如何通过引用传递一个子阵列是在动态分配的二维数组？ [英] How to pass by reference a sub-array which is in a dynamically allocated 2D array?

问题描述

我需要参考子阵列，它们是动态分配的二维阵列的一部分来传递。我曾尝试以下方法，这似乎并没有工作。如果有可能的任何想法？

 无效SET1（INT *一）{
    一个= malloc的（2 * sizeof的（INT））;
    一个[0] = 5;
    一个[1] = 6;
}无效SET2（INT *一）{
    一个= malloc的（2 * sizeof的（INT））;
    一个[0] = 7;
    一个[1] = 8;
}诠释主（）{
    INT ** X =的malloc（2 * sizeof的为（int *））;    SET1（X [0]）;
    SET2（X [1]）;    返回0;
}
 

解决方案

您需要通过的您的子阵列的地址的，就像这个例子：

 无效集（INT **一）
{
    * A =的malloc（2 * sizeof的（INT））;
    （* a）个[0] = 4;
    （* a）个[1] = 9;
}INT主要（无效）
{
    INT ** X =的malloc（2 * sizeof的为（int *））;
    集（安培; X [0]）;
    集（安培; X [1]）;
}
 

只要你有需要修改的东西在调用者范围的功能，你需要去与害羞;再与害羞; FE和害羞，伦斯在函数（如我们的 *一个）而在调用者的-的地址（如＆放大器; X [0] ）。

（您code，相比之下，指针分配的内存分配给一个局部变量（即 myArray的），当函数返回都将丢失。而内存与它失去了一起。）

I need to pass by reference sub-arrays which are part of a dynamically allocated 2D array. I have tried the following approach, which doesn't seem to work. Any ideas if it is possible?

void set1(int *a){
    a = malloc(2*sizeof(int));
    a[0] = 5;
    a[1] = 6;
}

void set2(int *a){
    a = malloc(2*sizeof(int));
    a[0] = 7;
    a[1] = 8;
}

int main(){
    int **x = malloc(2*sizeof(int*));

    set1(x[0]);   
    set2(x[1]);

    return 0;
}

解决方案

You need to pass the address of your sub-arrays, like in this example:

void set(int ** a)
{
    *a = malloc(2 * sizeof(int));
    (*a)[0] = 4;
    (*a)[1] = 9;
}

int main(void)
{
    int ** x = malloc(2 * sizeof(int*));
    set(&x[0]);
    set(&x[1]);
}

Whenever you have a function that needs to modify something in the caller's scope, you need a de­re­fe­rence in the function (like our *a) and an address-of in the caller (like &x[0]).

(Your code, by contrast, assigns the pointer to allocated memory to a local variable (namely myArray), which is lost when the function returns. And the memory is lost along with it.)

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