C++ std::stringstream 到 const char* 的转换

C++ std::stringstream to const char* conversion

我正在尝试将 stringstream 转换为 const char*,但我总是得到一个空字符串。

我的代码:

#include<string>
#include<sstream>
#include<stdio.h>

const char* test(std::string city, std::string street, int houseNumber, int postCode) {
std::stringstream ss;
ss << city << " " << street << " " << houseNumber << " " << postCode;
const std::string tmp =  std::string{ss.str()};
const char* str = tmp.c_str();
return str;
}

int main(){
printf(test("demo", "Demo", 1, 1234));
}

预期输出:demo Demo 1 1234

我尝试了 How to convert a std::string to const char* or char*? 中提到的所有增强功能,但没有成功。

问题是您的 const char* 指向的字符串数据在 return str; 之后已被释放。只要关联的 std::string 实例在范围内,const char* 指针将保持有效。这是假设您在此处显示的代码位于函数内。在这种情况下,字符串 tmp 将在 return str; 之后超出范围,并且将调用其析构函数,导致指向其字符数据的任何指针变为 dangling pointer.

Return 和 std::string,然后在返回的字符串上调用 c_str()。如前所述,只要关联的 std::string 实例在范围内,您就可以使用 const char* 指针。

您遇到的问题是,当此函数离开范围时(即当 return str; 执行时),str 指向的内存将被清理。您需要 return 一个对象来维护对数据的所有权。

//std::string_view is >=c++17, use std::string const& for older compilers
std::string get_formatted_address(
    std::string_view city, 
    std::string_view street, 
    std::string_view houseNumber, 
    std::string_view postCode
) {
    std::stringstream ss;
    ss << city << " " << street << " " << houseNumber << " " << postCode;
    return ss.str();
}

int main() {
    auto city = "New York", street = "Rockefeller Lane", houseNumber = "1234", postcode = "56789";
    auto formatted_address = get_formatted_address(city, street, houseNumber, postcode);
    //Here, formatted_address holds ownership over the memory, so calling c_str is safe
    printf("Address: %s\n", formatted_address.c_str());
}

您可以通过使用 static thread_local 存储临时对象将悬挂引用转换为非悬挂引用。

请注意,您一次只能有一个有效的 "non dangling" 参考:

#include <string>
#include <sstream>
#include <iostream>

const char* to_cstr(std::string && s)
{
    static thread_local std::string sloc;
    sloc = std::move(s);
    return sloc.c_str();
}

const char* make_address(std::string const& city, 
                         std::string const& street, 
                         std::string const& houseNumber, 
                         std::string const& postCode)
{
    std::stringstream ss;
    ss << city << " " << street << " " << houseNumber << " " << postCode;
    return to_cstr(std::move(ss).str());
}

int main()
{
    const char* p = make_address("London","Bob Street","42","W1ABC");
    std::cout << p << std::endl;
}

预期输出:

London Bob Street 42 W1ABC

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