c - 字符串的动态内存分配

Question

int ReadNames(char ***Names, int *n)    
{    
int i, k;    
char name[100];    
printf("Enter how many names\n");    
scanf("%d", n);       
/* Allocate memory and read names */
*Names=(char **)malloc((*n)*sizeof(char *));
for(i=0;i<(*n);i++)
{
    *(*Names+i)=(char*)malloc(sizeof(name));
    gets(name);
    strcpy(*(*Names+i),name);
}

for(i=0;i<(*n);i++)
    printf("%s\n",*(*Names+i));
return 1;
}    
void main()        
{
char **Names;
int n, i;
ReadNames(&Names, &n);
}

该程序运行良好...但与我的预期略有不同。问题是当我将“n”的值输入为 3 时，它只能读取 2 个字符串并打印这两个字符串....即。它读取 n-1 个字符串并打印 n-1 个字符串。我的代码有什么问题。

Answer 1

只需在 scanf() 之后添加 getchar()

以便每个 '\n' 在接受输入时被处理掉。您的代码将是

int ReadNames(char ***Names, int *n)
{
    int i, k;
    char name[100];
    printf("Enter how many names\n");
    scanf("%d", n);
    getchar(); // eats unnecessary '\n' in the buffer
    /* Allocate memory and read names */
    *Names=(char **)malloc((*n)*sizeof(char *));
    for(i=0;i<(*n);i++)
    {
            *(*Names+i)=(char*)malloc(sizeof(name));
                gets(name);
                    strcpy(*(*Names+i),name);
    }

    for(i=0;i<(*n);i++)
            printf("%s\n",*(*Names+i));
    return 1;
}
void main()
{
    char **Names;
    int n, i;
    ReadNames(&Names, &n);
}