#include <string.h>
#include <stdio.h>
#include <ctype.h>
#define SIZE 100
const char delim[] = ", . - !*()&^%$#@<> ? []{}";
int main(void)
{
int length[SIZE] = { 0 };
int name[SIZE];
int i = 0, ch;
int wordlength = 0;
int occurence = 0;
printf("Please enter sentence, end the setence with proper punctuation(! . ?) : ");
while (1)
{
ch = getchar();
if (isalpha(ch))
{
++wordlength;
}
else if (strchr(delim, ch))
{
if (wordlength)
{
length[wordlength - 1]++;
}
if(ch == '.' || ch == '?' || ch == '!')
{
break;
}
if(ch == '\'')
{
wordlength++;
}
wordlength = 0;
}
}
printf("Word Length \tOccurence \n");
for(i = 0; i<sizeof(length) / sizeof(*length); ++i)
{
if(length[i])
{
printf(" %d \t\t%d\n", i + 1, length[i]);
}
}
return 0;
}
所以程序应该计算一个句子,如......
"Hello how are you?"
然后输出:
5 Letter Words -> 1
3 Letter Words -> 3
但我希望它也将单引号计为字符,例如......
输入:
" 'Tis good. "
输出:
4 Letter words -> 2
我尝试过使用...
if(ch == '\'')
{
wordlength++;
}
但它只是不把单引号算作一个角色。
答案 0 :(得分:2)
您continue
遗失了if (ch =='\'') { ... }
;这就是为什么wordlength
之后立即归零。
另一种解决方案是在'
条件中包含if
:
if (isalpha(ch) || ch == '\'') {
wordlength ++;
}