Question

我要求用户输入两个整数， n 和 x 。之后，我需要向他们询问 a 变量n次的值。我需要为每个值创建一个新的变量。我该怎么做？我完全不知道。

另外，我需要在一行中进行，所以输入将是，例如50 30 21，而不是

50
30
21

感谢。

#include <stdio.h>

int main (void) {
  int a, n, x;
  int i = 0;

  scanf ("%d%d", &n, &x);

  scanf ("%d", &a); /* what should I do here? */
}

Answer 1

试试这个：

int arr[100]; // static allocation but you can also allocate the dynamically memory

printf("Enter the number for how many time repeat scanf()\n"); 
scanf("%d",&n);

for (int i = 0; i < n; i++)
{
    scanf("%d",&arr[i]);
}

Answer 2

请尝试以下代码：

#include <stdio.h>

    int main (void) {
    int a[100];
    int n, x;
    int i = 0;

    scanf ("%d%d", &n, &x);//n cannot be greater than 100 in this case.

    for(i = 0; i < n; i++)
    {
        scanf ("%d", &a[i]);
    }
    return 0;
}

Answer 3

使用dynamic memory：

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

int main (void) {
  int *a = NULL;
  int n, x, i;

  scanf("%d%d", &n, &x);

  if (n <= 0) {
    fprintf(stderr, "n must be > 0\n");
    return 1;
  }

  a = malloc(n * sizeof(int));
  if (a == NULL) {
    fprintf(stderr, "failed to allocate memory for "
        "%d integers\n", n);
    return 1;
  }

  /* reading the user input */
  for (i = 0; i < n; i++) {
    scanf ("%d", &a[i]);
  }

  /* usage */
  for (i = 0; i < n; i++) {
    printf("a[%d] = %d\n", i, a[i]);
  }
  printf("x = %d\n", x);

  free(a);

  return 0;
}

Answer 4

之后，我需要向他们询问变量的值n次。

高级编程语言甚至汇编语言（reference.child("restaurants").limitToFirst(200).addListenerForSingleValueEvent( new ValueEventListener() { @Override public void onDataChange(DataSnapshot dataSnapshot) { for (DataSnapshot child : dataSnapshot.getChildren()) { RestaurantEntity res = child.getValue(RestaurantEntity.class); rest_List.add(res); } //start list_more_res = rest_List.subList(0, 20 > rest_List.size() ? rest_List.size() : 20); LinearLayoutManager layoutManager = new LinearLayoutManager(HomeActivity.this, LinearLayoutManager.HORIZONTAL, false); hot_categories = (RecyclerView) findViewById(R.id.prd_categories_hot); hot_categories.setLayoutManager(layoutManager); productList = new ArrayList<>(); res_hot_adapter = new RestaurentAdapter(HomeActivity.this, list_more_res); hot_categories.setAdapter(res_hot_adapter); hot_categories.addOnScrollListener(new EndlessRecyclerViewScrollListener(layoutManager) { @Override public void onLoadMore(int page, int totalItemsCount) { tmp_more= rest_List.subList(20*count_loadmore ,20+ 20*count_loadmore>rest_List.size()? rest_List.size() : 20 + 20*count_loadmore); int curentSize = res_hot_adapter.getItemCount(); list_more_res.addAll(tmp_more); /*res_hot_adapter.notifyItemRangeInserted(curentSize, res_hot_adapter.getItemCount()-1);*/ Toast.makeText(HomeActivity.this,tmp_more.size()+" more",Toast.LENGTH_LONG).show(); count_loadmore++; } }); /* hot_categories.addOnItemTouchListener(new RecyclerItemClickListener(HomeActivity.this, new RecyclerItemClickListener.OnItemClickListener() { @Override public void onItemClick(View view, int position) { Intent i = new Intent(HomeActivity.this, RestaurentDetailActivity.class); startActivity(i); } }));*/ // end load from firebase } @Override public void onCancelled(DatabaseError databaseError) { Log.w(TAG, "getUser:onCancelled", databaseError.toException()); } }); }）为程序员提供Conditional/Unconditional Jump结构。 在计算机编程中，循环是一系列指令，不断重复，直到达到某个条件。在C / C ++中，我们有Loop，for，{{ 1}}循环结构。

因此，我要求用户提供n次值，您可以在程序中使用while。我在下面给出了一个例子：

do while

Answer 5

执行scanf时，它将输入直到空格字符，这是一个分隔符。您无需为此指定任何额外内容。如果您不需要再次使用n的值，则可以使用此代码。

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

int main (void) {
  int n, x, i;
  scanf ("%d%d", &n, &x);
  int *a = (int *) malloc (n * sizeof(int));    //This will allocate 'n' integer sized memory for 'a'
  for(i = 0; i < n; i++) {
        scanf ("%d", &a[i]);
  }
}

如何重复scanf（）n次？

5 个答案: