问strnstr的一个简单而安全的实现(在char数组的前n个字符中搜索子字符串)

EN

// Find an instance of substr in an array of characters
// the array of characters does not have to be null terminated
// the search is limited to the first n characters in the array.
char *strnstr(char *str, const char *substr, size_t n)
{
    char *p = str, *pEnd = str+n;
    size_t substr_len = strlen(substr);

    if(0 == substr_len)
        return str; // the empty string is contained everywhere.

    pEnd -= (substr_len - 1);
    for(;p < pEnd; ++p)
    {
        if(0 == strncmp(p, substr, substr_len))
            return p;
    }
    return NULL;
}

const char *strnstr(const char *str, const char *substr, size_t n)
{
    return strnstr((char *)str, substr, n);
}

#include <iostream>
#include <cstring>
#include <string>

int main()
{
    char s[] = "1234567890abcdefgh";
    size_t n = sizeof(s) - 1;

    const char *patterns[] = { "efgh", "0ab", "0b", NULL };
    const char *result = NULL;
    const char *pPattern = patterns[0];

    std::cout << "array of length " << n << " is: " << s << std::endl;

    for (int i = 0; pPattern; pPattern = patterns[++i])
    {
        result = strnstr(s, pPattern, n);
        std::cout << "finding " << pPattern << " n=" << n << ": "
            << (result ? result : "(null)") << std::endl;
    }

    pPattern = patterns[0];
    result = strnstr(s, pPattern, n-1);
    std::cout << "finding " << pPattern << " n=" << n-1 << ": "
        << (result ? result : "(null)") << std::endl;
    return 0;
}

array of length 18 is: 1234567890abcdefgh
finding efgh n=18: efgh
finding 0ab n=18: 0abcdefgh
finding 0b n=18: (null)
finding efgh n=17: (null)