输入一行字符，分别统计出其中英文字母、空格、数字和其他字符的个数。我的代码如下：

main()
{
    char c[100];
    printf("Please input a string less than 100 words:");
    scanf("%s",c);
    int i, alphabet=0, number=0, space=0, others=0;
    for ( i=0; i<100; i++ )
    {
        if (c[i]==32)
            space++;
         else if (c[i]>=48 && c[i]<=57)
            number++;
         else if(c[i]>=65 && c[i]<=90  ||  c[i]>=97 && c[i]<=122)
            alphabet++;
        else others++;
    }
    printf("The string has %d alphabets, %d numbers, % spaces and %d other character.\n", alphabet,number,space,others);
}

一编译，就出现如下错误信息：
expression syntax in function main
菜鸟请高手解决下，谢谢！看在我自学C、自主编的勇气上~~~

    char c[100];
    int i, alphabet=0, number=0, space=0, others=0;
    printf("Please input a string less than 100 words:");
    scanf("%s",c);
这样试试看看行不。

printf("The string has %d alphabets, %d numbers, %d spaces and %d other character.\n", alphabet,number,space,others);
楼主好粗心。

你不能用scanf("%s",c);来计入输入语句。应为scanf("%s",c);是以空格或回车结束的，即使他最后会计入最后的空格或回车，但你现在也应该知道如果你输入过程中出现了空格或回车 ，程序就误认为你要输入结束了

[ 本帖最后由 rock10822 于 2010-1-18 13:44 编辑 ]

再发一次贴，vc++下调试通过。
#include<stdio.h>
int main(void)
{
    char c[100];
    int i, alphabet=0, number=0, space=0, others=0;//要先定义，后写运行代码
    printf("Please input a string less than 100 words:");
    gets(c);//用scanf输入字符串时遇空格结束。按楼主的意思，这里应该用gets吧。
    for ( i=0; i<100; i++ )
    {
        if (c[i]==32)
            space++;
         else if (c[i]>=48 && c[i]<=57)
            number++;
         else if(c[i]>=65 && c[i]<=90  ||  c[i]>=97 && c[i]<=122)
            alphabet++;
        else others++;
    }
    printf("The string has %d alphabets, %d numbers, %d spaces and %d other character.\n", alphabet,number,space,others);
    return 0;
}

你好强哦

#include<stdio.h>
int main(void)
{
    char c[100];
    int i, alphabet=0, number=0, space=0, others=0；
    printf("Please input a string less than 100 words:");
    gets(c);
    for ( i=0; i<100; i++ )
    {
        if (c[i]==32)
            space++;
         else if (c[i]>=48 && c[i]<=57)
            number++;
         else if(c[i]>=65 && c[i]<=90  ||  c[i]>=97 && c[i]<=122)
            alphabet++;
        else others++;
    }
    printf("The string has %d alphabets, %d numbers, %d spaces and %d other character.\n", alphabet,number,space,others);
    return 0;
}

要先定义在写代码。。 c语言有这样严格的要求   LZ要切记哦

1
# include <stdio.h>
int main(void)
{
    int i,n,digit,letter,other;
    char ch;
    printf("Enter n:");
    scanf("%d",&n);
    printf("Enter %d characters:",n);
    digit=letter=other=0;
    for(i=1;i<=n;i++){
        ch=getchar();
        if((ch>='a'&&ch<='z')||(ch>='A'&&ch<='Z'))
            letter++;
        else if(ch>='0'&&ch<='9')
            digit++;
        else
            other++;
    }
    printf("letter=%d digit=%d other=%d\n",letter,digit,other);
    return 0;
}
2
# include <stdio.h>
int main(void)
{
    int i,n,digit,blank,other;
    char ch;
    printf("Enter n:");
    scanf("%d",&n);
    digit=blank=other=0;
    printf("Enter %d characters:",n);
    for(i=1;i<=n;i++){
        ch=getchar();
    switch(ch){
        case' ':
        case'\n':
            blank++;
            break;
        case'0':  case'1':  case'2':  case'3':  case'4':
        case'5':  case'6':  case'7':  case'8':  case'9':
            digit++;
            break;
        default:
            other++;
            break;
    }
     }
    printf("blank=%d digit=%d other=%d\n",blank,digit,other);
   
    return 0;
}

这个 不可以用 scanf