c - 低效地使用 strstr 和 strchr

Question

在审查我的代码时，我的教授说我使用 strstr 和 strchr 会导致大量资源浪费，因为它们中的每一个都会扫描字符串。

我可以减少函数的数量吗？

此代码扫描一个字符串，并根据设置的参数决定输入是否有效。
ch1 是 '@' 而 ch2 是 '.', (email[i]) 是字符串。

    for (i = 0; email[i] != 0; i++) {
        {
            if (strstr(email, "@.") ||
                strstr(email, ".@") ||
                strstr(email, "..") ||
                strstr(email, "@@") ||
                email[i] == ch1 ||
                email[i] == ch2 ||
                email[strlen(email) - 1] == ch1 ||
                email[strlen(email) - 1] == ch2) {
                printf("The entered e-mail '%s' does not pass the required parameters, Thus it is invalid\n", email);
            } else {
                printf("The email '%s' is a valid e-mail address\n",email);
            }
            break;
        }
    }

这就是我所说的片段。

我应该编写自己的代码来执行一次检查吗？如果是这样，你能给我一些这方面的指示吗？ 谢谢你。

编辑:非常感谢您的回复，我确实了解到我的代码中的错误，希望我能从中吸取教训。
再次感谢!

编辑:2:我想再次感谢您的回复，他们对我帮助很大，我相信我写出了更好的代码

int at_count = 0, dot_count = 0, error1 = 0, error2 = 0;
int i;
size_t length = strlen(email);
int ch1 = '@', ch2 = '.';

for ( i = 0; email[i] != '\0'; i++)  /* for loop to count the occurance of the character '@' */
    {
    if ( email[i] == ch1)
        at_count++;
    }

for ( i = 0; email[i] != '\0'; i++)  /* for loop to count the occurance of the character '.' */
    {
    if ( email[i] == ch2)
        dot_count++;
    }

if ( email[0] == ch1 || email[0] == ch2 || email[length-1] == ch1 || email[length-1] == ch2 )
        {
    error1++;
        }
else
        {
    error1 = 0;
        }


if ( strstr(email,".@") || strstr(email, "@.") || strstr(email, "..") || strstr(email, "@@"))
        {
    error2++;
        }
else
        {
    error2 = 0;
        }

if ( (at_count != 1) || (dot_count < 1) || (error1 == 1) || (error2 == 1))
    {
    printf("The user entered email address '%s' is invalid\n", email);
    }
else
    {
    printf("'%s' is a valid email address\n", email);
    }

我觉得这样的代码更优雅、更简单，也更高效。
我的主要灵感来自@chqrlie，因为我觉得他的代码非常好并且易​​于阅读。 无论如何，我可以改进吗？
(电子邮件检查只是为了练习，不要介意!) 非常感谢大家!

Answer 1

你的代码确实有多个问题:

for (i = 0; email[i] != 0; i++) {   // you iterate for each character in the string.
    {   //this is a redundant block, remove the extra curly braces
        if (strstr(email, "@.") ||  // this test only needs to be performed once
            strstr(email, ".@") ||  // so does this one
            strstr(email, "..") ||  // so does this one
            strstr(email, "@@") ||  // again...
            email[i] == ch1 ||      // this test is only performed once
            email[i] == ch2 ||      // so is this one
            email[strlen(email) - 1] == ch1 ||  // this test is global
            email[strlen(email) - 1] == ch2) {  // so is this one
            printf("The entered e-mail '%s' does not pass the required parameters, Thus it is invalid\n", email);
        } else {
            printf("The email '%s' is a valid e-mail address\n", email);
        }
        break;  // you always break from the loop, why have a loop at all?
    }
}

您扫描字符串 4 次以测试各种模式，另外扫描 2 次以测试 strlen()。应该可以在单次扫描过程中执行相同的测试。

另请注意，还有更多问题未被注意到:

• 应该有一个@存在
• 不能有空格
• 更一般地说，地址中允许的字符是有限的。

一些测试似乎有些矫枉过正:为什么拒绝 @ 之前的 ..，为什么拒绝 @ 之前的尾随 . ?

这是一个更高效的版本:

int at_count = 0;
int has_error = 0;
size_t i, len = strlen(email);

if (len == 0 || email[0] == ch1 || email[0] == ch2 ||
    email[len - 1] == ch1 || email[len - 1] == ch2) {
    has_error = 1;
}

for (i = 0; !has_error && i < len; i++) {
    if (email[i] == '.') {
        if (email[i + 1] == '.' || email[i + 1] == '@') {
            has_error = 1;
        }
    } else if (email[i] == '@') {
        at_count++;
        if (i == 0 || i == len - 1 || email[i + 1] == '.' || email[i + 1] == '@') {
            has_error = 1;
        }
    }
    // should also test for allowed characters         
}

if (has_error || at_count != 1) {
    printf("The entered e-mail '%s' does not pass the required tests, Thus it is invalid\n", email);
} else {
    printf("The email '%s' is a valid e-mail address\n", email);
}