Question

所以我有一个名为scanCode的函数，它扫描文本文件中的单词并将其存储在2D数组中。然后我想把这个数组返回到main函数中的一个数组变量，这是我到目前为止的代码

#include <stdio.h>

char **scanCode()
{
    FILE *in_file;
    int i = 0;
    static char scan[9054][6];

    in_file = fopen("message.txt", "r");
    while (!feof(in_file))
    {

        fscanf(in_file, "%s", scan[i]);
        i++;
    }
    return scan;
}

int main(void)
{

    int hi[9053];

    FILE *in_file;

    in_file = fopen("message.txt", "r");

    char **array = scanCode();

    printf("%c", array[0]);
    printf("%c", array[1]);
    printf("%c", array[2]);
    printf("%c", array[3]);
}

所以基本上从scanCode函数返回的数组我希望它存储在main函数的char数组中..在看了很多问题和答案之后，这就是我得到的但是指针等对我来说很难理解..有人能告诉我这里做错了吗？

Answer 1

按以下方式更改功能的返回类型

#include <stdio.h>

char ( *scanCode() )[6]
{
    FILE *in_file;
    int i = 0;
    static char scan[9054][6];

    in_file = fopen("message.txt", "r");
    while (!feof(in_file))
    {

        fscanf(in_file, "%s", scan[i]);
        i++;
    }
    return scan;
}

int main(void)
{

    int hi[9053];

    FILE *in_file;

    in_file = fopen("message.txt", "r");

    char ( *array )[6] = scanCode();

    printf("%s", array[0]);
    printf("%s", array[1]);
    printf("%s", array[2]);
    printf("%s", array[3]);
}

同样在printf语句中使用格式说明符%s

并改变函数中的循环，如

    while ( i < 9054 && fscanf(in_file, "%5s", scan[i]) == 1 ) ++i;

Answer 2

我更喜欢以这种方式简化代码：

#include <stdio.h>

#define NumLines   9054
#define NumCols    6

void freeMem(char **ele) {
    while (*ele != NULL) {
        free(*ele);
        ele++;
    }
}

char **scanCode(char *fileName)
{
    FILE *in_file;
    char readingFormat[128];
    int i = 0;

    /*
     * Instead to declare a static variable I prefer to allocate dynamically
     * the bidimensional array.
     * It is done in two steps:
     * 1. allocate the memory for the first dimension
     * 2. for each element in this dimension allocate the memory for each element in the second dimension
     *
    */
    char **scan = (char **)malloc((NumLines + 1) * sizeof(char *));
    if (scan == NULL) {
        return NULL;
    }
    for (int j = 0; j < NumLines; j++) {
        scan[j] = (char *)malloc(NumCols + 1);
        if (scan[j] == NULL) {
            freeMem(scan);
            return NULL;
        }
        scan[j][0] = NULL;
    }
    scan[NumLines] = NULL;  // define the end of memory

    in_file = fopen(fileName, "r");
    if (fopen == NULL) {
        freeMem(scan);
        return NULL;
    }
    sprintf(readingFormat, "%%%ds", NumCols);
    while (fscanf(in_file, readingFormat, scan[i]) == 1 && i < NumLines) {
        i++;
    }
    return scan;
}

int main(void)
{
    char **array = scanCode("message.txt");
    if (array == NULL) {
        printf("ERROR\n");
        exit(0);
    }
    for (char **tp = array; **tp != NULL; tp++) {
        printf("%s\n", *tp);
    }
}

Answer 3

数组不是指针（再次向我问好）。

此：

static char scan[9054][6];

有你想要的最明显的类型 - 'char [9054] [6]'而不是'char **'。它拼写为6个元素的数组，每个元素是9054个字符的另一个数组。另一方面，类型'char **'拼写为'指向char的指针'，你现在可能看到它们完全不同。

您的代码应如下所示：

#include <stdio.h>

typedef char yourArrayType[9054][6];

typedef struct { yourArrayType return_value; } letsReturnArraysType;

letsReturnArraysType scanCode()
{
    FILE *in_file;
    int i = 0;
    yourArrayType scan;

    in_file = fopen("message.txt", "r");
    while (!feof(in_file))
    {

        fscanf(in_file, "%s", scan[i]);
        i++;
    }
    return *(letsReturnArraysType*)scan;
}

int main(void)
{

    int hi[9053];

    FILE *in_file;

    in_file = fopen("message.txt", "r");

    letsReturnArraysType arrayStruct = scanCode();

    printf("%s", arrayStruct.return_value[0]);
    printf("%s", arrayStruct.return_value[1]);
    printf("%s", arrayStruct.return_value[2]);
    printf("%s", arrayStruct.return_value[3]);
}

如何使用指针返回2D数组函数？

3 个答案: