更多"以下程序运行后的输出结果是______。#includechar *s"的相关试题:
[填空题]以下程序运行后的输出结果是______。
#include
char *ss(char *s)
char *p, t;
p=s+1; t=*s;
while(*p) *(p-1) = *p; p++;
*(p-1)=t;
return s;main( )
char *p, str[10]="abcdefgh";
p = ss(str);
printf("%s/n",p);
[填空题]以下程序的输出结果是______。
#include
#include
char *fun(char *t)
char *p=t;
return (p+strlen(t)/2);main( )
char *str="abcdefgh";
str=fun(str);
puts(str);
[单项选择]以下程序运行后的输出结果是
void ss(char *s,char t)
while(*s)
if(*s==t) $s=t-’a’+’A’;
s++:
main( )
char str1[20]="abcddfefdbd",c=’d’;
ss(str1,c);
printf("%s/n",str1);
A. ABCDDEFEDBD
B. abcDDfefDbD
C. abcAAfefAbA
D. Abcddfefdbd
[单项选择]以下程序运行后的输出结果是()。
ss(char*s)
char*p=s;
while(*p)p++; return(p-s);
main( )
char*a="abded";int i;
i=ss(a);
printf("%d/n",i);
A. 8
B. 7
C. 6
D. 5
[填空题]以下程序运行后的输出结果是_________。
#include
char *ss(char *s)
{ char *p,t;
p=s+1;t=*s;
while(*p) {*(p-1)=*p; p++;}
*(p-1)=t;
return s;
}
main( )
{ char *p,str[10]="abcdefgh";
p=ss(str);
printf("%s/n",p);
}
[填空题]以下程序运行后的输出结果是______。
#include
char*ss(char*s)
char*p,t;
P=s+1;t=*s;
while(*p) *(p-1)=*p;p++;
*(p-1)=t;
return s;
main( )
char*p,str[10]="abcdefgh";
p=ss(str);
printf("%s/n",p);
[单项选择]以下程序运行后,消息框的输出结果是( )。 OPTION BASE 1 PRIVATE SUB COMMAND1_CLICK( ) DIM A(10) , P(3) AS INTEGER K=5 FOR I=1 TO 10 A(I) = I NEXT I FOR I=1 TO 3 P(I)=A(I*I) NEXT I FOR I=1 TO 3 K=K+P(I) ~ 2 NEXT I MSGBOX K END SUB
A. 33
B. 28
C. 35
D. 37
[单项选择]以下程序运行后,单击按钮输出结果是
Private Sub Commandl_Click( )
Dim x As Integer,y As Integer,z As Integer
X=4:y=2: Z=3
Call Gopd(x,x,z)
Print x;X;Z
Call Gopd(x,y,y)
Print x;y;y
End Sub
Private Sub Gopd(x As Integer,y As Integer,z As Integer)
X=3 * Z + 1
y=2 * z
z=x + y
End Sub
A. 6 6 12
B. 8 5 10
C. 9 6 12
D. 8 10 10
[单项选择]以下程序运行后,单击按钮输出结果是( )。
Private Sub Commandl_Click( )
Dim x As Integer,y As Integer,z As Integer
X=4:y=2: Z=3
Call Gopd(x,x,z)
Print x;X;Z
Call Gopd(x,y,y)
Print x;y;y
End Sub
Private Sub Gopd(x As Integer,y As Integer,z As Integer)
X=3 * Z + 1
y=2 * z
z=x + y
End Sub
[填空题]以下程序运行后的输出结果是【 】。
struct NODE
{int num;struct NODE *next;
};
main( )
{struct NODE s[3]={{1,’’/0’’},{2,’’/0’’},{3,’’0’’}},*p,*q,*r;
int sum=0;
s[0].next=s+1;s[1].next=s+2;s[2].next=s;
p=s; q=p->next; r=q->next;
sum+=q->next->num; sum+=r->next->next->num;
printf("%d/n",sum);
}