再次将 C 新手返回这里。我正在尝试 K 和 R C 中的练习,在尝试练习 1-14 的过程中,我真的被难住了。 我的解决方案有效,但并不总是正确的,我正在寻求帮助来完善我所写的内容,或者是否有更好(更容易理解!)的方法! 我的脚本:
#include <stdio.h>
/* How many times a character appears in an array */
main()
{
int c;
int count = 0;
int uniquecount = 0;
char array[20];
array[0] = '\0';
while((c = getchar()) != EOF)
{
array[++count] = c;
}
/* for each element in array,
* check if array[each] in newarray.
* if array[each] in newarray
* break and start checking again.
* if array[each] not in newarray
* add array[each] to end of newarray*/
printf("count = %d\n", count);
array[count] = '\0';
char newarray[count];
newarray[0] = '\0';
for(int a = 0; a < count; ++a)
{
for(int b = 0; b <= a; ++b)
{
if(newarray[b] == array[a])
break;
if(newarray[b] != array[b])
{
newarray[b] = array[b];
++uniquecount;
}
}
}
printf("uniquecount = %d\n", uniquecount);
newarray[uniquecount + 1] = '\0';
printf("array => ");
for(int i = 0; i < count; ++i)
printf("\'%c\'", array[i]);
printf("\n");
printf("newarray => ");
for(int i = 0; i < uniquecount + 1; ++i)
{
if(newarray[i] != '\0')
printf("\'%c\'", newarray[i]);
}
printf("\n");
}
当我尝试一些简单的字符串时,它可以工作,有时却不能:
./times_in_array
this is
count = 9
uniquecount = 5
array => '''t''h''i''s'' ''i''s'' '
newarray => 't''h''i''s'' '
./times_in_array
something comes
count = 16
uniquecount = 11
array => '''s''o''m''e''t''h''i''n''g'' ''c''o''m''e''s'
newarray => 's''o''m''e''t''h''i''n''g'' ''c'
./times_in_array
another goes
count = 13
uniquecount = 12
array => '''a''n''o''t''h''e''r'' ''g''o''e''s'
newarray => 'a''n''o''t''h''e''r'' ''g''o''e''s'
请有人指导我哪里错了?非常感谢!
最佳答案
我们最常写的是 int main(void)
而不是 main()
。
此外,了解当您像这样访问新数组时非常重要:
newarray[b]
您实际上正在访问未初始化内存。然而,由于匹配 array[a]
的随机字符的概率太小,看来您可以侥幸逃脱。
因此,我建议您初始化新数组,如下所示:
for(int i = 0; i < count; ++i)
newarray[i] = '\0';
现在,我不再提供我的解决方案,而是坚持让你明白你做错了什么;这就是你教育自己的方式。在内部循环开始时打印数据将非常有帮助,例如:
printf("b = %d, a = %d, newarray[b] = %c, array[a] = %c, array[b] = %c\n", b, a, newarray[b], array[a], array[b]);
并在增加唯一计数器时打印一条消息,如下所示:
printf("UNIQUE, %d\n", uniquecount);
当您执行程序时,您将看到:
...
b = 9, a = 12, newarray[b] = g, array[a] = s, array[b] = g
b = 10, a = 12, newarray[b] = , array[a] = s, array[b] = o
UNIEUQ, 10
b = 11, a = 12, newarray[b] = , array[a] = s, array[b] = e
UNIEUQ, 11
b = 12, a = 12, newarray[b] = , array[a] = s, array[b] = s
UNIEUQ, 12
uniquecount = 12
array => '''a''n''o''t''h''e''r'' ''g''o''e''s'
newarray => 'a''n''o''t''h''e''r'' ''g''o''e''s'
这强烈暗示你出了什么问题。在 's' (gos 的最后一个字母)第一次被发现是唯一的(这很好)之后,我们不会打破循环,因此您的代码将继续检查新数组并被愚弄,认为这个 's' 是再次独一无二。
因此,当您找到唯一元素时添加一个中断,此时应该没问题:
++uniquecount;
break;
关于c - 提取数组中的唯一元素(来自 K 和 R C ex1-14),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44225626/