用户1799795，
对于它的价值（现在你已经解决了你的问题），我冒昧地向你展示了在“使用数组”的限制下我是如何做到这一点的，并解释了我为什么要这样做。 . 请注意，虽然我是经验丰富的程序员，但我不是 C 大师……我曾与那些绝对让我陷入 C 杂草（双关语）的人一起工作。
#include <stdio.h>
#include <string.h>
#define LINE_SIZE 500
#define MAX_WORDS 50
#define WORD_SIZE 20
// Main function.
int main(int argc, const char * argv[])
{
int counter = 0;
// ----------------------------------
// Read a line of input from the user (ie stdin)
// ----------------------------------
char line[LINE_SIZE];
printf("Please type in a list of names then hit ENTER:\n");
while ( fgets(line, LINE_SIZE, stdin) == NULL )
fprintf(stderr, "You must enter something. Pretty please!");
// A note on that LINE_SIZE parameter to the fgets function:
// wherever possible it's a good idea to use the version of the standard
// library function that allows you specificy the maximum length of the
// string (or indeed any array) because that dramatically reduces the
// incedence "string overruns", which are a major source of bugs in c
// programmes.
// Also note that fgets includes the end-of-line character/sequence in
// the returned string, so you have to ensure there's room for it in the
// destination string, and remember to handle it in your string processing.
// -------------------------
// split the line into words
// -------------------------
// the current word
char word[WORD_SIZE];
int wordLength = 0;
// the list of words
char words[MAX_WORDS][WORD_SIZE]; // an array of upto 50 words of
// upto 20 characters each
int wordCount = 0; // the number of words in the array.
// The below loop syntax is a bit cyptic.
// The "char *c=line;" initialises the char-pointer "c" to the start of "line".
// The " *c;" is ultra-shorthand for: "is the-char-at-c not equal to zero".
// All strings in c end with a "null terminator" character, which has the
// integer value of zero, and is commonly expressed as '\0', 0, or NULL
// (a #defined macro). In the C language any integer may be evaluated as a
// boolean (true|false) expression, where 0 is false, and (pretty obviously)
// everything-else is true. So: If the character at the address-c is not
// zero (the null terminator) then go-round the loop again. Capiche?
// The "++c" moves the char-pointer to the next character in the line. I use
// the pre-increment "++c" in preference to the more common post-increment
// "c++" because it's a smidge more efficient.
//
// Note that this syntax is commonly used by "low level programmers" to loop
// through strings. There is an alternative which is less cryptic and is
// therefore preferred by most programmers, even though it's not quite as
// efficient. In this case the loop would be:
// int lineLength = strlen(line);
// for ( int i=0; i<lineLength; ++i)
// and then to get the current character
// char ch = line[i];
// We get the length of the line once, because the strlen function has to
// loop through the characters in the array looking for the null-terminator
// character at its end (guess what it's implementation looks like ;-)...
// which is inherently an "expensive" operation (totally dependant on the
// length of the string) so we atleast avoid repeating this operation.
//
// I know I might sound like I'm banging on about not-very-much but once you
// start dealing with "real word" magnitude datasets then such habits,
// formed early on, pay huge dividends in the ability to write performant
// code the first time round. Premature optimisation is evil, but my code
// doesn't hardly ever NEED optimising, because it was "fairly efficient"
// to start with. Yeah?
for ( char *c=line; *c; ++c ) { // foreach char in line.
char ch = *c; // "ch" is the character value-at the-char-pointer "c".
if ( ch==' ' // if this char is a space,
|| ch=='\n' // or we've reached the EOL char
) {
// 1. add the word to the end of the words list.
// note that we copy only wordLength characters, instead of
// relying on a null-terminator (which doesn't exist), as we
// would do if we called the more usual strcpy function instead.
strncpy(words[wordCount++], word, wordLength);
// 2. and "clear" the word buffer.
wordLength=0;
} else if (wordLength==WORD_SIZE-1) { // this word is too long
// so split this word into two words.
strncpy(words[wordCount++], word, wordLength);
wordLength=0;
word[wordLength++] = ch;
} else {
// otherwise: append this character to the end of the word.
word[wordLength++] = ch;
}
}
// -------------------------
// print out the words
// -------------------------
for ( int w=0; w<wordCount; ++w ) {
printf("Hi %s!\n", words[w]);
}
return 0;
}
在现实世界中，人们不能对单词的最大长度做出如此严格的假设，或者会有多少单词，如果给出这样的限制，它们几乎都是任意的，因此很快就会被证明是错误的......所以直接解决这个问题，我倾向于使用链表而不是“单词”数组......等到你得到“动态数据结构”......你会喜欢他们的;-)
干杯。基思。
PS：你做得很好......我的建议是“继续卡车”......通过练习这会变得容易得多。