为什么有些 glibc 函数包含在头文件中，有些包含在共享库中？

Question

Simple example:

#include <iostream>

int main(){
 float a = std::min(1.0, 2.0);
 printf("hello\n");
 return 0; 
}

Compiling the above with -g and using gdb we can see that std::min is part of the final binary, provided by <iostream> and not in a shared lib:

/usr/include/c++/6/bits/stl_algobase.h

template<typename _Tp>
_GLIBCXX14_CONSTEXPR
inline const _Tp&
min(const _Tp& __a, const _Tp& __b)
{
// concept requirements
  __glibcxx_function_requires(_LessThanComparableConcept<_Tp>)
//return __b < __a ? __b : __a;
  if (__b < __a)
    return __b;
  return __a;
}

However printf is linked dynamically and is part of the libc runtime, and does not exist within my final binary.

Why isn't std::min part of the runtime libc also?

Answer 1

Why isn't std::min part of the runtime libc also?

Firstly, libc is an implementation of the C standard library, and it doesn't have the namespace std because it is the standard library for the C language which is a language doesn't have namespaces.

Secondly, the C standard doesn't specify that the standard library has such function as min , so there is no reason for a C standard library implementation to provide such function.

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In case you are wondering why the std::min template from the C++ standard library is compiled into your binary, it is because the instance of that template is an inline function.