C++ 模板编译错误:“>"标记之前的预期主表达式 [英] C++ template compilation error: expected primary-expression before ‘>’ token

问题描述

此代码编译并按预期工作(它在运行时抛出，但没关系):

This code compiles and works as expected (it throws at runtime, but never mind):

#include <iostream>
#include <boost/property_tree/ptree.hpp>

void foo(boost::property_tree::ptree &pt) 
{
    std::cout << pt.get<std::string>("path"); // <---
}

int main()
{
    boost::property_tree::ptree pt;
    foo(pt);
    return 0;
}

但是一旦我添加模板并将 foo 原型更改为

But as soon as I add templates and change the foo prototype into

template<class ptree>
void foo(ptree &pt)

我在 GCC 中遇到错误:

I get an error in GCC:

test_ptree.cpp: In function ‘void foo(ptree&)’:
test_ptree.cpp:7: error: expected primary-expression before ‘>’ token

但 MSVC++ 没有错误！错误在标记的行 <--- 中.再一次，如果我将问题行更改为

but no errors with MSVC++! The error is in the marked line <---. And again, if I change the problem line into

--- std::cout << pt.get<std::string>("path"); // <---
+++ std::cout << pt.get("path", "default value");

错误消失(问题出在显式).

the error disappears (the problem is in explicit <std::string>).

Boost.PropertyTree 需要 Boost >= 1.41.请帮助我理解并修复此错误.

Boost.PropertyTree requires Boost >= 1.41. Please help me to understand and fix this error.

参见模板:模板函数不能很好地与类的模板成员函数配合使用——一个类似的流行问题，包含其他好的答案和解释.

See Templates: template function not playing well with class’s template member function — a similar popular question containing other good answers and explanations.

推荐答案

你需要做的:

std::cout << pt.template get<std::string>("path");

在与 typename 相同的情况下使用 template，除了模板成员而不是类型.

Use template in the same situation as typename, except for template members instead of types.

(也就是说，由于 pt::get 是模板成员依赖模板参数，你需要告诉编译器它是一个模板.)

(That is, since pt::get is a template member dependent on a template parameter, you need to tell the compiler it's a template.)

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