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反向字符数组中的单词

[英]Reverse characters word in array

for exemple i need to invers "Paris" to "siraP"... 例如,我需要将“巴黎”转换为“ siraP” ...

My main: 我的主要:

int main(void)
{
    char w1[] = "Paris";
    ReverseWord(w1);
    printf("The new word is: %s",w1);
    return0;
}

and my function: 和我的功能:

void ReverseWord(char *Str)
{
     int counter=0;
     for(int i=0; *(Str+i)!='\0'; i++)
          counter++;

     int length = counter-1;

     char temp[length];

     for(int j=0; temp[j]=='\0'; j++)
          temp[j]=Str[length-j];
}

Now I have my renverse word in temp[]. 现在,我在temp []中有一个单词。 I need to put it in my pointer *Str. 我需要将其放在指针* Str中。 How can I do it?? 我该怎么做??

Thanks 谢谢

If you want use temp must then your function like this 如果要使用temp必须像这样使用函数

void ReverseWord(char *Str)
{
    int i,j;

    if(str)
    {
      int length=strlen(Str);
      char temp[length+1];

      for( j=0; j<length; j++)
          temp[j]=Str[length-1-j];

      temp[j]='\0';

      strcpy(Str,temp);
   }

}

Without using temp as follows 不使用temp如下

void ReverseWord(char *Str)
{
    int end= strlen(Str)-1;
    int start = 0;

    while( start<end )
    {
        Str[start] ^= Str[end];
        Str[end] ^= Str[start];
        Str[start]^= Str[end];

        ++start;
        --end;
    }
}
void ReverseWord(char *Str)
{
    size_t len;
    char temp, *end;

    len = strlen(Str);
    if (len < 2)
        return;

    end = Str + len - 1;
    while (end > Str)
    {
        temp = *end;
        *end-- = *Str;
        *Str++ = temp;
    }
}

One more option, this time with dangerous malloc(3). 还有一个选择,这次是危险的malloc(3)。

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

char *rev(char s[]) {
  char *buf = (char *)malloc(sizeof(char) * strlen(s));
  int i, j;

  if(buf != NULL)
    for(i = 0, j = strlen(s) - 1; j >= 0; i++, j--)
      buf[i] = s[j];

  return buf;
}

int main(int argc, char **argv) {
  printf("%s\n", rev(argv[1]));
  return 0;
}

Run with "foo bar foobar baz" and get zab raboof rab oof back: 使用"foo bar foobar baz"运行,并返回zab raboof rab oof

~/tmp$ ./a.out "foo bar foobar baz"
zab raboof rab oof

Here I think you can study two algorithms: 我认为您可以在这里研究两种算法:

  1. C string length calculate: the end of the c string is '\\0' C字符串长度计算:C字符串的末尾为“ \\ 0”
  2. How to reverse ac string in place 如何将交流弦反向

And if you need to test the code, you should alloc testing strings in heap or strack. 而且,如果您需要测试代码,则应在堆或strack中分配测试字符串。 If you write a literal string, you may meet a bus error because of the literal string being saved in text-area which is a read only memory. 如果编写文字字符串,则可能会遇到总线错误,因为文字字符串已保存在只读存储器的文本区域中。

And the following is the demo: 以下是演示:

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

void reverse_string(char* str)
{
    size_t len;
    char tmp, *s;
    //Get the length of string, in C the last char of one string is \0
    for(s=str;*s;++s) ;
    len = s - str;

    //Here we use the algorithm for reverse the char inplace.
    //We only need a char tmp place for swap each char
    s = str + len - 1;
    while(s>str){
        tmp = *s;
        *s = *str;
        *str = tmp;
        s--;
        str++;
    }
}

int main()
{
    char* a = "abcd";
    //Here "abcd" will be saved in READ Only Memory. If you test code, you will get a bus error.
    char* b = (char*)calloc(1,10);
    strcpy(b,a);
    reverse_string(b);
    printf("%s\n",b);

    a = "abcde";
    strcpy(b,a);
    reverse_string(b);
    printf("%s\n",b);
}

you can do it simply by following code 您只需按照以下代码即可

for(int k=0;k<strlen(temp);k++)
     {
        Str[k]=temp[k];
     }

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