blocks|key|2675969|text|这就是我从你的问题中解释出来的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2675970|int+Array[20];
int+*p=Array;
int+no,i,len;
char+ch='+';+
len=sizeof(Array)/sizeof(int)
printf("Enter+no+of+elements+");+
scanf("%25d",&no);
for(i=0;i<no;i%2B%2B)
++++scanf("%25d",p%2Bi);+++
for(i=no;i<len;i%2B%2B)
++++p[i]=(int+)ch;|code-block|syntax|javascript|2675971|希望这能有所帮助。|2675972|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 what I interpreted from your question

<pre><code>int Array[20];
int *p=Array;
int no,i,len;
char ch=' '; 
len=sizeof(Array)/sizeof(int)
printf("Enter no of elements "); 
scanf("%d",&amp;no);
for(i=0;i&lt;no;i++)
 scanf("%d",p+i); 
for(i=no;i&lt;len;i++)
 p[i]=(int )ch;
</code></pre>

Hope this helps.

blocks|key|2676001|text|您可以使用memset(3)用空格填充内存区域：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2676002|size_t+total_size+=+get_total_size();+//+total+size+of+array,+in+bytes
size_t+len+=+get_len();+//+length+of+content,+in+bytes,+<=+total_size
assert(len+<=+total_size);
char+*array+=+malloc(total_size);
//+...+fill+the+first+len+bytes+with+your+data
memset(&array[len],+'+',+total_size+-+len);+//+and+the+rest+to+'+'+chars|code-block|syntax|javascript|2676003|不过，这种方法也有一些问题。首先，除非仔细检查该len+<+total_size，否则很容易受到缓冲区溢出的影响。第二，听起来像是要将它用作字符串，在这种情况下，您需要注意保留一个拖尾的空'\0‘字符。|offset|length|style|CODE|2676004|entityMap^0|0|0|O|G|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|$]]

You can use memset(3) to fill an area of memory with spaces:

<pre><code>size_t total_size = get_total_size(); // total size of array, in bytes
size_t len = get_len(); // length of content, in bytes, &lt;= total_size
assert(len &lt;= total_size);
char *array = malloc(total_size);
// ... fill the first len bytes with your data
memset(&amp;array[len], ' ', total_size - len); // and the rest to ' ' chars
</code></pre>

There are a couple problems with an approach like this though. The first is that you're vulnerable to buffer overflows unless you carefully check that <code>len &lt; total_size</code>. Second, it sounds like you're going to use this as a string, in which case you'll want to take care to leave a trailing null '\0' character.

blocks|key|3034093|text|我认为最干净的方法是在分配数组并将ptr分配到buf的开头之后。只需使用'+'填充所有内容，使用memset()即可。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3034094|+char+buf[50];
+char+*ptr+=+buf;
+memset+(ptr,+'+',+sizeof(buf));+//+like+this

+//+If+this+buffer+is+meant+to+be+printed+to+the+screen+or+used+as+a+string,
+//+it's+probably+better+to+write+the+end-of-string+character+at+the+last+
+//+position+of+the+buffer:
+buf[49]+=+0;

+//+operations+to+write+stuff+on+the+buffer|code-block|syntax|javascript|3034095|因此，在操作结束时，所有尚未使用的缓冲区的其余位置都将有'+'。|3034096|编辑：|BOLD|3034097|我一直在考虑您要求的内容，除非您有非常具体的理由来填充数组中的空空间，这通常不是我们处理这种情况的方法(请参阅下面的评论)。|3034098|char*+tmp_string+=+"**Copy+Me**";
printf("Size+of+string:%25d\n",+strlen(tmp_string));

char+buf[50];
printf("Size+of+buf:%25d\n",+sizeof(buf));
memset(buf,+0,+sizeof(buf));+//+cleaning+the+entire+buffer

char+*ptr+=+buf;
strncpy(ptr,+tmp_string,+strlen(tmp_string));
ptr+%2B=+strlen(tmp_string);
*ptr+=+'!';

printf("Size+of+buf+after+copy:+%25d\n",+strlen(buf));
printf("Result:+%25s\n",+buf);|3034099|输出：|3034100|Size+of+string:11
Size+of+buf:50
Size+of+buf+after+copy:+12
Result:+**Copy+Me**!|3034101|注意到，复制后buf的大小为12，这是因为我们在操作开始时对缓冲区进行了零化(即用\0填充缓冲区)。在缓冲区中\0的第一个符号时，printf()停止迭代数组和打印字符。|3034102|，但是，我可以想到一个很好的理由来解释为什么您不想要这种方法。下面是对上面代码的轻微修改。它用空空间填充缓冲区，并在缓冲区的最后一个位置添加字符串结束字符(\0)。注意到，最后字符串的大小不是12！它是49，因为您用有效字符填充了缓冲区。|3034103|char*+tmp_string+=+"**Copy+Me**";
printf("Size+of+string:%25d\n",+strlen(tmp_string));

char+buf[50];
printf("Size+of+buf:%25d\n",+sizeof(buf));
memset(buf,+'+',+sizeof(buf));+//+cleaning+the+entire+buffer+with+empty+spaces
buf[49]+=+0;+//+setting+a+\0+at+the+end+of+the+string

char+*ptr+=+buf;
strncpy(ptr,+tmp_string,+strlen(tmp_string));
ptr+%2B=+strlen(tmp_string);
*ptr+=+'!';

printf("Size+of+buf+after+copy:+%25d\n",+strlen(buf));

