blocks|key|10611|text|char+result[50];

strcpy(result,+ln);
strcat(result,+",+");
strcat(result,+fn);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|10612|他说得对，你说得太多了(而且浪费了太多的记忆)。|unstyled|10613|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>char result[50];

strcpy(result, ln);
strcat(result, ", ");
strcat(result, fn);
</code></pre>

He's right, you used way too many statements (and wasted too much memory doing it).

blocks|key|3650537|text|假设你知道字符串的长度，为什么不知道呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3650538|char+result[50];

sprintf(result,"%25s,+%25s",ln,+fn);|code-block|syntax|javascript|3650539|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Assuming you know length of strings, why not do

<pre><code>char result[50];

sprintf(result,"%s, %s",ln, fn);
</code></pre>

blocks|key|1792133|text|这就是你所需要的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1792134|strcpy+(str1,ln);++++//+Copy+the+last+name+to+str1
strcat+(str1,",+");++//+Now+append+",+"+to+the+last+name+in+str1
strcat+(str1,fn);++++//+Lastly,+append+the+first+name+to+str1|code-block|syntax|javascript|1792135|您必须确保str1足够大，足以容纳所有这些。一个由13个字符组成的数组可能不够。|offset|length|style|CODE|1792136|entityMap^0|0|0|5|4|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

This is all you need:

<pre><code>strcpy (str1,ln); // Copy the last name to str1
strcat (str1,", "); // Now append ", " to the last name in str1
strcat (str1,fn); // Lastly, append the first name to str1
</code></pre>

You must make sure that <code>str1</code> is large enough to hold all of that. An array of 13 characters may not be sufficient.

blocks|key|10517|text|您的strcpy调用实际上是不必要的，因为您已经在变量中有了名字和姓氏。您可以创建一个足够大的缓冲区来容纳整个字符串，然后使用strcat创建最终的“姓氏，名”字符串。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|10518|entityMap^0|2|6|1R|6|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

Your <code>strcpy</code> calls are really unnecessary since you already have first name and last name in variables. You can just create a buffer big enough to hold the entire string then use <code>strcat</code> to create the final "lastname, firstname" string.

blocks|key|3650563|text|我很惊讶你的老师接受了你的答案。您正在将结果累积到char数组str1中，您已经声明该数组只有11个字符。除非是strlen(ln)%2Bstrlen(fn)<=8，否则您将溢出为结果str1分配的空间。在过去糟糕的日子里，C程序员只会简单地分配一个看起来足够大的结果数组，而不会费心去检查。标准C库函数sprintf在stdio.h中是为这项工作设计的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3650564|#include+<stdio.h>
...fn,+ln+defined...
char+result[80];++/*+80+characters+is+probably+enough+*/
sprintf(result,+"%25s,+%25s",+ln,+fn);|code-block|syntax|javascript|3650565|现代C程序员绝不会认为ln和fn足够短，以至于不管时间多长，结果数组都不会溢出。安全、现代的替代方法是用sprintf替换snprintf调用。|3650566|snprintf(result,+80,+"%25s,+%25s",+ln,+fn);|3650567|您还可以使用strcpy/strcat系列函数，它们位于string.h中。(这是你的作业中指定的吗？)安全、现代的等价物是strncpy和strncat，它们允许您指定要复制的最大字符数。使用这些功能的快速、肮脏但安全的解决方案将是：|3650568|char+result[80];
result[0]+=+'\0';+++++/*+initialize+result+to+""+*/
strncat(result,+ln,+40);
strcat(result,+",+")++/*+you+know+this+adds+exactly+two+characters+*/
strncat(result,+fn,+38);|3650569|strncpy调用是危险的，因为它可能不会在result中留下一个以空结尾的字符串，这将导致后续strcat失败。这并不比您所做的更少，但是仅使用这些string.h函数很难做得更好，而且无论给出什么ln和fn，您都不能溢出结果缓冲区。|3650570|entityMap^0|1K|O|2I|4|46|7|4E|7|0|0|B|2|E|2|1G|7|1P|8|0|0|6|6|D|6|S|8|1Q|7|1Y|7|0|0|0|7|M|6|1C|6|23|8|2S|2|2V|2|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1R|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1S|8|@$9|1T|A|1U|B|C]|$9|1V|A|1W|B|C]|$9|1X|A|1Y|B|C]|$9|1Z|A|20|B|C]|$9|21|A|22|B|C]|$9|23|A|24|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|25|8|@]|D|@]|E|$]]]|V|$]]

I'm surprised your instructor accepted your answer. You are accumulating your result in the char array str1, which you've declared has only 11 characters. Unless <code>strlen(ln)+strlen(fn)&lt;=8</code>, you will overflow the space you've allocated for the result <code>str1</code>. In the bad old days, a C programmer would have simply allocated a result array that seemed big enough, and not bothered to check. The standard C library function <code>sprintf</code> in <code>stdio.h</code> is designed for this job:

<pre><code>#include &lt;stdio.h&gt;
...fn, ln defined...
char result[80]; /* 80 characters is probably enough */
sprintf(result, "%s, %s", ln, fn);
</code></pre>

A modern C programmer would never assume that <code>ln</code> and <code>fn</code> were short enough not to overflow a result array, no matter how long. The safe, modern alternative would be to replace the <code>sprintf</code> call by <code>snprintf</code>:

<pre><code>snprintf(result, 80, "%s, %s", ln, fn);
</code></pre>

You can also use the <code>strcpy</code>/<code>strcat</code> family of functions, which are in <code>string.h</code>. (Was this specified in your assignment?) The safe, modern equivalents are <code>strncpy</code> and <code>strncat</code>, which let you specify a maximum number of characters to copy. A quick and dirty but safe solution using these functions would be:

<pre><code>char result[80];
result[0] = '\0'; /* initialize result to "" */
strncat(result, ln, 40);
strcat(result, ", ") /* you know this adds exactly two characters */
strncat(result, fn, 38);
</code></pre>

The <code>strncpy</code> call is dangerous because it may not leave a null-terminated string in <code>result</code>, which will cause a subsequent <code>strcat</code> to fail. This isn't many fewer lines than what you did, but using just these <code>string.h</code> functions, it's tough to do much better and be guaranteed you can't overflow the result buffer no matter what <code>ln</code> and <code>fn</code> are given.

My homework requires that my name be displayed such that it looks like this: 'Lastname, Firstname'. Last name then [comma space] firstname. While not moving over the rest of the text after that name. This is my code:

<pre><code> char str1[11];
 char str2[3];
 char str3[16];
 strcpy (str1,fn);
 strcpy (str2,", ");
 strcpy (str3,ln);
 strncat (str1, str2, 14);
 strncat (str1, str3, 31);
</code></pre>

My teacher said that I did what he wanted, but he doesn't like how many lines of code I used and said I am doing extra work than I need.

Variables: <code>ln = last name</code>, <code>fn = first name</code> I made str2 for the ', ' comma space.

What is it that he wants me to do?

strncpy and strcat usage

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我的作业要求显示我的名字，使其看起来如下：'Lastname，Firstname‘。姓，然后是逗号，空格名。虽然没有移动的其余文本后，该名称。这是我的密码：  char str1[11];  char str2[3];  char str3[16];  strcpy (str1,fn);  strcpy (str2,", ");  strcpy (str3,ln);  strncat (str1,

问strncpy和strcat用法
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