C++ const到const char*的转换无效，如何修复？使用gcc7失败

这可以使用较旧版本的gcc进行编译，但在使用C++17的gcc 7时失败。 这是我得到的错误：

错误：从“char”到“const char*”[-fppermissive]end\u of_line='\0'的转换无效

似乎无法解决此编译错误

代码如下：

/*!
 * \brief Find the beginning of the next line in the given buffer.
 *
 * \param[in] str buffer to search for the beginning of the next line
 * \param[inout] ctx
 * \parblock
 * pointer to the end of the line (saved by this method)
 *
 * This pointer must be valid, and it must be set to NULL by the caller the first time.
 * \endparblock
 *
 * \return a pointer to the first character in the next line, or NULL if we have already hit EOF
 */
const char* Foo::GetNextLine(const char* str, const char** ctx)
{
    if(str == NULL)
    {
        return NULL;
    }

    // Save a pointer to the end of the line (first time, it should be set to NULL by the caller).
    const char*& end_of_line = *ctx;
    if(end_of_line && *end_of_line == '\0')
    {
        end_of_line = '\0';
        return NULL;
    }

    // The start of this line is one character after the end of the last one.
    const char* start_of_line = end_of_line ? end_of_line + 1 : str;

    // Are we at the end of the whole thing?
    if(*start_of_line == '\0')
    {
        end_of_line = '\0'; // Reset the context state to get ready for the next load!
        return NULL;
    }

    // Read forward to the end of the line.
    end_of_line = start_of_line;
    while(*end_of_line != '\n')
    {
        if(*end_of_line == '\0')
        {
            break;
        }
        ++end_of_line;
    }

    return start_of_line;
}

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由于C++11已被缺陷报告修改，因此不能将任意零值表达式指定给指针。它必须是常量0或类型为nullptr_t的对象，即nullptr。这可以隐藏在宏NULL后面

_行的表达式end_='\0'；尝试将字符常量指定给指针。这是不允许的。假设您的目的是清空原始指针，那么应该将该行更改为read

end_of_line = nullptr;

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你能发布命令和gcc调用的输出吗？