将NULL值从函数返回到C中的指针

Question

我有一个学校作业，我应该在其中创建三个函数。函数是 printFirstWord()、skipWords() 和 printWord()。

尽管写得并不完美，但我已经设法让 printFirstWord 函数正常工作，而其他两个函数大部分工作正常。

但是，在 skipWords() 中，如果您希望跳过的单词数量大于您输入的字符串中的单词数量，我应该返回一个指针值 NULL。这是我的代码：

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

//Call the functions
void printFirstWord(char inputString[]);
char* skipWords(char sentence[], int words);
void printWord(char sentence[], int wordNumber);

int main()
{
    printWord("I like to eat bunnies", 0); //Prints I
    printWord("I like to eat bunnies", 1); //Prints like
    printWord("I like to eat bunnies", 2); //etc
    printWord("I like to eat bunnies", 3);
    printWord("I like to eat bunnies", 4);
    printWord("I like to eat bunnies", 5); //This should return NULL
    printWord("I like to eat bunnies", 6); //This should return NULL

    return 0;
}

//Function that prints only the first word of a string
void printFirstWord(char inputString[])
{
    int i = 0;

    //Removes initial non-alpha characters, if any are present
    while (!isalpha(inputString[i]))
        i++;

    //Checks if the next input is alphabetical or is the character '-'
    while (inputString[i] != ' ')
    {
        printf("%c", inputString[i]);
        i++;
    }

}

char* skipWords(char sentence[], int words)
{
    int i = 0, wordCount = 0;

    for(i = 0; wordCount < words; i++)
    {
        if(sentence[i] == ' ')
        {
            wordCount++;
        }
    }

    //Can't get this to work, not sure how to return NULL in a function
    if (words >= wordCount)
        return NULL;
    else
        return &sentence[i];
}

void printWord(char sentence[], int wordNumber)
{
    char *sentencePointer;
    sentencePointer = skipWords(sentence, wordNumber);

    if (sentencePointer != NULL)
        printFirstWord(sentencePointer);
    else if (sentencePointer == NULL)
        printf("\nError. Couldn't print the word.\n");
}

最初我的问题是程序经常崩溃，但我添加了 printWord 函数的最后一部分并且它停止了崩溃。我期望这个输出：

Iliketoeatbunnies

Error. Couldn't print the word.

Error. Couldn't print the word.

这是我收到的输出：

Error. Couldn't print the word.

Error. Couldn't print the word.

Error. Couldn't print the word.

Error. Couldn't print the word.

Error. Couldn't print the word.

Error. Couldn't print the word.

指针是我的弱点，我觉得我错过了一些重要的东西，我一直在网上寻找，但我还没有找到任何适合我的解决方案，或者至少我觉得不适合我。

Answer 1

您的代码中几乎没有错误。更正将是

for(i = 0; sentence[i] && wordCount < words; i++)

另一个是

while (inputString[i] !=' ' && inputString[i]!='\0')

最后一个

if (words > wordCount)

第一个解释是 - 你不会检查字符串的末尾。否则你有未定义的行为。

当您到达字符串末尾但仍然没有空格时，可能会出现这种情况。为避免这种情况，您还需要考虑 \0 情况。

如果是words > wordCount，那么只有你应该抛出错误。如果它们相等，则应打印该值。

Answer 2

skipWords 中的这个循环是问题所在。它将遍历字符串并超出，因为您没有检查字符串的结尾。

for(i = 0; wordCount < words; i++)
{
    if(sentence[i] == ' ')
    {
        wordCount++;
    }
}

由于退出循环的唯一方法是让它找到与words 一样多的空格，因此它将始终返回NULL。

可以说wordCount 也应该从 1 开始，因为如果字符串中没有空格，您将始终至少有一个单词……除非字符串为空。

Answer 3

我们初学者应该互相帮助。:)

你来了

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

char * skipWords( const char *s, size_t n )
{
    if ( n )
    {
        while ( isblank( ( unsigned char )*s ) ) ++s;

        do
        {
            while ( *s && !isblank( ( unsigned char )*s ) ) ++s;
            while ( isblank( ( unsigned char )*s ) ) ++s;
        } while ( *s && --n );
    }

    return ( char * ) ( *s ? s : NULL );
}

void printFirstWord( const char *s )
{
    while ( isblank( ( unsigned char )*s ) ) ++s;
    while ( *s && !isblank( ( unsigned char )*s ) ) putchar( *s++ );
}

void printWord( const char *s, size_t n )
{
    const char *word = skipWords( s, n );

    if ( word )
    {
        printFirstWord( word );
    }
    else
    {
        printf( "%s", "Error. Could not print the word." );
    }

    putchar( '\n' );
}

int main(void) 
{
    printWord("I like to eat bunnies", 0); //Prints I
    printWord("I like to eat bunnies", 1); //Prints like
    printWord("I like to eat bunnies", 2); //etc
    printWord("I like to eat bunnies", 3);
    printWord("I like to eat bunnies", 4);
    printWord("I like to eat bunnies", 5); //This should return NULL
    printWord("I like to eat bunnies", 6); 

    return 0;
}

程序输出是

I
like
to
eat
bunnies
Error. Could not print the word.
Error. Could not print the word.

至于你的代码，你通常不会像例子那样检查终止零

while (inputString[i] != ' ')
{
    printf("%c", inputString[i]);
    i++;
}

并忽略多个空格相互跟随的情况。

Answer 4

你在skipWords()的算法坏了，我改算法实现你的功能，把int改成size_t：

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

int printFirstWord(char const *inputString);
char const *skipWords(char const *sentence, size_t words);
int printWord(char const *sentence, size_t wordNumber);

int main(void) {
  printWord("I like to eat bunnies", 0); // Prints I
  printf("\n");
  printWord("I like to eat bunnies", 1); // Prints like
  printf("\n");
  printWord("I like to eat bunnies", 2); // etc
  printf("\n");
  printWord("I like to eat bunnies", 3);
  printf("\n");
  printWord("I like to eat bunnies", 4);
  printf("\n");
  printWord("I like to eat bunnies", 5); // This should return NULL
  printf("\n");
  printWord("I like to eat bunnies", 6); // This should return NULL
  printf("\n");
}

int printFirstWord(char const *inputString) {
  int ret = 0;
  while (isblank(*inputString)) {
    inputString++;
  }
  while (!isblank(*inputString) && *inputString) {
    int tmp = printf("%c", *inputString++);
    if (tmp < 0) {
      return -1;
    }
    ret += tmp;
  }
  return ret;
}

char const *skipWords(char const *inputString, size_t words) {
  while (isblank(*inputString)) {
    inputString++;
  }
  while (words > 0) {
    words--;
    while (!isblank(*inputString)) {
      if (!*inputString) {
        return NULL;
      }
      inputString++;
    }
    while (isblank(*inputString)) {
      inputString++;
    }
  }
  return *inputString ? inputString : NULL;
}

int printWord(char const *sentence, size_t wordNumber) {
  char const *sentencePointer = skipWords(sentence, wordNumber);
  if (sentencePointer != NULL) {
    return printFirstWord(sentencePointer);
  } else {
    fprintf(stderr, "Error. Couldn't print the word.\n");
    return -1;
  }
}