c - 将函数返回的数组分配给二维数组的一行 C 编程

Question

我有一个返回缓冲区数组的函数。调用函数有一个二维数组，我需要将返回的数组仅分配给二维数组的特定行。我的代码不起作用。请帮助我理解我的错误。

在调用函数中:

char buff[5][BUFF_SIZE] = {{0}};
char statement[] = "USE THIS STATEMENT;";
char name[] = "some_name";
int value1 = 90;
double value2 = 20;

/**The statement below causes the error that array is not assignable **/
buff[0] = function(statement, name, value1, value2));   
if( buff[0] != NULL)
   // do stuff

功能:

char *function(char *statement, char *name, int value1, double value2)
{
    size_t size = BUFF_SIZE;

    char *buff = malloc (size);
    int nchars;
    if (buff == NULL)
        return NULL;

    /* Try to print in the allocated space. */
    nchars = snprintf(
                      buff,
                      BUFF_SIZE,
                      statement,
                      value1,
                      value2,
                      name
                      );
    if (nchars >= size)
    {
        /* Reallocate buffer now that we know
         how much space is needed. */
        size = nchars + 1;
        buff = realloc (buff, size);

        if (buff != NULL)
              /* Try again. */
              snprintf(
                      buff,
                      size,
                      statement,
                      value1,
                      value2,
                      name
                    );
    }
    /* The last call worked, return the string. */
    return buff;
}

非常感谢!编辑:更正了第二次调用 snprintf 时的错误

Answer 1

您有以下声明:

char buff[5][BUFF_SIZE] = {{0}};

这是一个字符的二维数组。但是您的函数返回一个 char *，它与 buff 的行不兼容，因为它们没有 char * 类型。因此，将您的声明替换为:

char *buff[5] = {NULL};

这是一个指向 char 的指针数组(在本例中为 5 个)。然后为其分配内存:

for (i = 0; i < 5; i++)
{
    buff[i] = malloc (size*sizeof(char));    //or just malloc(size);
    if (buff[i] == NULL)
        printf ("Error allocating memory\n");
}