
Look at stackoverflow.com/questions/64782/… for various ways to fix #2 — Karthik T
– Karthik T, Commented Nov 22, 2012 at 7:46

Angew is no longer proud of SO · Accepted Answer · 2012-11-22 07:48:21Z

1

You could use a string stream to construct output, like this:

std::wostringstream output; output << deg << L"º " << min << L"' " << sec << L"\""; return output.str();

answered Nov 22, 2012 at 7:48

Angew is no longer proud of SO

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Angew is no longer proud of SO Over a year ago

@Victor What error? BTW, of course you have to remove your original declaration of output as a wstring, and you have to #include <sstream>.

spin_eight · Accepted Answer · 2012-11-22 07:42:32Z

1

sec = Math::Round(static_cast<int>(sec_all) % 60, 3);

answered Nov 22, 2012 at 7:42

spin_eight

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spin_eight Over a year ago

output = std::to_string(deg) + L"º " +std::to_string(min) + L"' " + std::to_string(sec) + L"\"";

ISTB · Accepted Answer · 2012-11-22 07:44:56Z

You cannot use modulo on doubles. Modulo on doubles:

 int result = static_cast<int>( a / b ); return a - static_cast<double>( result ) * b;

Strike · Accepted Answer · 2012-11-22 07:45:27Z

For the first error try this:

 min = (static_cast<int>(min_all) ) % 60;

This should make sure the cast will be completed first, before trying to do any other calculating.

If your other error is not an error that was caused by the first error, you may want to try using stringstream. It behaves like a normal I/O stream and is thus ideal for formatting strings.

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