我正在处理某个程序,但目前要求您输入值。
ex.
How many number you want to enter?
5
Type in the numbers.
1 2 3 4 5
但我想预先定义数字;
int i [5] = {1,2,3,4,5}
怎么做?
这是代码。
#include <stdio.h>
#include <conio.h>
void main ()
{
int number[30];
int i,n,a,j;
printf ("Enter the value of n\n");
scanf ("%d",&n);
printf ("Enter the numbers\n");
for (i=0; i<n; ++i)
scanf ("%d", &number[i]);
printf ("Enter the position of the element to split the array \n");
scanf ("%d",&a);
for (i=0; i<a; ++i)
{
number[n] = number[0];
for (j=0; j<n; ++j)
{
number[j] = number[j+1];
}
}
printf("The resultant array is\n");
for (i=0; i<n; ++i)
{
printf ("%d\n",number[i]);
}
getch();
}
答案 0 :(得分:0)
您是否有要插入的数字的最大限制? 如果是,您可以使用预定义的数字初始化您的使用数组,例如:
int predefined[5] = {1, 2, 3, 4, 5};
并在需要时替换值。
答案 1 :(得分:-1)
#include <stdio.h>
#include <conio.h>
int main ()
{
int number[5]={1,2,3,4,5};
int i,n=5,a,j;
printf ("Enter the position of the element to split the array \n");
scanf ("%d",&a);
for (i=0; i<a; ++i)
{
number[n] = number[0];
for (j=0; j<n; ++j)
{
number[j] = number[j+1];
}
}
printf("The resultant array is\n");
for (i=0; i<n; ++i)
{
printf ("%d\n",number[i]);
}
getch();
return 0;
}