Question

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

main()
{
  int i;
  int *b, *z;
  char name[30];
  char vowel[5] = {'A', 'E', 'I', 'O', 'U'};
  char consonants[23] = {'B','C','D','F','G','H','J','K','L','M','N','P','Q','R','S','T','V','W','X','Y','Z'};

  printf ("input the string: ");
  scanf  ("%s", name);
  printf ("The string is %s\n", name);

  for (i=0; name[i]!='\0'; i++){
    if
      (b=strchr(vowel, name[i]) != NULL) {
      printf ("The vowels are:  %s\n", b); }
    else if
      (z=strchr(consonants, name[i]) != NULL) {
      printf ("The consonants are:  %s\n", z);
    }
  }
}

我想找出阵列中有多少个元音和辅音。这是我们老师向我们展示的唯一算法，但它不起作用。任何人都可以指出我的错误吗？

我只是再试一次，提出所有建议，

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

int main()
{
int vow, cons, i;
char *s, *s1;
char name[30];
char vowel[6] = "AEIOU";
char consonants[21] = "BCDFGHJKLMNPQRSTVWXYZ";

printf ("input the string: ");
scanf  ("%s", name);
printf ("The string is %s\n", name);
for (i=0; name[i]!='\0'; i++)
s = strchr(vowel, name[i]);
printf ("The vowels are:  %s\n", s);

s1 =strchr(consonants, name[i])) {
printf ("The consonants are:  %s\n", s1);
}

return 0;

}

这就是我改变它的方式，你的所有建议，我的其他问题是什么？因为仍然不能正常工作。感谢。

这是我的另一个程序版本

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

 int main()
 {
 int i;
 int counter=0, counter2=0;
 char *s;
 char name[30];
 char vowel[6] = "AEIOU";
 char consonants[21] = "BCDFGHJKLMNPQRSTVWXYZ";

 printf ("input the string: ");
 scanf  ("%s", name);
 printf ("The string is %s\n", name);
 for (i=0; name[i]!='\0'; i++) {
 if (s = strchr(vowel, name[i])) {
 counter++;
 }
 else if (s =strchr(consonants, name[i])) {
 counter2++;
 }
 printf ("First counter is %d\n", counter);
 printf ("The second counter is %d\n", counter2);
 return 0;
 }
  }

我添加了计数器来计算元音和辅音的数量，但仍然不起作用。

Answer 1

strchr()用于搜索字符串。

char vowel[] = "AEIOU";
char consonants[] = "BCDFGHJKLMNPQRSTVWXYZ";

Answer 2

#include< stdio.h>
int main()
{
    int vowel=0,consonant=0;
    printf ("input the string: ");
    scanf  ("%s", name);
    printf ("The string is %s\n", name);
    for(int i=0;name[i] !='\0';i++)
    {
        if( name[i] == 'A' || name[i] == 'E' || name[i] == 'I' || name[i] == 'O' || name[i] == 'U' )    
        {
            vowel++;
        }
        else
            consanant++;
    }
    printf("%d %d",vowel,consonant);
    return 0;
}

Answer 3

当我编译它时，我收到以下消息：

$ gcc -Wall vc.c
vc.c:4:1: warning: return type defaults to ‘int’ [-Wreturn-type]
vc.c: In function ‘main’:
vc.c:17:8: warning: assignment makes pointer from integer without a cast [enabled by default]
vc.c:17:3: warning: suggest parentheses around assignment used as truth value [-Wparentheses]
vc.c:18:4: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int *’ [-Wformat]
vc.c:20:13: warning: assignment makes pointer from integer without a cast [enabled by default]
vc.c:20:3: warning: suggest parentheses around assignment used as truth value [-Wparentheses]
vc.c:21:4: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int *’ [-Wformat]
vc.c:24:1: warning: control reaches end of non-void function [-Wreturn-type]

因此，首先要确保main的返回类型是'int'

int main(){

并在函数底部添加一个返回值

return 0;

此后，将b和z设置为char * s，以便它们与strchr的返回类型匹配

char *b, *z;

这将消除所有警告。

$ gcc -Wall vc.c
$

优异。现在，当我们运行你的程序时：

$ ./a.out 
input the string: aaa
The string is aaa
Segmentation fault

“分段错误”意味着您正在运行数组的末尾并读取您不拥有的内存。现在实施Ignacio Vazquez-Abrams的解决方案

char vowel[] = "AEIOU";
char consonants[] = "BCDFGHJKLMNPQRSTVWXYZ";

现在你的程序将运行完毕。

$ ./a.out 
input the string: AAA   
The string is AAA
The vowels are:  AEIOU
The vowels are:  AEIOU
The vowels are:  AEIOU

但它做得不多，是吗？

所以，如果你只想计算有多少元音和辅音，你可以为每一个添加一个整数，每次找到正确的类型时都会增加，并在结尾输出：

printf("Vowels:\t%d\nConsonants:\t%d", vowelsFound, consonantsFound);

但是，如果您尝试将它们作为列表输出，那么您将需要执行更多数据操作。一些链接要检查：

Linux Man Page for printf

Linux Man Page for String functions

Answer 4

您已将return语句放在for循环内，这阻止了它扫描整个name数组。

使用strchr时，您还需要将当前循环字符转换为大写，以使其正确匹配，因为您已经以大写形式定义了vowels。要使用toupper()，您需要添加ctype.h。

您也无需定义consonants。什么不是元音是辅音。

这是代码。我测试了它并且它有效：

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

int main()
{
    int i;
    int counter=0, counter2=0;
    char *s;
    char name[30];
    char vowel[6] = "AEIOU";

    printf ("input the string: ");
    scanf  ("%s", name);
    printf ("The string is %s\n", name);
    for (i=0; name[i]!='\0'; i++) {
        if (strchr(vowel, toupper(name[i])) != NULL) {
            counter++;
        }
        else {
            counter2++;
        }
    }
    printf ("First counter is %d\n", counter);
    printf ("The second counter is %d\n", counter2);
    return 0;
}

Answer 5

可替换地。

#include <stdio.h>
#include <string.h>
int main() {
 int t [256];
 int i,c;
 int cntw = 0;
 int cntc = 0;
 const char * vowel="AEIOUaeiou";
 const char * consonants="BCDFGHJKLMNPQRSTVWXYZbcdfghjklmnpqrstvwxyz";
 memset(t,0,256);
 while (*vowel) { t[*vowel] = 1; ++vowel;}
 while (*consonants) { t[*consonants] = 2; ++consonants;}
 printf ("Input the text:  CTRL-D to end\n");
 c = getchar();
 while(c >=0) {
  switch(t[c]) {
   case 1: ++cntw; break;
   case 2: ++cntc; break;
  }
  c=getchar();
 }
 printf ("Text has %d vowel%s and %d consonant%s\n",
  cntw,(cntw>1)?"s":"",
  cntc,(cntc>1)?"s":"");
 return 0;
}

用C语言在字符串中查找字符的堆栈

5 个答案: