Question

我正在尝试将stringstream转换为const char*，但是我总是得到一个空白字符串。

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。

Answer 1

问题是您的const char*指向的字符串数据已在return str;之后释放。只要关联的const char*实例在范围内，std::string指针将保持有效。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，并且其析构函数将被调用，导致指向其字符数据的所有指针都变为dangling pointer。

相反，返回std::string，然后对返回的字符串调用c_str()。如前所述，只要关联的const char*实例在范围内，就可以使用std::string指针。

Answer 2

您可以使用<a href='http://localhost:3000/quotes/' + @quote.id>Show</a>存储临时对象，从而将悬空引用转换为非悬空引用。

请注意，您一次只能有一个活动的“非悬挂”参考：

static thread_local

预期输出：

#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;
}

http://coliru.stacked-crooked.com/a/228e6c0cb8359222

Answer 3

您遇到的问题是，当此函数离开作用域（即执行return str;时），指向str的内存将被清除。您需要返回一个对象，该对象将保持对数据的所有权。

//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());
}

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

3 个答案: