s = "ABCDEF"那么s[::-1]应该输出什么？
A: 'ABCDEF'
B: 'FEDCBA'
C: 'F'
D: 'A'
A: 'ABCDEF'
B: 'FEDCBA'
C: 'F'
D: 'A'
举一反三
- 以下程序运行后的输出结果是( )。#include int main(){char s[]="abcdef";s[3]= '\0';printf("%s\n",s);return 0;} A: abcd B: abcde C: abc D: abcdef
- 下面程序段的运行结果是()。#includeintmain(void){chars[]="abcdef";s[3]=s[6];printf("%s",s);return0;} A: abcdef B: abcd C: abc D: a
- s="abcdef",s[1:5]的结果是_.
- 下面程序的输出结果是（ ）Private Sub proc(ch As String) s = "" For k = Len(ch) To 1 Step -1 s = s & Mid(ch, k, 1) Next k ch = sEnd SubPrivate Sub Command1_Click() ch$ = "ABCDEF" proc ch Print ch End Sub A: ABCDEF B: A C: FEDCBA D: F
- 若有:char s[ ]="abcdef"; 则数组s包含______ 个元素。