C++ syntax error: When a function returns a pointer or reference, it cannot return a local variable or temporary object. What should I do?

C is an object-oriented programming language, and its flexibility and power often provide programmers with great help. However, precisely because of its flexibility, it is difficult to avoid various small errors when programming. One of the most common mistakes is that when a function returns a pointer or reference, it cannot return a local variable or temporary object. So how to deal with this problem? This article will introduce the relevant content in detail.

1. Cause of the problem

In C language, local variables and temporary objects are dynamically allocated during the running of the function. When the function ends, the memory space of these local variables and temporary objects will be reclaimed. Therefore, if a function returns a pointer or reference to a local variable or temporary object, using this pointer or reference outside the function will cause unpredictable behavior, such as segfaults, crashes, and so on.

2. Operation method

In order to avoid this situation from happening, there are several ways to deal with this problem.

(1) Pass pointer or reference

If the function needs to return a pointer or reference type data, the data can be passed outside the function in the form of parameters instead of returning a pointer or reference. This ensures that the data returned is not a local variable or temporary object, but data that already exists outside the function.

(2) Use static local variables

Declaring local variables as static can make them exist during the execution of the function. This way, you can return a pointer or reference to a static local variable because it will not be recycled at the end of the function.

For example, the following code will return a pointer to a static local variable:

int* GetStaticPtr()
{
    static int value = 42;
    return &value;
}

(3) Using dynamic memory allocation

Using dynamic memory allocation can be done during function running Allocate heap space so that a pointer or reference points to data that still exists outside the function. Something to keep in mind is that this approach requires manual freeing of memory. If the memory is not released correctly, it can cause a memory leak.

For example, the following code will return a pointer to a dynamically allocated memory block:

int* GetDynamicPtr()
{
    int* ptr = new int(42);
    return ptr;
}

(4) Return value copy

If the function returns a value type, A copied object can be returned directly. This avoids problems with pointers or references not pointing to the correct data.

For example, the following code will return a copied object:

typedef struct Point
{
    int x;
    int y;
} Point;

Point GetPoint()
{
    Point p;
    p.x = 10;
    p.y = 20;
    return p;
}

3. Summary

In C programming, when a function returns a pointer or reference, The problem of not being able to return local variables or temporary objects is a very common one. In order to avoid this situation, you can pass pointers or references, use static local variables, use dynamic memory allocation, and return value copies. It needs to be used flexibly according to the specific situation.

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