我有一个分配，用户在其中输入一个字符串，然后在一个函数中输入一个模式，然后必须检查该模式是否存在于字符串中，它出现了多少次以及偏移了多少。我很沮丧，我的同学不断给我一些神秘的提示。下面是我的get函数
int getNums()
{
printf("Please enter a number: "); //Initial printf
int count, patcount;
int torf;
char len_num[31]; //The character array for the initial entered string
char pat_num[6]; //The character array for the entered pattern after initial string
char *lenptr = len_num; //pointer to the address of the first element of len_num
char *patptr = pat_num; //pointer to the address of the first element of len_num
scanf("%s", len_num); //Where the user scans in their wanted number, which is treated as a string
printf("\n");
printf("%s\n", lenptr);
int len = stringLength(lenptr); //Checks how long string is
int valid = isValid(len_num); //Checks if string is valid
for(count=0; count<len_num[count]; count++) //Checks if length of string is within appropriate range
{
if(len>=10 && len<=30) //Continues to pattern get if within range
{
torf=1;
}
else //Denies continuation if string is outside of range
{
torf=0;
printf("Not within range! Try again!\n");
return (1);
}
}
printf("Please enter a pattern: "); //Initial entry statement for pattern
scanf("%s", pat_num); //User scans in pattern
printf("\n");
printf("%s\n", pat_num);
len = stringPattern(patptr); //Check how long pattern is
valid = isValid(pat_num); //Checks if pattern is valid
for(patcount=0; patcount<pat_num[patcount]; patcount++) //Checks if length of pattern is within appropriate range
{
if(len>=2 && len<=5) //Continues to pattern check if within range
{
torf=1;
}
else //Denies continuation if pattern is outside of range
{
torf=0;
printf("Pattern not within range! Try again!\n");
return (1);
}
}
checkPattern();
}
我不知道如何启动检查功能。更不用说我必须通过引用传递指针，并且我也坚持这样做
由于您要求使用模式匹配功能，因此我没有检查您的字符串输入功能。您可以使用以下简单的驱动程序代码来测试我的解决方案：
#include <stdio.h>
void findPattern(char* input, char* pattern);
int main()
{
char input[31], pattern[6];
printf("Enter the string: ");
scanf("%s", input);
printf("Enter the pattern: ");
scanf("%s", pattern);
findPattern(input, pattern);
return 0;
}
[findPattern胜过checkPattern。您应根据自己的方便将其重命名。根据您的要求，除stdio.h中的库函数外，我没有使用任何库函数。以下是我承担的这项任务，我在注释中解释了逻辑。基本上，它只遍历整个输入字符串一次，在此检查与模式中的初始字符是否匹配。如果是这样，它将标记偏移量并在模式中进一步搜索以找到完全匹配。
void findPattern(char* input, char* pattern)
{
int i = 0; // iterator for input
int j = 0; // iterator for pattern
// solution variables
int offset = 0;
int occurrence = 0;
// Search the entire input string
while (input[i] != '\0')
{
// Mark the offset whenever the first character of the pattern matches
if (input[i] == pattern[j])
{
offset = i;
// I didn't quite get the relativity of your offset
// Maybe you need: offset = i + 1;
}
// Search for complete pattern match
while (input[i] != '\0' && pattern[j] == input[i])
{
// Go for the next character in the pattern
++j;
// The pattern matched successfully if the entire pattern was searched
if (pattern[j] == '\0')
{
// Display the offset
printf("\nPattern found at offset %d", offset);
// Increment the occurrence
++occurrence;
// There are no more characters left in the pattern
break;
}
else
{
// Go for the next character in the input
// only if there are more characters left to be searched in the pattern
++i;
}
}
// Reset the pattern iterator to search for a new match
j = 0;
// Increment the input iterator to search further down the string
++i;
}
// Display the occurrence of the pattern in the input string
printf("\nThe pattern has occurred %d times in the given string", occurrence);
}
我必须通过引用传递指针，但我也坚持这样做
如果是这种情况，请以findPattern(input, pattern);代替：
findPattern(&input, &pattern);
您可能会想不通解决方案。您有一个字符串input，其中包含许多字符，您想计算其中的pattern多字符匹配数。关于字符串的一件好事是，您不需要知道它们要迭代多长时间。 ，因为根据定义，C中的字符串以nul-terminated字符结尾。
这可以让您简单地将索引保留在findpattern函数中，并且每次input中的字符与pattern中的字符匹配时都可以增加索引（否则将索引清零）。如果到达pattern[index] == '\0'的位置，则表示您已匹配模式中的所有字符。
您必须始终使用type声明一个函数，如果需要其余代码，则该有意义的返回值将指示该函数执行的任何操作的成功/失败（如果该函数仅打印输出） -那么void就可以了）。
否则，您需要选择一个合理的返回类型以指示在pattern中是否找到input匹配项（以及匹配多少）。在这里，简单的int类型即可。 （这限制了可以返回2147483647的匹配数，应该足够了。
将这些部分放在一起，您可以将功能简化为类似于：
int findpattern (const char *input, const char *ptrn)
{
int n = 0, idx = 0; /* match count and pattern index */
while (*input) { /* loop over each char in s */
if (*input == ptrn[idx]) /* if current matches pattern char */
idx++; /* increment pattern index */
else /* otherwize */
idx = 0; /* zero pattern index */
if (!ptrn[idx]) /* if end of pattern - match found */
n++; /* increment match count */
input++; /* increment pointer */
}
return n; /* return match count */
}
添加一个简短的示例程序，该程序允许您将pattern和input输入为程序的前两个参数（或使用未显示的默认值）；
int main (int argc, char **argv) {
char *pattern = argc > 1 ? argv[1] : "my",
*input = argc > 2 ? argv[2] : "my dog has fleas, my cat has none";
int n;
if ((n = findpattern (input, pattern)))
printf ("'%s' occurs %d time(s) in '%s'\n", pattern, n, input);
else
puts ("pattern not found");
}
注意提供有意义的回报如何使您同时（1）validate是否找到匹配项； （2）提供通过返回找到的匹配数。完整的代码只需要标题stdio.h，例如
#include <stdio.h>
int findpattern (const char *input, const char *ptrn)
{
int n = 0, idx = 0; /* match count and pattern index */
while (*input) { /* loop over each char in s */
if (*input == ptrn[idx]) /* if current matches pattern char */
idx++; /* increment pattern index */
else /* otherwize */
idx = 0; /* zero pattern index */
if (!ptrn[idx]) /* if end of pattern - match found */
n++; /* increment match count */
input++; /* increment pointer */
}
return n; /* return match count */
}
int main (int argc, char **argv) {
char *pattern = argc > 1 ? argv[1] : "my",
*input = argc > 2 ? argv[2] : "my dog has fleas, my cat has none";
int n;
if ((n = findpattern (input, pattern)))
printf ("'%s' occurs %d time(s) in '%s'\n", pattern, n, input);
else
puts ("pattern not found");
}
示例使用/输出
检查多个匹配项：
$ ./bin/findpattern
'my' occurs 2 time(s) in 'my dog has fleas, my cat has none'
单场比赛：
$ ./bin/findpattern fleas
'fleas' occurs 1 time(s) in 'my dog has fleas, my cat has none'
未找到模式
$ ./bin/findpattern gophers
pattern not found
仔细检查，如果还有其他问题，请告诉我。