相关疑难解决方法(0)

我找到了以下用于将文件读入字符串的习惯用法:

std::ifstream file("path/to/some/file.ext");
std::string contents(
    std::istreambuf_iterator<char>(file),
    (std::istreambuf_iterator<char>())
);

哪个工作得很好.但是,如果我删除第二个迭代器参数周围的括号,那就是:

std::string contents(
    std::istreambuf_iterator<char>(file),
    std::istreambuf_iterator<char>()
);

一旦我尝试在字符串对象上调用任何方法,例如:

const char *buffer = contents.c_str();

我收到表单的编译错误:

error: request for member 'c_str' in 'contents', which is of non-class type 'std::string(std::istreambuf_iterator<char, std::char_traits<char> >, std::istreambuf_iterator<char, std::char_traits<char> > (*)()) {aka std::basic_string<char>(std::istreambuf_iterator<char, std::char_traits<char> >, std::istreambuf_iterator<char, std::char_traits<char> > (*)())}'

此外,如果我尝试将该字符串分配给另一个:

std::string contents2 = contents;

我收到表单错误:

error: conversion from 'std::string(std::istreambuf_iterator<char, std::char_traits<char> >, std::istreambuf_iterator<char, std::char_traits<char> > (*)()) {aka std::basic_string<char>(std::istreambuf_iterator<char, std::char_traits<char> >, std::istreambuf_iterator<char, std::char_traits<char> > (*)())}' to non-scalar type 'std::string {aka std::basic_string<char>}' requested

为什么是这样？我看不出需要这些括号的原因,更不用说影响 …