在C中如何strcmp仅开头2个字符然后连接？

发布于 2024-11-07 08:03:56 字数 215 浏览 1 评论 0原文

在C中如何strcmp仅开头2个字符？然后与另一个字符串连接？像这样的东西：

char s[10];
scanf("%s",s);

/* if i input "cs332" or "cs234", anything start with cs */

if (strcmp("cs",???)==0)
    strcat(s,"by professor");

原文

In C how do I strcmp just the beginning 2 characters? Then concatenate with another string? Something like this:

char s[10];
scanf("%s",s);

/* if i input "cs332" or "cs234", anything start with cs */

if (strcmp("cs",???)==0)
    strcat(s,"by professor");

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我早已燃尽 2024-11-14 08:03:56

if (strncmp("cs",???, 2)==0) strcat(s,"by professor");

使用strncmp

if (strncmp("cs",???, 2)==0) strcat(s,"by professor");

Use strncmp

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忘你却要生生世世 2024-11-14 08:03:56

您正在寻找 strncmp 函数，该函数在功能上与 strcmp 相同，但限制检查的字符数。因此，您将使用长度为 2 的比较字符串和 "cs"。但是，这里还有一些其他问题。

首先，你的缓冲区不够大。当您将文本“by Professor”附加到十个字符的缓冲区时，没有任何字符串适合该缓冲区。

其次，健壮的代码永远不会将scanf与无界字符串格式说明符一起使用：这会导致缓冲区溢出问题。 scanf 系列适用于格式化输入，并且未格式化比用户输入多一点:-)

如果您想要一个强大的输入解决方案，请参阅我之前的答案之一。

第三，您应该始终假设连接字符串可能会溢出缓冲区，并引入代码来防止这种情况发生。您需要添加：

用户输入的字符串的当前长度。
附加字符串的长度（“by Professor”）。
还有一个用于空终止符。

并确保缓冲区足够大。

我将使用的方法是拥有一个（例如）200 字节的缓冲区，使用链接答案中的 getLine() （在下面复制以使该答案独立）且大小足够小（比如 100），那么您可以放心，附加“by Professor”不会溢出缓冲区。

功能：

#include <stdio.h>
#include <string.h>

#define OK       0
#define NO_INPUT 1
#define TOO_LONG 2
static int getLine (char *prmpt, char *buff, size_t sz) {
    int ch, extra;

    // Get line with buffer overrun protection.
    if (prmpt != NULL) {
        printf ("%s", prmpt);
        fflush (stdout);
    }
    if (fgets (buff, sz, stdin) == NULL)
        return NO_INPUT;

    // If it was too long, there'll be no newline. In that case, we flush
    // to end of line so that excess doesn't affect the next call.
    if (buff[strlen(buff)-1] != '\n') {
        extra = 0;
        while (((ch = getchar()) != '\n') && (ch != EOF))
            extra = 1;
        return (extra == 1) ? TOO_LONG : OK;
    }

    // Otherwise remove newline and give string back to caller.
    buff[strlen(buff)-1] = '\0';
    return OK;
}

测试代码：

// Test program for getLine().

int main (void) {
    int rc;
    char buff[10];

    rc = getLine ("Enter string> ", buff, sizeof(buff));
    if (rc == NO_INPUT) {
        // Extra NL since my system doesn't output that on EOF.
        printf ("\nNo input\n");
        return 1;
    }

    if (rc == TOO_LONG) {
        printf ("Input too long [%s]\n", buff);
        return 1;
    }

    printf ("OK [%s]\n", buff);

    return 0;
}

You are looking for the strncmp function which is functionally identical to strcmp but limits the number of characters checked. So you would use it with a length of two and the comparison string of "cs". But, you have a few other problems here.

First, your buffer is not big enough. There is no string that will fit into a ten-character buffer when you append the text "by professor" to it.

Secondly, robust code will never use scanf with an unbounded-string format specifier: that's asking for a buffer overflow problem. The scanf family is meant for formatted input and there is little more unformatted than user input :-)

If you want a robust input solution, see one of my previous answers.

Thirdly, you should always assume that concatenating a string may overflow your buffer, and introduce code to prevent this. You need to add up:

the current length of the string, input by the user.
the length of the appending string ("by professor").
one more for the null terminator.

and ensure the buffer is big enough.

