如何在字符串char []中的兩個字母之間交換？

Question

對不起，標題，但是我不知道這個程序叫什么。

我正在嘗試編寫一個帶有int參數的程序，並輸出一個長度為n的單詞，該單詞由字母“ a”和“ b”組成，例如：

n= 3 
the result =>: `aaa,baa,aba,aab,bba,bab,abb,bbb`

這是我的代碼。 里面有問題：

#include <stdio.h>
#include <stdlib.h>

#define n  5

char word[n] = "aaa";
char noob[n];

void
myfunction(void)
{
    int x,
     i,
     j;

    for (i = 0; i < n - 1; i++) {
        // I did this to convert first from a char to int and then add one in
        // order to change a to b
        x = word[i];
        x++;
        word[i] = x;
        puts(word);

        // here is the problem after going through the first loob (i) we have
        // the word baa and it should go through the second loop j but it
        // doesn't
        for (j = 0; j < n - 1; j++) {
            noob[j] = word[j];
            word[j + 1] = word[j];
            word[j] = noob[j];
            puts(word);
        }

    }
}

int
main(void)
{
    myfunction();

    return 0;
}

有人能幫助我嗎 ？？？

Answer 1

只是做char word[n] = "aaaaa";

完整代碼：

#include <stdio.h>
#include <stdlib.h>

#define n  5

char word[n] = "aaaaa";
char noob[n];

void
myfunction(void)
{
    int x, i, j;

    for (i = 0; i < n - 1; i++) {
        // I did this to convert first from a char to int and then add one in
        // order to change a to b
        x = word[i];
        x++;
        word[i] = x;
        puts(word);

        // here is the problem after going through the first loob (i) we have
        // the word baa and it should go through the second loop j but it
        // doesn't
        for (j = 0; j < n - 1; j++) {
            noob[j] = word[j];
            word[j + 1] = word[j];
            word[j] = noob[j];
            puts(word);
        }

    }
}

int
main(void)
{
    myfunction();

    return 0;
}

當然，當您在編譯時知道單詞的長度時，這項工作就可以了。 該長度僅在運行時才知道（例如，在用戶輸入時），您需要動態分配（例如，使用malloc ）並刪除。