将指向指针数组的指针理解为函数中的参数 [英] Understanding pointer to pointer arrays as arguments in a function

问题描述

在尝试自己学习C的同时，我遇到了一个我想开发的简单程序.它只是试图利用指向指针数组的指针来制作类似于矩阵的东西.我在Windows上编译，运行时它崩溃了，与此同时，在Linux上尝试此代码时显示segmentation fault，这是因为函数参数是数组吗?我在这里做什么错了?

While trying to learn C by myself, I came across this simple program that I want to develop. It just tries to make use of pointer to pointer arrays to make something that resembles matrices. I'm compiling on Windows and when I run it, it just crashes, meanwhile, trying this code on Linux it says segmentation fault, is this because of the function arguments that are arrays? What am I doing wrong here?

#include <stdio.h>
#include <stdlib.h>

void initializeArray(float** array, int size);
void printArray(float** array, int size);

int main()
{
    float** array_1 = NULL;
    int array_size = 3;

    initializeArray(array_1, array_size);

    // Free memory from array
    for (int i = 0; i < array_size; i++)
    {
        free(array_1[i]);
    }

    free(array_1);

    return 0;
}

void initializeArray(float** array, int size)
{
    array = malloc(size * sizeof(float*));

    if (array)
    {
        for (int i = 0; i < size; i++)
        {
            array[i] = malloc(size * sizeof(float));
            if (!array[i])
            {
                exit(0);
            }
        }
    }

    for (int i = 0; i < size; i++)
    {
        for (int j = 0; j < size; j++)
        {
            array[i][j] = 0;
        }
    }
}


void printArray(float** array, int size)
{
    for (int i = 0; i < size; i++)
    {
        for (int j = 0; j < size; j++)
        {
            printf("%f\t", array[i][j]);
        }

        printf("\n");
    }
}

推荐答案

执行时:

void initializeArray(float** array, int size)
{
    array = malloc(size * sizeof(float*));

您没有在函数外更改array，因此array_1在调用之后(如之前)指向NULL(并导致内存泄漏).您需要返回它(或将其作为三重***指针传递并将其用作*array，但这不太方便).

you're not changing array outside the function so array_1 points to NULL after (like before) the call (and creates a memory leak). You need to return it (or to pass it as triple *** pointer and use it as *array, but that's less convenient).

float **initializeArray(int size)
{
    float** array = malloc(size * sizeof(float*));
   ...
    return array;
}

和来自主站:

array_1 = initializeArray(array_size);

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