使用strstr和strchr效率不高

发布时间：2020-12-16 09:22:36 所属栏目：百科来源：网络整理

导读：在查看我的代码时,我的教授说我使用strstr和strchr会导致大量浪费的资源,因为每个人都会扫描字符串. 我能以一种好的方式减少功能吗？此代码扫描字符串,并根据设置的参数决定输入是否有效. ch1是’@’,ch2是’.’,(email [i])是字符串. for (i = 0; email[i]

在查看我的代码时,我的教授说我使用strstr和strchr会导致大量浪费的资源,因为每个人都会扫描字符串.

我能以一种好的方式减少功能吗？

此代码扫描字符串,并根据设置的参数决定输入是否有效.
ch1是’@’,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 invalidn",email);
            } else {
                printf("The email '%s' is a valid e-mail addressn",email);
            }
            break;
        }
    }

这是我正在谈论的片段.

我应该编写自己的代码进行一次检查吗？如果是的话,你能否就这方面给我一些指示？
谢谢.

编辑：非常感谢您的回复,我确实了解了我的代码中的错误,希望我从中学习.
再次感谢！

编辑：2：我想再次感谢你的回复,他们给了我极大的帮助,我相信我写了更好的代码

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

for ( i = 0; email[i] != ''; 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 invalidn",email);
    }
else
    {
    printf("'%s' is a valid email addressn",email);
    }

我觉得这是更优雅和更简单的代码,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,易于阅读.
反正我还能提高吗？
(电子邮件检查仅用于练习,不要介意！)
?非常感谢大家！

解决方法

您的代码确实存在多个问题：

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,email);
        } else {
            printf("The email '%s' is a valid e-mail addressn",email);
        }
        break;  // you always break from the loop,why have a loop at all?
    }
}

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

还要注意更多问题不被注意：

>应该有一个@在场
>不应该有任何空格
>更一般地说,地址中允许的字符是有限的.

有些测试似乎有些过分：为什么拒绝..在@之前,为什么拒绝尾随.之前 @？

这是一个更有效的版本：

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,email);
} else {
    printf("The email '%s' is a valid e-mail addressn",email);
}

（编辑：李大同）

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