以下程序的输出结果是
void reverse(int a[],int n)
{ int i,t;
for(i=0;i
A.22
B.10
C.34
D.30
[试题]以下程序的输出结果是______nt fun(int*x,int n){if(n==0)return x[0];else return x[0]+fun(x+1,n-1);}main(){int a[]={1,2,3,4,5,6,7};printf(“%d/n”,fun(a,3));}
[单选题]有以下程序 int a=2; int f(int n) { static int a=3; int t=0; if(n%2) {static int a=4; t+=a++;} else {static int a=5; t+=a++;} return t+a++; } main ( ) { int s=a, i; for (i=0;i<3; i++) s+=f (i) print f ("%d/n" , s ); } 程序运行后的输出结果是A.26B.28C.29D.24
[单选题]下面程序的运行结果是#includeint fun(int a[]. int n){int result=l:for(int i=1;iresult=result*a[i];return result;}void main(){int a[3]={3,4,5};cout<<}A.12;B.15;C.20;D.60
[单选题]有以下程序struct S{ int n; int a[20]; }; void f(int *a,int n) {int i; for(i=0;i a[i]+=i; } main() {int i; struct S s={10,{2,3,1,6,8,7,5,4,10,9}}; f(s.a, s.n); for(i=0;i printf(“%d”,s.a[i]); } 程序运行后的输出结果是( )。A.2,4,3,9,12,12,11,11,18,9,B.3,4,2,7,9,8,6,5,11,
[主观题]以下程序的输出结果是( )。main(){int a=0;a+=(a=8);printf("%d\n",a);}
[单选题]有以下程序: void sort(int a[],int n) { int i,j,t; for(i=0;i<n-1;i+=2) for(j=i+2;j<n;j+=2) if(a[i]<a[j]){t=a[i];a[i]=a[j];a[j]=t;} } main() { int aa[10]={1,2,3,4,5,6,7,8,9,10},i; sort(aa,10); for(i=0;i<10;i++) printf("%d",aa[i]); printf("/n"); } 其输出结果是
[单选题]以下程序的输出结果是()。includevoid prt(int *x,int *y,int *z){printf("%d,%d,%d/n"以下程序的输出结果是( )。 #include<stdio.h> void prt(int *x,int *y,int *z) {printf("%d,%d,%d/n",++*x,++*y,*(z++));} main() {inta=10,b=40,c=20; prt(&a,&b,&C); prt(&a,&b
[主观题]以下程序的输出结果是______。 include int add(int x, int y) { return x£«y; } doub以下程序的输出结果是______。include<iostream.h>int add(int x, int y){return x+y;}double add(double x, double y){return x+y;}void main(){int a=3, b=3;double c=6.5, d=1.5;cout<<add(a, b)<<","<
[单选题]以下程序的输出结果是()。includestruct st{int x;int*y;}*p; int dt[4] ={ 10,20,30,4以下程序的输出结果是( )。 #include<stdio.h> struct st { int x; int *y;} *p; int dt[4] ={ 10,20,30,40 }; struct st aa[4]={ 50,&dt[0],60,&dt[0],60,&dt[0],60,&dt[0]}; main() { p=
[单选题]以下程序的输出结果是()。includevoid prt(int*x,int*y,int*z){printf("%d,%d,%d/n",£«£«以下程序的输出结果是( )。 #include<stdio.h> void prt(int *x,int *y,int *z) { printf("%d,%d,%d/n",++ *x,++ *y,*(z++));} main() { int a=10,b=40,c=20; prt(&a,&b,&c); prt(&a