实验4
任务一
应该是不能的,return返回来是一个值,而根可能有2个
任务2
#include <stdio.h> long long fac(int n); int main() { int i,n; printf("Enter n: "); scanf("%d", &n); for(i=1; i<=n; ++i) printf("%d! = %lld\n", i, fac(i)); return 0; } long long fac(int n) { static long long p = 1; p = p*n; printf("p = %lld\n", p); return p; }
#include<stdio.h> int func(int, int); int main() { int k=4,m=1,p1,p2; p1 = func(k,m) ; p2 = func(k,m) ; printf("%d,%d\n",p1,p2) ; return 0; } int func(int a,int b) { static int m=0,i=2; i += m+1; m = i+a+b; return (m); }
static是静态变量,数据储存再静态储物区,它与局部变量的区别在于函数退出时,变量始终存在,但不能被其他函数使用,当再次进入该函数的时候,将保存上次的结果,其他与局部变量一样。
#include <stdio.h> #define N 1000 int fun(int n,int m,int bb[N]) { int i,j,k=0,flag; for(j=n;j<=m;j++) { flag=1 ; for(i=2;i<j;i++) if(j%i==0) { flag=0; break; } if(flag==1) bb[k++]=j; } return k; } int main(){ int n=0,m=0,i,k,bb[N]; scanf("%d",&n); scanf("%d",&m); for(i=0;i<m-n;i++) bb[i]=0; k=fun(n,m,bb); for(i=0;i<k;i++) printf("%4d",bb[i]); return 0
}
#include <stdio.h> long long fun(int n); int main() { int n; long long f; while(scanf("%d", &n) != EOF) { f = fun(n); printf("n = %d, f = %lld\n", n, f); } return 0; } long long fun (int n){ int i,sum=1 ; for (i=1;i<=n;i++){ if (n==0) { return 0;break;} else sum = 2*sum ; } return sum-1; }
#include <stdio.h> void draw(int n, char symbol); #include <stdio.h> int main() { int n, symbol; while(scanf("%d %c", &n, &symbol) != EOF) { draw(n, symbol); printf("\n"); } return 0; } void draw (int n, char symbol){ int line,b,c; for (line=1;line<=n;line++) { for(b=1;b<=n-line;b++)printf (" "); for (c=1;c<=2*line-1;c++) printf ("%c",symbol); printf ("\n"); } }