为什么编译器不解析对模板函数的调用？
在下面的程序中，为什么编译器会为调用 printMax 模板函数而不是调用 printMaxInts 函数生成错误？
#include <iostream>
template<class A>
void printMax(A a,A b)
{
A c = a>b?a:b;
std::cout<<c;
}
void printMaxInts(int a ,int b)
{
int c = a>b?a:b;
std::cout<<c;
}
int main()
{
printMax(1,14.45);
printMaxInts(1,24);
}
In below program why does the compiler generate an error for the call to the printMax template function and not the call to the printMaxInts function?
#include <iostream>
template<class A>
void printMax(A a,A b)
{
A c = a>b?a:b;
std::cout<<c;
}
void printMaxInts(int a ,int b)
{
int c = a>b?a:b;
std::cout<<c;
}
int main()
{
printMax(1,14.45);
printMaxInts(1,24);
}
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为了让编译器从传递给函数模板的参数中推导出模板参数
A，两个参数a和b都必须具有相同类型。您的参数的类型为
int和double，因此编译器无法推断出它应实际用于A的类型。A应该是int还是应该是double？您可以通过使两个参数具有相同的类型来解决此问题：
或者通过显式指定模板参数：
可以调用对非模板函数的调用的原因是编译器不需要推导参数的类型：它知道参数的类型，因为您在函数声明中说了它们的内容：
a和b都是int类型。当您将double作为参数传递给此函数时，double ->对参数执行 int标准转换，并使用结果int调用该函数。In order for the compiler to deduce the template parameter
Afrom the arguments passed to the function template, both arguments,aandbmust have the same type.Your arguments are of type
intanddouble, and so the compiler can't deduce what type it should actually use forA. ShouldAbeintor should it bedouble?You can fix this by making both arguments have the same type:
or by explicitly specifying the template parameter:
The reason that the call to the non-template function can be called is that the compiler does not need to deduce the type of the parameters: it knows the type of the parameters because you said what they were in the function declaration:
Both
aandbare of typeint. When you pass adoubleas an argument to this function, thedouble -> intstandard conversion is performed on the argument and the function is called with the resultingint.以下代码基于詹姆斯的答案。您会注意到我已经取出了条件表达式：我这样做是因为该表达式的结果子句必须具有相同的类型，这对 A 和 B 施加了额外的限制。
对 A 和 B 的唯一要求此版本的代码是有一个与它们相关的运算符<()（或者可以将一个运算符转换为另一个），并且存在必需的运算符<<()函数。
The following code builds on James's answer. You'll notice that I've taken out the conditional expression: I've done this because the result clauses to that expression must have the same type, which imposes an additional restriction on A and B.
The only requirements on A and B in this version of the code is that there's an operator<() that related them (or one can be converted to the other), and that the requisite operator<<() functions exist.