交换字符串中两个字符的C++程序

Question

我是 C++ 的初学者，我想编写一个简单的程序来交换字符串中的两个字符。

例如；我们输入这个字符串:“EXAMPLE”，我们给它交换这两个字符:“E”和“A”，输出应该类似于“AXEMPLA”。

我在编写以下代码时考虑的算法；很简单，希望你能通过引用代码得到它，但最后我被弄糊涂了! (我在这个网站上搜索并发现了类似的问题，但它们要么是困难和复杂的语法，要么是另一种语言)。任何建议和帮助表示赞赏。

#include <iostream>
#include <stdio.h>
#include <conio.h>
#include <cctype> // is this necessary?! I doubt between <stodio.h> and <cctype>
// By the way! when do we use double quotations instead of <> ?

using namespace std;

char array1[30], char1, char2;
int i, j, char1count = 0, char2count = 0, locofchar1[15], locofchar2[15], n1 = 0, n2 = 0;

int main()
{
    cout << "Enter your string: " << endl;
    gets(array1);

    cout << "\nEnter the 2 characters you want to swap." << endl
         << "Character #1: ";
    cin  >> char1;
    cout << "\nCharacter #2: ";
    cin  >> char2;

      for (i = 0; array1[i]; i++) // A "for" loop for counting the number of repetitions of char1
// and saving the locations of char1 in a new array called "locofchar1"
           if (array1[i] == char1){
              char1count++;
              for (j = n1; j <= char1count; j++)
                   locofchar1[j] = i;
                   n1++;
     }

      for (i = 0; array1[i]; i++) // Another "for" loop for counting the number of repetitions of char2
// and saving the locations of char1 in a new array called "locofchar2"
           if (array1[i] == char2)
              char2count++;
              for (j = n2; j <= char2count; j++)
                  locofchar2[j] = i;
                  n2++;

/* 

I'm already stuck at here! and I think I have some problems in the above code... We assume that the program determined the number of repetitions and their element address/location in the char1count and char2count arrays, and we want to use this informations to swap them correctly. 

*/

getch();
return 0;
}

Answer 1

您忘记在多个位置放置括号。

  for (i = 0; array1[i]; i++) // Another "for" loop for counting the number of repetitions of char2
// and saving the locations of char1 in a new array called "locofchar2"
       if (array1[i] == char2)
          char2count++;
          for (j = n2; j <= char2count; j++)
              locofchar2[j] = i;
              n2++;

如果没有大括号，在C/C++中，只有if之后的第一个过程才符合条件。其余无条件执行。所以每次都会执行上面代码中的for循环。所以加上大括号:

       if (array1[i] == char2) {
          char2count++;
          for (j = n2; j <= char2count; j++) {
              locofchar2[j] = i;
              n2++;
          }
       }

为每个条件和每个循环添加大括号是一种很好的做法，这样您就不会忘记哪些行属于这些条件/循环。