how come an undeclared variable is outputting a value

Question

In my function sumofrange I decided to output an undeclared variable just to learn the different compiler errors in C++. To my surprise, time seems to output 1 even though it is not declared anywhere.

#include <iostream>
#include <cmath>

using namespace std;

int  sumOfrange( int   lower, int  upper){
   cout<<time<<endl;
   return ((( (pow(upper,2)) + upper) - ((pow(lower,2)) + lower)) / 2);
}

int main(){  
    cout<<sumOfrange(7,100)<<endl;
    return 0;
}

This is why namespaces are important and `using namespace std;` is sometimes dangerous. — ForceBru, Aug 31 '17 at 14:51
You should read [this](https://stackoverflow.com/questions/1452721/why-is-using-namespace-std-considered-bad-practice) — NathanOliver, Aug 31 '17 at 14:53
btw there is no such thing as an "undeclared variable". If it was undeclared you could not use it — 463035818_is_not_an_ai, Aug 31 '17 at 14:54
@ForceBru Remove `using namespace std;`, and you'll get the same output for `std::cout << time << '\n';` on a lot of implementation. — Holt, Aug 31 '17 at 15:09
Even if `using namespace std;` should probably be removed here, the real solution would be to enable your compiler warning (`-W -Wall` with clang / gcc) to obtain a warning for such code. — Holt, Aug 31 '17 at 15:11
I disagree with the dupe target. [Removing `using namespace std;`](http://coliru.stacked-crooked.com/a/fd9c9486c617ce8d) has no effect on the issue. — nwp, Aug 31 '17 at 15:11
BTW, the [`std::pow` function](http://en.cppreference.com/w/cpp/numeric/math/pow) is for floating point. To square numbers, use multiplication, e.g. `upper * upper` and `lower * lower`. The multiplication is more efficient since it doesn't require a function call. — Thomas Matthews, Aug 31 '17 at 15:18

Ron · Answer 1 · 2017-08-31T15:37:39.387

4

You are outputting the address of a std::time function declared in a <ctime> header. You are also using a using namespace std; statement. Why that should be avoided is explained in this SO post. Depending on the compiler and the platform you might get the hexadecimal output similar to (0x)00DC52E0 if using a VC++ compiler on Windows or a number 1 if using a g++ compiler on Linux.

edited Aug 31 '17 at 15:37

answered Aug 31 '17 at 14:50

Ron

14,674
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Yet another example of the subtle pitfalls of using `using namespace std;`. – François Andrieux Aug 31 '17 at 14:51
3

not sure if a function has a value. Anyhow its the adress of the function – 463035818_is_not_an_ai Aug 31 '17 at 14:51
3

I believe it is being cast to a `bool` and returned as 1. `cout<<(void*)time< – Persixty Aug 31 '17 at 14:53
@Persixty You should not mix function pointers and `void *`. Use `&time` instead. – nwp Aug 31 '17 at 14:57
@nwp I agree. But `&time` just outputs 1 as well. Forcing a cast makes it output some kind of address... – Persixty Aug 31 '17 at 15:03
2

@Persixty Interesting. Apparently there is no overload for function pointers on `std::cout` which only leaves the implicit conversion to `bool`. Someone should fix that in a standards proposal. – nwp Aug 31 '17 at 15:07
@nwp Crazy, huh? – Persixty Aug 31 '17 at 15:10

how come an undeclared variable is outputting a value

1 Answers1