Question

用C编写一个程序：声明一个名为buffer的字符串变量，最大大小为80 从键盘输入一个字符串到缓冲区通过替换任何元音来修改缓冲区中包含的字符串（大写或小写＆＃39; a＆＃39;＆＃39; e＆＃39;＆＃39; i＆＃39;，＆＃39; o＆＃39;或者＆＃39; u＆＃39;加号（+）打印出修改过的字符串打印出被替换的元音总数

我这样做了：

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

int main(){
    char buffer[80];
    char word[80];
    char word2[80];
    scanf("%s",word);
    strcpy(buffer, word);

   int i;
   int counter=0;
   for(i=0;i<80;i++)
   {
      word2[i]="";
   }

   for(i=0;i<strlen(buffer);i++)
   {
   if(buffer[i]=="a" || buffer[i]=="e" || buffer[i]=="i" || buffer[i]=="o" || buffer[i]=="u" || buffer[i]=="A" || buffer[i]=="E" || buffer[i]=="I" || buffer[i]=="O" || buffer[i]=="U")
  //   if(strcmp(buffer[i],"a")==0)
      {
          strcat(word2,"+");
          counter++;
      }else{
       strcat(word2,buffer[i]);
      }
   }

   printf("The modified string is %s",word2);
   printf("in total there was %d vowels.",counter);

   return 0;
}

但我一直在收到错误，请帮助我。

Answer 1

您需要为字符使用单引号：

buffer[i]=='a'

"a"表示一个指向字符串文字的指针，该字符串文字将存储在程序的某个地址空间段中。

Answer 2

试试这个

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

int main()
{
    char buffer[80];
    char word[80];
    char word2[80];
    scanf("%s",word);
    strcpy(buffer, word);

    int i;
    int counter=0;


    for(i=0;i<strlen(buffer);i++)
    {
        if(buffer[i]=='a'|| buffer[i]=='e' || buffer[i]=='i' || buffer[i]=='o' || buffer[i]=='u' || buffer[i]=='A' || buffer[i]=='E' || buffer[i]=='I' || buffer[i]=='O' || buffer[i]=='U')
        {
           word[i] = '+';
           counter++;
        }

    }

    printf("The modified string is %s",word);

    printf("in total there was %d vowels.",counter);

    return 0;

}

有人可以用C帮我

2 个答案: