确定一个字符串是否发生在另一个字符串的末尾

Question

这是来自K&R的练习5-4，我花了几个小时对它进行微调，但现在它似乎起作用了。我是新手，我欢迎任何关于如何做得更好的评论。

/* Function strend(s, t), which returns 1 if the string t 
 * occurs at the end of string s and zero otherwise */
int strend(const char *s, const char *t)
{   
    const char *s0, *t0;

    if (s == 0) {
        printf("s is NULL pointer\n");
        return (-1);
    }
    if (t == 0) {
        printf("t is NULL pointer\n");
        return (-1);
    }
    s0 = s;
    t0 = t;

    while (*s++)
        ;
    s -= 2;         /* *s points to last real char in s */

    while (*t++)
        ;
    t -= 2;         /* *t points to last real char in t */

    if ((t-t0) > (s-s0))
            return (0);  /* t is longer than s */

    while (t>=t0) {
        if (*s-- != *t--) 
            return (0);  /* Mismatch */
    }

    return (1); /* Match */
}

以下是main()：

/* Test the function strend(s, t), which returns 1 if the string t 
 * occurs at the end of string s and zero otherwise */
#include "jim.h"
#include "subs.c"

char a[MAXLINE], b[MAXLINE];

int main (void)

{
    printf("Return = %1d, Expect = 1\n", strend("12345", "45"));
    printf("Return = %1d, Expect = 0\n", strend("12345", "35"));
    printf("Return = %1d, Expect = 0\n", strend("45", "345"));
    printf("Return = %1d, Expect = 1\n", strend("12345", "12345"));
    printf("Return = %1d, Expect = 1\n", strend("12345", "5"));
    printf("Return = %1d, Expect = 0\n", strend("12345", "4"));
    printf("Return = %1d, Expect = 1\n", strend("12345", ""));
    printf("Return = %1d, Expect = 0\n", strend("12345", "+"));
    printf("Return = %1d, Expect = 0\n", strend("12345", "a2345"));
    printf("Return = %1d, Expect = 1\n", strend("", ""));
    printf("Return = %1d, Expect = 0\n", strend("", "Z"));
    printf("Return = %1d, Expect = 1\n", strend("1", "1"));
    printf("Return = %1d, Expect = 0\n", strend("1", "1A"));
    printf("Return = %1d, Expect = -1\n", strend(0, "1A"));
    printf("Return = %1d, Expect = -1\n", strend("1", 0));
}

这是输出：

返回= 1，Expect =1返回= 0，Expect =0返回= 0，Expect =0返回= 1，Expect =1返回= 1，Expect =1返回= 1，Expect =1返回= 0，Expect =0返回= 0，Expect =0返回= 1，Expect =0返回= 1，Expect =1返回= 0，Expect =0是空指针返回= -1，Expect = -1 t是空指针返回= -1，Expect = -1

Answer 1

修正指针。这个函数没有改变内存中的任何东西，没有理由它们不应该是const。与您在函数中声明的变量相同；使它们为const

int strend(const char* s, const char* t)

在输入指针中检查NULL。

尽可能使用库字符串函数。而不是做：

while(*s++) ;
s-=2;

调用strlen(s);，然后根据需要在algo中减去1。由于您正在迭代字符串，所以用函数调用替换它具有相同的效率，并使其更具可读性。

在while循环周围添加{。这是未来的证明。

单元测试以确认正确性。

Answer 2

我会改变它更像是..。

char* start_of_s = s;
char* start_of_t = t;
if(s == NULL || t==NULL) return 0;
while(*s) s++;
while(*t) t++;
while(*s == *t && s > start_of_s && t> start_of_t)
{
    s--; t--;
}
return (t == start_of_t) && (*s == *t);

此外，单元测试与您的测试(和我的)，除了我标出了"string_is_at_end“。我使用http://code.google.com/p/seatest/进行单元测试。(主要是因为我写的，虽然我有一个很晚的版本，但我还没有发布)

    assert_true(string_is_at_end("12345",""));
    assert_true(string_is_at_end("123","23"));
    assert_true(string_is_at_end("123","3"));
    assert_false(string_is_at_end("","123"));
    assert_false(string_is_at_end("123","1"));
    assert_false(string_is_at_end("123","2"));
    assert_true(string_is_at_end("12345", "45"));
    assert_false(string_is_at_end("12345", "35"));
    assert_false(string_is_at_end("45", "345"));
    assert_true(string_is_at_end("12345", "12345"));
    assert_true(string_is_at_end("12345", "5"));
    assert_false(string_is_at_end("12345", "4"));
    assert_true(string_is_at_end("12345", ""));
    assert_false(string_is_at_end("12345", "+"));
    assert_false(string_is_at_end("12345", "a2345"));
    assert_true(string_is_at_end("", ""));
    assert_false(string_is_at_end("", "Z"));
    assert_true(string_is_at_end("1", "1"));
    assert_false(string_is_at_end("1", "1A"));
    assert_false(string_is_at_end(0, "1A"));
    assert_false(string_is_at_end("1", 0));