#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=aa;
printf("%d/n",++(p->x));}
A.10
B.11
C.51
D.60
[单选题]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=aa;printf("%d/n",++(p->x));}A.10B.11C.51D.60
[单选题]以下程序输出的结果是#include<stdio.h>#include<string.h>main(){ char w[][10]={"ABCD","EFGH","IJKL","MNOP"},k;for(k=1;k<3;k++) printf("%s/n",&w[k][k]);}A.ABCD FGH KLB.ABC EFG IJ MC.EFG JK 0D.FGH KL
[单选题]以下程序输出的结果是#include<stdio.h>#include<string.h>main(){ char w[][10]={"ABCD","EFGH","IJKL","MNOP"},k;for(k=1;k<3;k++) printf("%s/n",&w[k][k]);}A.ABCD FGH KLB.ABC EFG IJ MC.EFG JK 0D.FGH KL
[单选题]下面程序的输出结果是#include<stdio.h>#include<string.h>main(){ char *p1="abc",*p2="ABC",str[50]= "xyz";strcpy(str+2,strcat(p1,p2));printf("%s/n",str);}A.xyzabcABCB.zabcABCC.xyabcABCD.yzabcABC
[主观题]以下程序的输出结果是_[13]_______#include <stdio.h>#include <string.h>char *fun(char *t){ char *p=t;return(p+strlen(t)/2);}main(){ char *str="abcdefgh";str=fun(str);puts(str);}
[试题]以下程序的输出结果是_[19]_______#include <stdio.h>#define M 5#define N M+Mmain(){ int k;k=N*N*5; printf("%d/n",k);}
[试题]以下程序的输出结果是_[9]_______#include <stdio.h>main(){ int n=12345,d;while(n!=0){ d=n%10; printf("%d",d); n/=10;}}
[试题]以下程序的输出结果是 ( 8 ) 。#include<stdio.h>main(){int i,j,sum;for(i=3;i>=1;i--){sum=0;for(j=1;j<=i;j++) sum+=i*j;}printf("%d/n",sum);}
[试题]以下程序的输出结果是 ( 9 ) 。#include<stdio.h>main(){ int j,a[]={1,3,5,7,9,11,13,15},*p=a+5;for(j=3;j;j--){ switch(j){ case 1:case 2:printf( " %d " ,*p++); break;case 3:printf( " %d " ,*(--p));}}}
[单选题]以下程序的输出结果是#include<stdio.h>main(){int a=15,b=21,m=0;switch(a%3){case 0:m++;break;case 1:m++;switch(b%2){ default:m++;case 0:m++;break; } }printf("%d/n",m);}A.1B.2C.3D.4