答:void StrAnalyze(Stringtype S) //统计串S中字符的种类和个数 { mytype T[MAXSIZE]; //用结构数组T存储统计结果 for(i=1;i<=S[0];i++) { c=S[i];j=0; while(T[j].ch T[j].ch!=c) j++; //在结构数组T中逐元素查找当前字符c是否已记录过. //当循环停止时,再看是什么原因造成的...
7. `replace(&s, 'STUDENT', q)`:将字符串 `s` 中的 "STUDENT" 替换为 "WORKER",结果为 "I AM A WORKER"。 8. `concat(&temp1, t, sub2)`:将字符串 `t` 和 `sub2` 连接起来,并将结果存入 `temp1` 中。结果为 "GOODA STUDEN"。 9. `concat(&temp2, sub1, temp1)`:将字符串...
设s=”I AM A STUDENT”,t=”GOOD”,q=”WORKER”,下面各个操作的结果是什么? (1) StrLength(s);(2) StrLength(t);(3) SubString(s,8,7);(4) SubString(t,2,1);(5) Index(s,”A”); (6) Replace(s,”STUDENT”,q);(7) Concat(SubString(s,6,2),Concat(t,SubString(s,7,8)))。
设s='I AM A STUDENT’, t='GOOD’, q='WORKER’。给出下列操作的结果: StrLength(s); SubString(sub1,
解析 解:① Replace(s, ? STUDENT’,q)=’ I AM A WORKER' ②因为 Substring (s, 6, 2) = ‘A Substring(s,7, 8) = ‘ STUDENT' Concat (t, Substring(s, 7, 8)) = GOOD STUDENT’ 所以Concat (Substring(s, 6, 2), Concat (t, Substring(s, 7, 8))) = A GOOD STUDENT?
【题目】设s='I AM A STUDENT', t='GOOD' , q=lWORKER' 。求:StrLength(s), StrLength(t), SubString(s,8,7), SubString(t,2,1),Index(s,'A'),, Index(s,t), Replace(s,'STUDENT',q),Concat(SubString(s,6,2),Concat(t,SubString(s,7,8)) ...
解析 这是计算机题目呀:Strlength (S) = 14SubString(⊂,s,8,7) = 'student'SubString(⊂,T,2,1) = 'o'Index(S,'a',1) = 3Index(S,T,1) = 0Replace(&s,'student',Q) = 'I am a worker'Concat(&N,Substring(&V,...反馈 收藏 ...
设s='I AM A STUDENT', t=lGOOD' , q=lWORKER' 。求:StrLength(s), StrLength(t), SubString(s,8,7), SubString(t,2,1),Index(s,'A'), Index(s,t),Replace(s,'STUDENT',q),Concat(SubString(s,6,2),Concat(t,SubString(s,7,8))。 相关知识点: 试题来源: 解析 14,4,'S...
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给定字符串S为'I am a student'，T为'good'，Q为'worker'，我们可以进行一系列的字符串操作。首先，计算S的长度，即Strlength(S) = 14。接着，通过SubString函数获取子串，如'SubString(S, 8, 7)'得到'student'，'SubString(T, 2, 1)'得到'o'。查找字符'a'在S中的索引，Index(S, 'a'...