blocks|key|1539854|text|这是如此简单，似乎我一定遗漏了什么。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1539855|lengthName+=+strlen(student[i].name);
while+(lengthName+<+10)
++student[i].name[lengthName%2B%2B]+=+'-';
student[i].name[lengthName]+=+'\0';|code-block|syntax|javascript|1539856|也许你被C+#(假定)拥有一流的字符串类型弄糊涂了?在C中没有这样的东西，只有字符的裸数组；这既乏味(你必须自己做所有的内存管理)又解放(正如你在上面看到的)。|1539857|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This is so simple that it seems like I must be missing something.

<pre><code>lengthName = strlen(student[i].name);
while (lengthName &lt; 10)
 student[i].name[lengthName++] = '-';
student[i].name[lengthName] = '\0';
</code></pre>

Perhaps you are confused by C#'s (presumed) possession of a first-class string type? No such thing in C, only bare arrays of characters; which is at once tedious (you have to do all the memory management yourself) and liberating (as you see above).

blocks|key|4040763|text|这个怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4040764|//+Pre-fills+with+hyphens.
memset(student[i].name,+'-',+sizeof(student[i].name)+-+1);
scanf("%2510s",+student[i].name);|code-block|syntax|javascript|4040765|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|$]]

How about this?

<pre><code>// Pre-fills with hyphens.
memset(student[i].name, '-', sizeof(student[i].name) - 1);
scanf("%10s", student[i].name);
</code></pre>

blocks|key|2444580|text|while(lengthName+<+10)++
{++
++++student[i]->name[lengthName%2B%2B]+=+'-';+//+add+this+line++
}++
student[i]->name[lengthName]+=+0;++//+and+this+line+|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2444581|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>while(lengthName &lt; 10) 
{ 
 student[i]-&gt;name[lengthName++] = '-'; // add this line 
} 
student[i]-&gt;name[lengthName] = 0; // and this line 
</code></pre>

blocks|key|376549|text|尝试以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|376550|while+(lengthName+<+10)
{
++++(student%2Bi)->name[lengthName%2B%2B]+=+'-';
}|code-block|syntax|javascript|376551|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|$]]

Try the following:

<pre><code>while (lengthName &lt; 10)
{
 (student+i)-&gt;name[lengthName++] = '-';
}
</code></pre>

blocks|key|4040812|text|您的char+name[10+%2B+1];变量显然有11个字节长。添加超过11个字节的任何内容都将导致缓冲区溢出。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4040813|这是缓冲区溢出可能发生的地方：|4040814|scanf("%25s",&(student%2Bi)->name);|4040815|您的连字符功能可以在您修复此可能的缓冲区溢出问题后出现。|4040816|修复缓冲区溢出|4040817|解决此类问题的一种简单方法是让scanf本身限制要读取的字符数：|4040818|//+read+only+10+chars,+since+`name`+can+only+
//+hold+10+chars,+plus+the+extra+NULL+char
scanf("%2510s",+&(student%2Bi)->name);|code-block|syntax|javascript|4040819|预填入连字符|4040820|最后，在您解决了内存问题之后，您可以继续并在用实际的学生数据覆盖字符串之前预先填充字符串。|4040821|//+pre-fill+your+data+with+hypens.
memset((student%2Bi)->name,+'-',+sizeof((student%2Bi)->name)+-+1);

//+read+only+10+chars,+since+`name`+can+only+
//+hold+10+chars,+plus+the+extra+NULL+char
scanf("%2510s",+&(student%2Bi)->name);|4040822|entityMap^0|2|I|0|0|0|V|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|16|8|@$9|17|A|18|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|19|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|1A|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|1B|8|@]|D|@]|E|$]]|$1|P|3|Q|5|R|7|1C|8|@]|D|@]|E|$S|T]]|$1|U|3|V|5|6|7|1D|8|@]|D|@]|E|$]]|$1|W|3|X|5|6|7|1E|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|R|7|1F|8|@]|D|@]|E|$S|T]]|$1|10|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|11|$]]

Your <code>char name[10 + 1];</code> variable is obviously 11 bytes long. Adding anything more than 11 bytes will cause a buffer overflow.

This is where the buffer overflow might happen: 

<code>scanf("%s",&amp;(student+i)-&gt;name);</code>

Your hyphens feature can come after you fixed this possible buffer overflow issue. 

<h2>Fixing the buffer overflow</h2>

One easy way to fix this sort of problem is to have scanf itself limit the number of chars to read: 

<pre><code>// read only 10 chars, since `name` can only 
// hold 10 chars, plus the extra NULL char
scanf("%10s", &amp;(student+i)-&gt;name);
</code></pre>

<h2>Pre-filling with hyphens</h2>

Finally, after you took care of your memory issue, you can go ahead and pre-fill your string before overwriting it with the actual student data.