The method I would use would be to have a (for example) 200-byte buffer, use getLine() from the linked answer (reproduced below to make this answer self-contained) with a sufficiently smaller size (say 100), then you can be assured that appending "by professor" will not overflow the buffer.

Function:

#include <stdio.h>
#include <string.h>

#define OK       0
#define NO_INPUT 1
#define TOO_LONG 2
static int getLine (char *prmpt, char *buff, size_t sz) {
    int ch, extra;

    // Get line with buffer overrun protection.
    if (prmpt != NULL) {
        printf ("%s", prmpt);
        fflush (stdout);
    }
    if (fgets (buff, sz, stdin) == NULL)
        return NO_INPUT;

    // If it was too long, there'll be no newline. In that case, we flush
    // to end of line so that excess doesn't affect the next call.
    if (buff[strlen(buff)-1] != '\n') {
        extra = 0;
        while (((ch = getchar()) != '\n') && (ch != EOF))
            extra = 1;
        return (extra == 1) ? TOO_LONG : OK;
    }

    // Otherwise remove newline and give string back to caller.
    buff[strlen(buff)-1] = '\0';
    return OK;
}

Test code:

// Test program for getLine().

int main (void) {
    int rc;
    char buff[10];

    rc = getLine ("Enter string> ", buff, sizeof(buff));
    if (rc == NO_INPUT) {
        // Extra NL since my system doesn't output that on EOF.
        printf ("\nNo input\n");
        return 1;
    }

    if (rc == TOO_LONG) {
        printf ("Input too long [%s]\n", buff);
        return 1;
    }

    printf ("OK [%s]\n", buff);

    return 0;
}

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空袭的梦i 2024-11-14 08:03:56

为什么不直接比较字符而不是调用strcmp？

例如

if(s[0]=='c' && s[1]=='s'){
...
}

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廻憶裏菂餘溫 2024-11-14 08:03:56

有几种方法可以做到这一点。

字符串比较：

if ( s[0] == 'c' && s[1] == 's' )

这是一种简单的方法，因为您无法轻松地将其扩展到稍长的代码（例如长度为 3/4 个字符）。

我想您已经知道应该使用 strncmp() 对吗？

字符串连接

不要使用 strcat。真的。如果您连接两个长度大于 s （目标）大小的字符串，您就会遇到麻烦。考虑使用 snprint() 代替，例如这个：

char str[80];
snprintf(str, 80, "%s by professor", s);

或者，您可以使用 strncat() 正如 Heath 指出的：

char s[80];
strncat(s, " by professor", 80);

Several ways to do this.

String comparison:

if ( s[0] == 'c' && s[1] == 's' )

Is the naive way, as you can't expand this easily to slightly longer codes (say 3/4 characters in length).

I guess you've gathered you should be using strncmp() right?

String Concaternation

Don't use strcat. Really. If you concatenate two strings whose length is greater than the size of s (the destination) you're in trouble. Consider using snprint() instead, like this:

char str[80];
snprintf(str, 80, "%s by professor", s);

Or, you could use strncat() as Heath points out:

char s[80];
strncat(s, " by professor", 80);

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绝對不後悔。 2024-11-14 08:03:56

您可以使用 strncmp。

编辑：

strcat(s,"by professor");

// s is an array of 10 characters. You need to make sure s is big enough  
// to hold the string that needs to be concatenated + to have a terminating 
// character '\0'.

You can use strncmp.

Edit:

strcat(s,"by professor");

// s is an array of 10 characters. You need to make sure s is big enough  
// to hold the string that needs to be concatenated + to have a terminating 
// character '\0'.

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全部不再 2024-11-14 08:03:56

是的，如上所述，strcmp 是首选方法。这里只是用不同的方法来做同样的事情。

#define CS 29539
char s[80];

scanf("%60s", s);
if( *(short *)s == CS )
    if( strlcat(s, " by professor", sizeof(s)) >= sizeof(s) )
        fprintf(stderr, "WARNING: truncation detected: %s", s);

Yes, as said, strcmp is the preferred method. Here's just a different way to do the same.

#define CS 29539
char s[80];

scanf("%60s", s);
if( *(short *)s == CS )
    if( strlcat(s, " by professor", sizeof(s)) >= sizeof(s) )
        fprintf(stderr, "WARNING: truncation detected: %s", s);

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