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I have a vector that looks like this.

std::vector<std::pair<int,int> > v;

and I want to pass it into a function and make it so it's constant.

first I tried

void fun(const std::vector<std::pair<int,int> > v)

but that function allowed me to have the line

std::pair<int,int> p = v[i];

which I thought should have thrown an error since pair is not of type const. Then I realized that it's only the pointers that are being declared constant in vector so I tried 

void fun(const std::vector<const std::pair<int,int> > v)

but that throws the error about there being no conversion.

I'm sure there's some inner workings that I'm not understanding that makes this illegal but would love some insight. Thanks.

financial_physician
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2 Answers2

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I think you confuse reference and value semantics here. Lets the first function signature you tested:

void fun(const std::vector<std::pair<int,int> > v)

This copies the function argument into a new object v, which is additionaly const-qualified. Such signatures seldomly make sense: either you want to pass it as a const reference (to avoid copying) or you want to pass it by non-const value, because the function body will mutate the argument, but shall operate on its only copy. Besides, this snippet:

std::pair<int,int> p = v[i];

compiles just fine with the above function signature, because it copies the vector element at position i into a new pair object. The latter can be mutated, but this doesn't affect the vector at all.

Let's now consider the second function signature:

void fun(const std::vector<const std::pair<int,int> > v)

The same as before still applies here, and in addition, std::vector<const T> isn't useful, see this thread for an explanation.

How to fix it? If you don't want to copy the argument, but the function doesn't modify the vector, pass it by const reference. If the function modifies the vector and these modifications shall be visible at the call side, pass it as non-const reference. If the function modifies the vector but that shall happen independently of the call side vector, pass by non-const value.

lubgr
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    Another reason it's inappropriate: The declaration `void fun(const std::vector > v)` is **the same** as `void fun(std::vector > v)`, top level cv qualification is ignored. – Caleth Jun 07 '19 at 08:40
  • @Caleth Sorry I don't get that. Did you forget a `const` in the template type? – lubgr Jun 07 '19 at 08:47
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    No, I mean "Top-level cv-qualifiers are dropped from the parameter type" https://en.cppreference.com/w/cpp/language/function – Caleth Jun 07 '19 at 08:49
  • @Caleth Interesting, I wasn't aware of that. Thanks for the hint! – lubgr Jun 07 '19 at 08:52
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void fun(const std::vector<std::pair<int,int> > v)
{
   ////
   std::pair<int,int> p = v[i];
}

https://en.cppreference.com/w/cpp/container/vector/operator_at

No error will happen here because std::vector has overloaded operator[] both for const and non-const instances. In your case it will be resolved to const version and return const & to underlying i element.
Consider following code:

void fun(const std::vector<std::pair<int,int> > v)
{
   ////
   std::pair<int,int> p = v[i]; //OK
   v[i] = std::pair<int,int>{10,20}; //Compile error, trying to modify const data!
}
b2mb0e
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