0 
int ReadNames(char ***Names, int *r,int *c)    
{
int i, j, k;
char name[100];
printf("Number of Rows: ");
scanf("%d", r);

printf("Number of Columns: ");
scanf("%d", c);

Names=(char ***)malloc(sizeof(char **)*(*r));
for(i=0;i<(*r);i++)
*(Names+i)=(char **)malloc(sizeof(char *)*(*c));

for(i=0;i<(*r);i++)
for(j=0;j<(*c);j++)
{
fflush(stdin);
gets(name);
strcpy(*(*(Names+i)+j),name);
}
return 1;
}

I am trying to allocate the memory to a 2-D array of strings. Later on i want to sort them row wise and column wise, but while allocating the memory , the program is not responding. Is there something i am doing in my code.

in main function readname is called as 

    ReadNames(&p,&r,&c)

where r and c are the no. of rows and columns.

4

2 回答 2

1

你需要：

*Names = (char **)malloc(sizeof(char **) * (*r));

和相应的变化。

您正在传递一个三指针，以便能够返回一个双指针。您正在做的是丢失有关在何处存储双指针的信息。

被删除的评论有些道理；还有一个失误。二维字符串数组意味着您在基本数据中具有三个级别的指针。并且您需要第四级指针才能传递给函数。

此外，使用gets()是灾难的秘诀。永远不要（如never，如never ever）使用该gets()函数。甚至在玩具程序中也没有。它会让你养成坏习惯。第一个互联网蠕虫通过使用的程序传播gets()（谷歌搜索“莫里斯互联网蠕虫”）。

在 Unix 和其他基于 POSIX 的系统上，使用fflush(stdin)会导致未定义的行为。在 Windows 上，行为由 Mi​​crosoft 定义。如果您在 Windows 上运行，那么您可以；如果不是，你不是。

而且我认为三星级编程很糟糕！

这可能不是我这样做的方式，但它是将您编写的内容直接转换为有效的内容，以及一个main()测试它并释放所有分配内存的程序。它假定strdup()可用；如果没有，写它是微不足道的。

样本输出：

Number of Rows: 2
Number of Columns: 3
R0C0: Row 1, Column 1.
R0C1: Ambidextrous Armless Individual.
R0C2: Data for the third column of the first row.
R1C0: Row 2, Column 1.
R1C1: Row 2, Column 2.
R1C2: Given that the number of rows is 2 and the number of columns is 3, this should be the last input!
Rows = 2, cols = 3.
[0,0] = <<Row 1, Column 1.>>
[0,1] = <<Ambidextrous Armless Individual.>>
[0,2] = <<Data for the third column of the first row.>>
[1,0] = <<Row 2, Column 1.>>
[1,1] = <<Row 2, Column 2.>>
[1,2] = <<Given that the number of rows is 2 and the number of columns is 3, this should be the last input!>>

工作四星代码：

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

static void ReadNames(char ****Names, int *rows, int *cols)
{
    char name[100];
    printf("Number of Rows: ");
    scanf("%d", rows);

    printf("Number of Columns: ");
    scanf("%d", cols);

    int c;
    while ((c = getchar()) != EOF && c != '\n')
        ;

    *Names = (char ***)malloc(sizeof(char ***)*(*rows));
    for (int i = 0; i < (*rows); i++)
        (*Names)[i] = (char **)malloc(sizeof(char **)*(*cols));

    for (int i = 0; i < (*rows); i++)
    {
        for (int j = 0; j < (*cols); j++)
        {
            printf("R%dC%d: ", i, j);
            if (fgets(name, sizeof(name), stdin) == 0)
            {
                fprintf(stderr, "Unexpected EOF\n");
                exit(1);
            }
            name[strlen(name)-1] = '\0';    // Zap newline
            (*Names)[i][j] = strdup(name);
        }
    }
}

int main(void)
{
    int rows;
    int cols;
    char ***data = 0;

    ReadNames(&data, &rows, &cols);
    printf("Rows = %d, cols = %d.\n", rows, cols);
    for (int i = 0; i < rows; i++)
    {
        for (int j = 0; j < cols; j++)
            printf("[%d,%d] = <<%s>>\n", i, j, data[i][j]);
    }

    for (int i = 0; i < rows; i++)
    {
        for (int j = 0; j < cols; j++)
            free(data[i][j]);
        free(data[i]);
    }
    free(data);
    return 0;
}

