Question

我在此网站上阅读了“Passing multi-dimensional arrays in C”的示例。

这是一个使用char数组的好例子，我从中学到了很多东西。我想通过创建一个处理动态分配的一维整数数组的函数来做同样的事情，然后创建另一个函数来处理一个多维整数数组。我知道如何将它作为函数的返回值。但是在这个应用程序中，我需要在函数的参数列表上进行。

就像我上面提到的例子一样，我想将指向整数数组的指针传递给函数，以及元素数量“num”（或者用于2D数组函数的“col”和“col”）等）。我在这里得到了另一个例子的重写版本，但我不能让它工作，尽可能尝试（标记的代码行是新的或修改过的）。有谁知道如何解决这个问题？

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define ELEMENTS 5
void make(char **array, int **arrayInt, int *array_size) { 
    int i;
    char *t = "Hello, World!";
    int s = 10; // new
    array = malloc(ELEMENTS * sizeof(char *));
    *arrayInt = malloc(ELEMENTS * sizeof(int *));  // new
    for (i = 0; i < ELEMENTS; ++i) {
        array[i] = malloc(strlen(t) + 1 * sizeof(char));
        array[i] = StrDup(t);
        arrayInt[i] = malloc( sizeof(int)); // new
        *arrayInt[i] = i * s; // new
    }
}
int main(int argc, char **argv) {
    char **array;
    int  *arrayInt1D; // new
    int size;
    int i;
    make(array, &arrayInt1D, &size); // mod
    for (i = 0; i < size; ++i) {
        printf("%s and %d\n", array[i], arrayInt1D[i]); // mod
    }
    return 0;
}

Answer 1

该代码中存在很多问题。看看以下内容：

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

#define ELEMENTS 5

/*
 * A string is an array of characters, say char c[]. Since we will be creating
 * an array of those, that becomes char *(c[]). And since we want to store the
 * memory we allocate somewhere, we must be given a pointer. Hence char
 * **(c[]).
 *
 * An int doesn't require a complete array, just int i. An array of those is
 * int i[]. A pointer to those is then int *(i[]).
 */
void
make(char **(chars[]), int *(ints[]), size_t len)
{
    static char hw[] = "Hello, World!";
    size_t i = 0;

    /*
     * Allocate the memory required to store the addresses of len char arrays.
     * And allocate the memory required to store len ints.
     */
    *chars = malloc(len * sizeof(char *));
    *ints = malloc(len * sizeof(int));

    /* Fill each element in the array... */
    for (i = 0; i < ELEMENTS; i++) {
        /* ... with a *new copy* of "Hello world". strdup calls malloc under
         * the hood! */
        (*chars)[i] = strdup(hw);
        /* ...with a multiple of 10. */
        (*ints)[i] = i * 10;
    }
}

int
main(void)
{
    /* A string c is a character array, hence char c[] or equivalently char *c.
     * We want an array of those, hence char **c. */
    char **chars = NULL;
    /* An array of ints. */
    int *ints = NULL;
    size_t i = 0;

    /* Pass *the addresses* of the chars and ints arrays, so that they can be
     * initialized. */
    make(&chars, &ints, ELEMENTS);
    for (i = 0; i < ELEMENTS; ++i) {
        printf("%s and %d\n", chars[i], ints[i]);
        /* Don't forget to free the memory allocated by strdup. */
        free(chars[i]);
    }

    /* Free the arrays themselves. */
    free(ints);
    free(chars);

    return EXIT_SUCCESS;
}

Answer 2

您在这里缺少一行的大小：


arrayInt[i] = malloc( sizeof(int)); // new

应该是这样的：


arrayInt[i] = malloc( row_len * sizeof(int)); // new

在您使用给定字符串的长度作为行大小之前（strlen(t)+1也应该在括号中，但效果是相同的，因为sizeof(char)是1）

在C中传递动态分配的整数数组

2 个答案: