Why c++ lambda capture needn't type declaration?

Question

A lambda function:

auto wc = find_if(words.begin(), words.end(),
[sz](const string &a)    //sz does not require type declaration
{
   return a.size() >= sz;
})

is equal to

class SizeComp {
    SizeComp(size_t n): sz(n) { }  // Type required here.
    bool operator()(const string &s) const { return s.size() >= sz; }
private:
    size_t sz; 
};

Why does the lambda capture of sz not require a type declaration?

Answer 1

由于捕获的变量是在父作用域中声明的，因此其类型是已知的。