通过随机播放用户输入的字符串 [英] Shuffle a string input by user

问题描述

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;
＃包括LT＆;＆string.h中GT;
焦炭兰特（字符x）的;
诠释的main（）
{
字符输入[80] = {0};
焦兰多[80] = {0};
字符选择= 0;
焦炭rando2 [80] = {0};
如果（strlen的（输入）GT; = 10＆放大器;＆放大器; strlen的（输入）下; = 80）; {
    的printf（请输入一个字符串的字符10-80：）;
    scanf函数（％S，输入）;
    的printf（原单字符串：％S \\ n，输入）;
    兰多= my_rand（输入）;
    的printf（新的字符串：％S \\ n，兰多）; }
其他{
    返回0; }
的printf（你想重新洗牌这个字符串（Y或N）？）;
scanf函数（％C \\ n，＆安培;选择）;
如果（选择=='Y'）{
    rando2 = my_rand（兰多）;
    的printf（新的字符串：％S \\ n，rando2）;
        }
否则，如果{
    的printf（你想洗牌另一个字符串（Y或N）？）;
    scanf函数（％C \\ n，＆安培;选择2）; }
    如果（选择2 =='Y'）{
        的printf（请输入一个字符串的字符10-80：）;
        scanf函数（％S，输入2）;
        的printf（原始字符串：％S \\ n，输入2）;
        焦炭rando3 = my_rand（rando2）;
        的printf（新的字符串：％S \\ n，rando3）; }
    其他：        返回0;
返回0;
}
 

您好，大家好，我的目标是重新洗牌的用户输入的字符串很多次，因为他们想，促使是否继续还是没有。我有一个艰难的时间搞清楚如何洗牌字符串，任何人都可以伸出援助之手？

这是示例输出：

 请输入字符串的字符10-80：
initialnaivepassword原始的字符串：initialnaivepassword新的字符串：ntlvdiepnaaorsiiiwas你想重新洗牌这个字符串：Y新的字符串：saiiwndrvpaiioneslat你想重新洗牌这个字符串：N你想洗牌另一个字符串？ ：Y
请10-80个字符输入一个字符串：
anothernaivepassword原始的字符串：anothernaivepassword新的字符串：svdoanoprhsterneaaiw你想重新洗牌这个字符串：Y新的字符串：eaapnrtwhrosvidosaen你想重新洗牌这个字符串：N你想洗牌另一个字符串？ ：N
 

解决方案

这是不是一个解决方案（但你最欢迎给予好评它:-D），只是一个长注释。
更改

1

 字符输入[80] =;
焦兰多[80] =;
字符的选择[1] =;
炭rando2 [80] =;
 

要

 字符输入[80] = {0}; //将初始化所有80字符纪念品为0
焦兰多[80] = {0};
字符选择= 0; //当你只需要一个字符作为输入，无需使用数组。
                      //初始化刚刚炭为0
焦炭rando2 [80] = {0};
 

2

 其他：//是不是在C支持
 

要

 其他{/ * code这里* /}
 

3。

  scanf函数（％C \\ N的选择）;
如果（选择== Y）
 

要

  scanf函数（％C \\ n，＆安培;选择）; //选择不再是一个阵列，
                        //所以我们要通过选择的地址
如果（选择=='Y'）//'Y'不是一个int值，它的一个char字面
 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char rand(char x);
int main()
{
char input[80] = {0};
char rando[80] = {0};
char choice = 0;
char rando2[80] = {0};
if(strlen(input) >= 10 && strlen(input) <= 80); {
    printf("Please enter a string with 10-80 characters: ");
    scanf("%s", input);
    printf("Orginal string: %s\n", input);
    rando = my_rand(input);
    printf("New string: %s\n", rando);  }
else{
    return 0; }
printf("Would you like to shuffle this string again?(y or n): ");
scanf("%c\n", &choice);
if( choice == 'y') {
    rando2 = my_rand(rando);
    printf("New string: %s\n", rando2);
        }
else if {
    printf("Would you like to shuffle another string?(y or n): ");
    scanf("%c\n", &choice2); }
    if(choice2 == 'y') {
        printf("Please enter a string with 10-80 characters: ");
        scanf("%s", input2);
        printf("Original string: %s\n", input2);
        char rando3 = my_rand(rando2);
        printf("New string: %s\n", rando3); }
    else:

        return 0;
return 0;
}

Hello guys, my goal is to shuffle a user input string as many times as they would like, prompting whether to keep going or not. I am have a tough time figuring out how to shuffle the string, can anyone lend a hand?

This is the sample output:

Please enter a string with 10-80 characters:
initialnaivepassword

Original string: initialnaivepassword

New string: ntlvdiepnaaorsiiiwas

Would you like to shuffle this string again:y

New string: saiiwndrvpaiioneslat

Would you like to shuffle this string again:n

Would you like to shuffle another string? :y
Please enter a string with 10-80 characters:
anothernaivepassword

Original string: anothernaivepassword

New string: svdoanoprhsterneaaiw

Would you like to shuffle this string again:y

New string: eaapnrtwhrosvidosaen

Would you like to shuffle this string again:n

Would you like to shuffle another string? :n

解决方案

This is not a solution ( but you are most welcome to upvote it :-D ), just a long comment. Change your

1.

char input[80] = "";
char rando[80] = "";
char choice[1] = "";
char rando2[80] = "";

To

char input[80] = {0}; // Will initialize all 80 char mem to 0
char rando[80] = {0};
char choice = 0;      // As you want only one char as input, no need to use an array.
                      // Just initialize the char to 0
char rando2[80] = {0};

2.

else: // Is not supported in C

To

else {/* code here*/}

3.

scanf("%c\n", choice);
if( choice == y)

To

scanf("%c\n", &choice); // choice is no longer an array, 
                        // so we have to pass the address of choice
if( choice == 'y' ) // 'y' is not an int value, its a char literal

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