[英]Address of stack memory returned when overloading operator [ ]
I'm encountering some issues with overloading the [] operator for a problem in class.由于 class 中的问题,我在重载 [] 运算符时遇到了一些问题。 This is the function for the overloading:
这是用于过载的 function:
const char* Person::operator[](const char* str)
{
if (strcmp(str, "name") == 0)
return reinterpret_cast<const char *>atoi(name);
if (strcmp(str, "age") == 0)
{
char temp[4];
_itoa(age, temp, 10);
//cout << temp;
return temp;
}
}
The class defition looks like this class 定义看起来像这样
class Person
{
private:
const char* name;
unsigned int age;
double height;
int gradeNo;
int grades[10];
public:
Person();
Person(const char* name, int age, double height);
void addGrade(int grade);
const char* operator [] (const char* str);
operator int();
};
The problem I'm getting is with the return temp;
我遇到的问题是
return temp;
line from the operator overload function.来自运算符重载 function 的行。 CLion returns the following warning:
Address of stack memory associated with local variable 'temp' returned
CLion 返回以下警告:
Address of stack memory associated with local variable 'temp' returned
Sure enough, when trying to use the operator, the value returned is a memory address.果然,在尝试使用操作符的时候,返回的值是一个memory地址。 How can I go about fixing this?
我该如何解决这个问题? Is it related to the return type of the function?
是不是和function的返回类型有关?
You are taking an address to a temporary (that is located on the stack), that will leave the returned pointer dangling almost immediately.您正在获取一个临时地址(位于堆栈上),这将使返回的指针几乎立即悬空。
I would very strongly suggest using std::string
for strings, do not write C++ just as C with classes.我强烈建议使用
std::string
作为字符串,不要像 C 那样写 C++ 和类。
Then return by std::string
by value here.然后在这里按值通过
std::string
返回。 Bad C-like alternative is to allocate the string on the heap and return that, preferably as std::unique_ptr
at least.糟糕的类似 C 的替代方法是在堆上分配字符串并返回,最好至少作为
std::unique_ptr
。
Since you are required to convert an integer to string and return the value, you cannot return a temporary variable, the result must outlive the method.由于您需要将 integer 转换为字符串并返回值,因此您不能返回临时变量,结果必须比方法更有效。 There are basically two bad options:
基本上有两个不好的选择:
temp
static, this way the pointer remains valid.temp
static,这样指针保持有效。 Downside is the function is no longer re-entrant.const char* foo(int age){
static char temp[4];
_itoa(age, temp, 10);
return temp;
}
const char* foo(int age){
char* temp = new char[256];
_itoa(age, temp, 10);
return temp;
}
I believe you also have a typo in your code:我相信您的代码中也有错字:
return reinterpret_cast<const char *>atoi(name);
The atoi
should not be there, right? atoi
不应该在那里,对吧? reinterpret_cast
should not be needed.不需要
reinterpret_cast
。
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