获取：错误 C2668：'sqrt'：对重载函数的不明确调用

Getting: error C2668: 'sqrt' : ambiguous call to overloaded function

本文关键字：函数 不明确 调用 重载 错误 获取 sqrt C2668      更新时间：2023-10-16

尝试根据教科书中给出的示例构建下面的源代码。我使用的是visualstudio2008。

编译器似乎不知道如何处理sieve:

1>------ Rebuild All started: Project: fig21.40, Configuration: Debug Win32 ------
1>Deleting intermediate and output files for project 'fig21.40', configuration        'Debug|Win32'
1>Compiling...
1>fig21_40.cpp
1>c:usersocukdocumentsc++chapter 21fig21.40fig21.40fig21_40.cpp(27) : error C2668: 'sqrt' : ambiguous call to overloaded function
1>        c:program files (x86)microsoft visual studio 9.0vcincludemath.h(581): could be 'long double sqrt(long double)'
1>        c:program files (x86)microsoft visual studio 9.0vcincludemath.h(533): or       'float sqrt(float)'
1>        c:program files (x86)microsoft visual studio 9.0vcincludemath.h(128): or       'double sqrt(double)'
1>        while trying to match the argument list '(const int)'
1>Build log was saved at "file://c:UsersocukDocumentsC++Chapter 21fig21.40fig21.40DebugBuildLog.htm"
1>fig21.40 - 1 error(s), 0 warning(s)
========== Rebuild All: 0 succeeded, 1 failed, 0 skipped ==========

代码：

//Fig. 21.40: fig21_40.cpp
//Using a bitset to demonstrate the Sieve of Eratosthenses
#include <iostream>
using std::cin;
using std::cout;
using std::endl;
#include <iomanip>
using std::setw;
#include <bitset>   //bitset class definition
#include <cmath>    //sqrt prototype
int main()
{
const int size=1024;
int value;
std::bitset<size> sieve;
sieve.flip();
//perform Sieve of Eratosthenses
int finalBit = sqrt( sieve.size() ) + 1;
for(int i=2; i<finalBit; ++i)
    if(sieve.test(i))
        for(int j=2*i; j<size; j+=i)
            sieve.reset(j);
cout << "The prime numbers in the range 2 to 1023 are:n";
//display prime numbers in range 2-1023
for(int k=2, counter=0; k<size; ++k)
    if(sieve.test(k)){
        cout <<setw(5)<<k;
        if(++counter % 12 ==0)
            cout << 'n';
    }//end outer if
cout << endl;
//get value from user to determine whether value is prime
cout << "nEnter a value from 1 to 1023(-1 to end): ";
cin>>value;
while(value != -1){
    if(sieve[value])
        cout << value << " is a prime numbern";
    else
        cout << value << " is not a prime numbern";
    cout << "nEnter a value from 2 to 1023 (-1 to end): ";
    cin >> value;
}//end while

return 0;
}

我认为这是因为在C++的较新版本中，sqrt重载（参数可以是double、float或long double），并且您传递了一个int。只需将参数强制转换为double即可明确：

int finalBit = sqrt( (double) (sieve.size()) ) + 1;

sqrt函数有以下重载：

float       sqrt(float arg);
double      sqrt(double arg);
long double sqrt(long double arg);

由于sieve.size()返回size_t，编译器无法选择进行哪种转换—"float"、double或long double。

添加强制转换以解决此问题。考虑到你的尺码范围，你选择什么类型并不重要。

int finalBit = sqrt( (double)sieve.size() ) + 1;
//                   ^^^^^^^^
// float or long double would work as well.
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