Can I use `distance` as my class member variable name? - c++

Question

It would be ok to define a class like this (ignore its implmentation):
class MyEngine {
private:
int* params;
int param_len;
public:
void set_params(int* _params, int _len);
float get_result(); // relies on `distance` member
float distance; // people can modify this
};
However, if using vector, assume it implicitly includes <iterator> which contains std::distance, how do compiler distinguish std::distance and distance member? (Or will it cause un-expected crash when run?). Say, the function get_result() relies distance member value.
#include <vector>
using namespace std;
class MyEngine {
private:
vector<int> params;
public:
void set_params(int* _params, int _len);
float get_result(); // relies on `distance` member
float distance; // people can modify this
};
update
As people mentioned, using namespace std is bad practice; however, there are still people writing code with using namespace std, and if we are collaborate with them, using their code, is there any concreate example that demonstrate the badness of using namespace std, expecially severe run error? This, would be my real purpose.
There is an answer, saying distinguish the two distance by type. Let's just try this snippet:
#include <stdio.h>
#include <stdlib.h>
#include <vector>
#include <iterator>
using namespace std;
class MyEngine {
private:
vector<int> params;
float* distance; // people can modify this
int len;
public:
void setup();
};
void MyEngine::setup()
{
len = 100;
distance = (float*)malloc(sizeof(float)*len);
for(int i=0; i<len; i++) {
distance[0] = len - i;
params.push_back(len-i);
}
int num = distance(params.begin(), params.end());
printf("distance is: %d\n", num);
}
int main(){
MyEngine engine;
int len = 10;
engine.setup();
return 0;
}
Which, would cause compile error saying:
main.cpp:25:23: error: called object type 'float *' is not a function or function pointer
int num = distance(params.begin(), params.end());
~~~~~~~~^
1 error generated.
Demonstrates that it can't distinguish the two distance from their types.

Answer 1

Well, one is MyEngine::distance, and the other is std::distance. They're different names. This is the whole point of scopes.
There is only a problem if you use the unqualified name distance and leave the compiler to figure it out, but if that's not going to work then it'll be because the type of the one chosen doesn't match your usage, so your program won't compile.
If you ever did put things in std then the name can clash in potentially undiagnosable ways, which can cause crashes, and has undefined behaviour per the standard.

Answer 2

The function distance has a different type than the member distance, so the compiler figures that out by the type.
You can even have a function distance(...). As long as parameters are different, they have different type and the compiler figures that out.
Also note that you should not use using namespace std because that also irritates the human reader.
In your updated example, the compiler is indeed confused. You can specify that you want to use the distance function in this case by changing it to int num = std::distance(...);