从函数 c 返回二维数组的正确方法 [英] correct way to return two dimensional array from a function c

问题描述

我已经试过了，但是不行:

I have tried this but it won't work:

#include <stdio.h>

    int * retArr()
    {
    int a[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
    return a;
    }

    int main()
    {
    int a[3][3] = retArr();
    return 0;
    }

我收到这些错误:

错误 3 错误 C2075:'a':数组初始化需要花括号
4 IntelliSense:返回值类型与函数类型不匹配

Error 3 error C2075: 'a' : array initialization needs curly braces
4 IntelliSense: return value type does not match the function type

我做错了什么?

推荐答案

结构是一种方法:

struct t_thing { int a[3][3]; };

然后按值返回结构体.

完整示例:

struct t_thing {
    int a[3][3];
};

struct t_thing retArr() {
    struct t_thing thing = {
        {
            {1, 2, 3},
            {4, 5, 6},
            {7, 8, 9}
        }
    };

    return thing;
}

int main(int argc, const char* argv[]) {
    struct t_thing thing = retArr();
    ...
    return 0;
}

<小时>

你面临的典型问题是int a[3][3] = {{1,2,3},{4,5,6},{7,8,9}}; 指的是在函数返回后回收的内存.这意味着您的调用者阅读(未定义行为)是不安全的.

---

The typical problem you face is that int a[3][3] = {{1,2,3},{4,5,6},{7,8,9}}; in your example refers to memory which is reclaimed after the function returns. That means it is not safe for your caller to read (Undefined Behaviour).

其他方法包括将数组(调用者拥有的)作为参数传递给函数，或创建新的分配(例如使用 malloc).该结构体很好，因为它可以消除许多陷阱，但它并不适用于所有场景.当结构的大小不是恒定的或非常大时，您将避免按值使用结构.

Other approaches involve passing the array (which the caller owns) as a parameter to the function, or creating a new allocation (e.g. using malloc). The struct is nice because it can eliminate many pitfalls, but it's not ideal for every scenario. You would avoid using a struct by value when the size of the struct is not constant or very large.

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