Why does sqrt() work fine on an int variable if it is not defined for an int?

问题

In chapter 3 of Programming: Principles and Practice using C++ (sixth printing), Stroustrup states (p.68): "Note that sqrt() is not defined for an int".

Here is a simple C++ program based on that chapter:

#include "std_lib_facilities.h"

int main()
{
    int n = 3;
    cout << "Square root of n == " << sqrt(n) << "\n";
}

Given the quote above, I would expect the process of compiling or running this program to fail in some way.

To my surprise, compiling it (with g++ (GCC) 4.2.1) and running it succeeded without errors or warnings, and produced the following perfectly decent output:

Square root of n == 1.73205

My question therefore is: if sqrt() really is not defined for an int, then why doesn't the program above fail somehow?

回答1:

Update 2

This question was merged with an exact duplicate, on taking a look at this, the actual answer is much simpler than anyone originally thought. The current version of std_lib_facilities.h includes the following line:

inline double sqrt(int x) { return sqrt(double(x)); }   // to match C++0x

which creates a specific overload for the int case to match what modern compilers should be be doing which is cast integer arguments to double, although this version does not cover all the cases.

If std_lib_facilities.h was not being used than the original logic still applies, although gcc-4.2 is rather old compared to Visual Studio 2012 from the original question but a 4.1.2 version have uses __builtin_sqrt specifically for the integer case.

Original

Since around 2005 the draft standard required integer argument to be cast to double this is covered in the draft C++ standard. If we look in section 26 Numerics library and then go to section 26.8 C library which covers the <cmath> header, it specifies overloads of the math functions for float, double and long double which is covered in paragraph 8:

In addition to the double versions of the math functions in , C++ adds float and long double overloaded versions of these functions, with the same semantics.

which would be ambiguous for the int case but the standard requires that sufficient overload are provided so that integer arguments are cast to double. It is covered in paragraph 11 which says(emphasis mine):

Moreover, there shall be additional overloads sufficient to ensure:

1. If any arithmetic argument corresponding to a double parameter has type long double, then all arithmetic arguments corresponding to double parameters are effectively cast to long double.
2. Otherwise, if any arithmetic argument corresponding to a double parameter has type double or an integer type, then all arithmetic arguments corresponding to double parameters are effectively cast to double.
3. Otherwise, all arithmetic arguments corresponding to double parameters have type float.

Update

As @nos points out it is possible that the version of sqrt being called is from math.h header as opposed to the overloads from cmath, if that is the case and there is likely a implementation defined caveat here then we may be reverting to old C style behavior if the only version available is sqrt(double) which would mean that int would be converted to double implicitly.

One way I found to test this on gcc and clang would be to use long type for a which along with -Wconversion flag triggers a warning for a potentially value altering conversion on my platform if we only have sqrt(double) available. Indeed if I include math.h instead of cmath we can produce this warning. Although I can not trigger this behavior in clang which seems to indicate this is implementation dependent.

回答2:

The 10 is being implicitly converted to a double. This will happen automatically as long as you have the correct function prototype for sqrt.

Edit: beaten by comments

回答3:

Because of implicit conversions. sqrt is defined for double, and an int value can be (and is) converted implicitly to a value of type double.

(It is in fact pretty hard to prevent a function that takes a double from being called with an int. You may get your compiler to emit a warning, but since this is typically a value-preserving conversion, even that may be hard. C++ inherits from C the design to try as hard as possible to make code work, even if it requires contortions. Other languages are much stricter about this sort of thing.)

回答4:

sqrt is defined for double. And C++ allows you to convert int to double implicitly.

int n = 3;
double x = sqrt(n);    // implicit conversion of n from int to double

Implicit conversions may happen when you use the value as a function parameter or assign it to a variable.

An example for the second case would be:

int n = 3;
double x = n;          // implicit conversion of n from int to double

Note that operators are also simply functions. Thus, you can also add an int to a double, which converts the int to a double before invoking the actual addition:

int n = 3;
double x = 1.0;
double sum = n + x;    // implicit conversion of n from int to double

回答5:

Because there's an implicit conversion from int to double.

With the conversion, your code would look like this:

cout << "Square root of n == " << sqrt((double)n) << "\n";

回答6:

Because the compiler is actually automatically (i.e. "implicitly") converting the integer to a double (or maybe long double) and sending that value to sqrt(). This is completely normally and completely legal.

来源：https://stackoverflow.com/questions/19613191/why-does-sqrt-work-fine-on-an-int-variable-if-it-is-not-defined-for-an-int