<pre><code>// pre-fill your data with hypens.
memset((student+i)-&gt;name, '-', sizeof((student+i)-&gt;name) - 1);

// read only 10 chars, since `name` can only 
// hold 10 chars, plus the extra NULL char
scanf("%10s", &amp;(student+i)-&gt;name);
</code></pre>

blocks|key|1539904|text|@kotlinski想法的修改版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1539905|#define+NAME_SIZE+10

fgets(student[i].name,+NAME_SIZE,+stdin);
lengthName+=+strlen(student[i].name);
if+(lengthName+>+0+&&+student[i].name[lengthName-1]+==+'\n')+{
++++--lengthName;
}
if+(lengthName+<+NAME_SIZE)+{
++++memset(student[i].name+%2B+lengthName,+'-',+NAME_SIZE+-+lengthName);
++++student[i].name[NAME_SIZE]+=+'\0';
}|code-block|syntax|javascript|1539906|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|$]]

Modified version of @kotlinski's idea:

<pre><code>#define NAME_SIZE 10

fgets(student[i].name, NAME_SIZE, stdin);
lengthName = strlen(student[i].name);
if (lengthName &gt; 0 &amp;&amp; student[i].name[lengthName-1] == '\n') {
 --lengthName;
}
if (lengthName &lt; NAME_SIZE) {
 memset(student[i].name + lengthName, '-', NAME_SIZE - lengthName);
 student[i].name[NAME_SIZE] = '\0';
}
</code></pre>

blocks|key|376742|text|使用fgets()读取输入，如果有空间，它会将换行符保存到缓冲区；为此，您必须添加一些特殊处理。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|376743|请注意，对于输入流，fflush()的行为是未定义的；您不能通过调用fflush(stdin);来清除不需要的输入。|376744|然后，需要将其余字符设置为连字符。|376745|if+(fgets(students[i].name,+sizeof+students[i].name,+stdin)+!=+NULL)
{
++size_t+max+=+sizeof+students[i].name+-+1;
++size_t+len+=+0;

++char+*newline+=+strchr(students[i].name,+'\n');
++if+(newline)
++++*newline+=+0;+//+clear+out+newline
++else
++{
++++while+(fgetc(stdin)+!=+'\n')+//+clear+out+remaining+input;+
+++++++++++++++++++++++++++++++++//+name+potentially+truncated
++++++;
++}

++//+pad+remainder+with+hyphens
++len+=+strlen(students[i].name);
++while+(len+<+max)
++++students[i].name[len%2B%2B]+=+'-';
++students[i].name[max]+=+0;
}|code-block|syntax|javascript|376746|entityMap^0|2|7|0|A|8|Y|E|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|Y|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|Z|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

Use <code>fgets()</code> to read the input, with the understanding that it will save the newline character to the buffer if there's room; you'll have to add some special handling for that. 

Note that the behavior of <code>fflush()</code> is not defined for input streams; you cannot clear unwanted input by calling <code>fflush(stdin);</code>. 

Then it's a matter of setting the remaining characters to hyphens.

<pre><code>if (fgets(students[i].name, sizeof students[i].name, stdin) != NULL)
{
 size_t max = sizeof students[i].name - 1;
 size_t len = 0;

 char *newline = strchr(students[i].name, '\n');
 if (newline)
 *newline = 0; // clear out newline
 else
 {
 while (fgetc(stdin) != '\n') // clear out remaining input; 
 // name potentially truncated
 ;
 }

 // pad remainder with hyphens
 len = strlen(students[i].name);
 while (len &lt; max)
 students[i].name[len++] = '-';
 students[i].name[max] = 0;
}
</code></pre>

I have the following part of code :

<pre><code>for(int i=0;i&lt;n;i++)
{
 printf("Student %d\n",i+1);
 printf("Enter name : ");
 scanf("%s",&amp;(student+i)-&gt;name);
 fflush(stdin);

 lengthName = strlen((student+i)-&gt;name);
 while(lengthName !='\0')
 {

 }} 
</code></pre>

when the length is shorter than 10, it will add hyphens until reaching the maximum size.
Ex : John =>> 6 hyphens will be added

I know how to do it in csharp but can't figure it out in c. 
Could some of you give me some lights please?

PS : Oh yes the variable name is char name[10+1] and it a part of the structure called student.

How to add characters to reach the maximum size of a char[]

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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我有以下部分代码：for(int i=0;i<n;i++){ printf("Student %d\n",i+1); printf("Enter name : "); scanf("%s",&(student+i)->name); fflush(stdin); lengthName = strlen((student+i)->name); while(lengt

问如何添加字符以达到char[]的最大大小
EN

回答 7

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问如何添加字符以达到char[]的最大大小
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