C++ 对函数的引用不明确
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Reference to function is ambiguous
提问by MarJamRob
Why is this cause a compiler error, stating that my references are ambiguous? I have a float, intand string, which should all create separate function signatures, right?
为什么这会导致编译器错误，指出我的引用不明确？我有一个float, intand string，它们都应该创建单独的函数签名，对吗？
Here is what I have so far:
这是我到目前为止所拥有的：
#include <iostream>
#include <string>
using namespace std;
int plus(int a, int b);
float plus(float a, float b);
string plus(string a, string b);
int main(void)
{
int n = plus(3, 4);
double d = plus(3.2, 4.2);
string s = plus("he", "llo");
string s1 = "aaa";
string s2 = "bbb";
string s3 = plus(s1, s2);
}
int plus(int a, int b) {
return a+b;
} // int version
float plus(float a, float b) {
return a+b;
} // float version
string plus(string a, string b) {
return a+b;
} // string version
回答by John Calsbeek
First, don't use using namespace std;. In this case, there happens to be a struct called std::plus—oh, wait, never mind, that's actually called plusand its constructor is thrown into the bucket for overload resolution with your function called plus.
首先，不要使用using namespace std;. 在这种情况下，恰好有一个结构称为 -std::plus哦，等等，别介意，它实际上被调用了，plus并且它的构造函数被扔进了存储桶中，用你的函数调用了重载解析plus。
Second, you have an ambiguity because 3.2and 4.2are of type doubleand can convert equally well to floator int.
其次，您有歧义，因为3.2and4.2是类型double并且可以同样好地转换为floator int。
This is a simplification, but when it comes to passing numbers to an overloaded function, C++ basically uses these rules:
这是一种简化，但在将数字传递给重载函数时，C++ 基本上使用以下规则：
- If all the arguments exactly match one of the overloads, then use that one.
- If all the arguments can be promoted to match one of the overloads, then use that one.
- If all the arguments can be numerically converted to match one of the overloads, then use that one.
- 如果所有参数与重载之一完全匹配，则使用该重载。
- 如果可以提升所有参数以匹配重载之一，则使用该重载。
- 如果所有参数都可以进行数字转换以匹配重载之一，则使用该重载。
If you have multiple candidates at a specific level, then that is an ambiguity. Crucially, a doubledoes not promote to a float—that would be a downgrade. So it has to use a standard conversion to a float, which ties with the standard conversion to an int, so those two overloads are ambiguous.
如果您在特定级别有多个候选人，那么这是一个歧义。至关重要的是，adouble不会升级为floata——那将是降级。所以它必须使用到 a 的标准转换float，这与到 an 的标准转换有联系int，所以这两个重载是不明确的。
回答by user1610015
Call the float overload like this:
像这样调用浮点重载：
double d = plus(3.2f, 4.2f);
A constant like 3.2 actually has type double.
像 3.2 这样的常量实际上具有double类型。