替代 3 星代码

使用三级指针已经够糟糕的了；四是可怕的。此代码将自身限制为三个级别的指针。我假设 C99 兼容，因此可以在函数中方便时声明变量。使用 C89/C90 编译器（现在已经倒退了 14 年）的更改非常简单。

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

static char ***ReadNames(int *r, int *c)
{
    int i, j;
    char name[100];
    printf("Number of Rows: ");
    scanf("%d", r);

    printf("Number of Columns: ");
    scanf("%d", c);

    int x;
    while ((x = getchar()) != EOF && x != '\n')
        ;

    char ***Names = (char ***)malloc(sizeof(char ***)*(*r));
    for (i = 0; i < (*r); i++)
        Names[i] = (char **)malloc(sizeof(char **)*(*c));
    for (i = 0; i < (*r); i++)
    {
        for (j = 0; j < (*c); j++)
        {
            if (fgets(name, sizeof(name), stdin) == 0)
            {
                fprintf(stderr, "Unexpected EOF\n");
                exit(1);
            }
            name[strlen(name)-1] = '\0';
            Names[i][j] = strdup(name);
        }
    }
    return Names;
}

static void PrintNames(char ***Names, int r, int c)
{
    int i, j;
    for (i = 0; i < r; i++)
    {
        for (j = 0; j < c; j++)
            printf("%s ", Names[i][j]);
        printf("\n");
    }
}

int main(void)
{
    int rows;
    int cols;
    char ***data = ReadNames(&rows, &cols);

    PrintNames(data, rows, cols);

    printf("Rows = %d, cols = %d.\n", rows, cols);
    for (int i = 0; i < rows; i++)
    {
        for (int j = 0; j < cols; j++)
            printf("[%d,%d] = <<%s>>\n", i, j, data[i][j]);
    }

    for (int i = 0; i < rows; i++)
    {
        for (int j = 0; j < cols; j++)
            free(data[i][j]);
        free(data[i]);
    }
    free(data);
    return 0;
}

示例输出

Number of Rows: 3
Number of Columns: 4
R1C1
R1C2
R1C3
R1C4-EOR
R2C1
R2C2
R2C3
R2C4-EOR
R3C1
R3C2
R3C3
R3C4-EOR
R1C1 R1C2 R1C3 R1C4-EOR 
R2C1 R2C2 R2C3 R2C4-EOR 
R3C1 R3C2 R3C3 R3C4-EOR 
Rows = 3, cols = 4.
[0,0] = <<R1C1>>
[0,1] = <<R1C2>>
[0,2] = <<R1C3>>
[0,3] = <<R1C4-EOR>>
[1,0] = <<R2C1>>
[1,1] = <<R2C2>>
[1,2] = <<R2C3>>
[1,3] = <<R2C4-EOR>>
[2,0] = <<R3C1>>
[2,1] = <<R3C2>>
[2,2] = <<R3C3>>
[2,3] = <<R3C4-EOR>>

这两个程序都在valgrind.

于 2013-11-05T04:52:59.307 回答
0

所以，这是代码工作...... 

    *Names=(char **)malloc(sizeof(char **)*(*r));    
for(i=0;i<(*r);i++)    
    *(*Names+i)=(char*)malloc(sizeof(char *)*(*c));    
for(i=0;i<(*r);i++)
for(j=0;j<(*c);j++)
{
fflush(stdin);
gets(name);
strcpy((*(*Names+i)+j),name);
}

但是在打印存储的这些名称时...我提供了一个功能..

int PrintNames(char **Names, int r, int c)
{

int i,j;
for(i=0;i<r;i++)
{   printf("\n");
    for(j=0;j<c;j++)
        printf("%s ",*(*(Names+i)+j));
}
return 1;

}

现在这个 PrintNames 也通过 main 调用...作为 "PrintNames(p, r, c);"....但是程序在打印时停止...可能有什么问题？

于 2013-11-05T05:49:34.653 回答