printf("Result:+%25s\n",+buf);|3034104|产出：|3034105|Size+of+string:11
Size+of+buf:50
Size+of+buf+after+copy:+49
Result:+**Copy+Me**!|3034106|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/clibrary/cstring/memset/^0|10|3|1C|8|0|0|0|S|3|0|0|3|0|0|0|0|3|0|0|0|3|15|2|1J|2|0|0|3|1B|3|2A|3|2T|2|26|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1F|8|@$9|1G|A|1H|B|C]]|D|@$9|1I|A|1J|1|1K]]|E|$]]|$1|F|3|G|5|H|7|1L|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1M|8|@$9|1N|A|1O|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1P|8|@$9|1Q|A|1R|B|O]]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1S|8|@]|D|@]|E|$]]|$1|R|3|S|5|H|7|1T|8|@]|D|@]|E|$I|J]]|$1|T|3|U|5|6|7|1U|8|@$9|1V|A|1W|B|O]]|D|@]|E|$]]|$1|V|3|W|5|H|7|1X|8|@]|D|@]|E|$I|J]]|$1|X|3|Y|5|6|7|1Y|8|@$9|1Z|A|20|B|O]|$9|21|A|22|B|C]|$9|23|A|24|B|C]]|D|@]|E|$]]|$1|Z|3|10|5|6|7|25|8|@$9|26|A|27|B|O]|$9|28|A|29|B|O]|$9|2A|A|2B|B|O]|$9|2C|A|2D|B|O]|$9|2E|A|2F|B|C]]|D|@]|E|$]]|$1|11|3|12|5|H|7|2G|8|@]|D|@]|E|$I|J]]|$1|13|3|14|5|6|7|2H|8|@]|D|@]|E|$]]|$1|15|3|16|5|H|7|2I|8|@]|D|@]|E|$I|J]]|$1|17|3|-4|5|6|7|2J|8|@]|D|@]|E|$]]]|18|$19|$5|1A|1B|1C|E|$1D|1E]]]]

I think that the cleanest way to do this is right after you allocate the array and assign ptr to the beginning of buf. Just fill everything with <code>' '</code> using <a href="http://www.cplusplus.com/reference/clibrary/cstring/memset/" rel="nofollow">memset()</a>.

<pre><code> char buf[50];
 char *ptr = buf;
 memset (ptr, ' ', sizeof(buf)); // like this

 // If this buffer is meant to be printed to the screen or used as a string,
 // it's probably better to write the end-of-string character at the last 
 // position of the buffer:
 buf[49] = 0;

 // operations to write stuff on the buffer
</code></pre>

So at the end of your operations, all the remaining locations of the buffer which haven't been used will have <code>' '</code>.

EDIT:

I've been thinking about what you asked and unless you have very specific reasons to fill the array with empty spaces that's not usually how we deal with this situation (check my comment below this answer).

<pre><code>char* tmp_string = "**Copy Me**";
printf("Size of string:%d\n", strlen(tmp_string));

char buf[50];
printf("Size of buf:%d\n", sizeof(buf));
memset(buf, 0, sizeof(buf)); // cleaning the entire buffer

char *ptr = buf;
strncpy(ptr, tmp_string, strlen(tmp_string));
ptr += strlen(tmp_string);
*ptr = '!';

printf("Size of buf after copy: %d\n", strlen(buf));
printf("Result: %s\n", buf);
</code></pre>

Outputs:

<pre><code>Size of string:11
Size of buf:50
Size of buf after copy: 12
Result: **Copy Me**!
</code></pre>

Notice that the size of the buf after copy is 12. That's because we have zeroed the buffer in the beginning of the operations (i.e. filled the buffer with <code>\0</code>). At the first sign of <code>\0</code> in the buffer, printf() stops iterating on the array and printing characters.

HOWEVER, I can think of one good reason of why you don't want this method. The following is a slight modification of the code above. It fills the buffer with empty spaces and adds the end-of-string character (<code>\0</code>) at the last position of the buffer. Notice that the size of the string in the end is not 12! It's 49 because you filled the buffer with valid characters.

<pre><code>char* tmp_string = "**Copy Me**";
printf("Size of string:%d\n", strlen(tmp_string));

char buf[50];
printf("Size of buf:%d\n", sizeof(buf));
memset(buf, ' ', sizeof(buf)); // cleaning the entire buffer with empty spaces
buf[49] = 0; // setting a \0 at the end of the string

char *ptr = buf;
strncpy(ptr, tmp_string, strlen(tmp_string));
ptr += strlen(tmp_string);
*ptr = '!';

printf("Size of buf after copy: %d\n", strlen(buf));

printf("Result: %s\n", buf);
</code></pre>

Outputs:

<pre><code>Size of string:11
Size of buf:50
Size of buf after copy: 49
Result: **Copy Me**!
</code></pre>

I have a pointer pointing to an array which is filled dynamically at run time. After collecting and storing the array, I want all the remaining buffer locations of the array to be filled with empty space. How can I do that?

<hr>

From some comments:

<blockquote>
 Here is what I have:
</blockquote>

<pre><code> char buf[50];
 char *ptr = buf;
 strncpy(ptr, info.a, strlen(info.a));
 ptr += strlen(info.a);
 strncpy(ptr, info.b, strlen(info.b));
 ptr += strlen(info.b);
 strncpy(ptr, info.c, strlen(info.c));
 ptr += strlen(info.c);
</code></pre>

<blockquote>
 How do I fill the remaining pointer locations with <code>' '</code>?
</blockquote>

dynamic pointer value allocation

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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我有一个指针，指向在运行时动态填充的数组。在收集和存储数组之后，我希望数组的所有剩余缓冲区位置都填充空空间。我怎么能这么做？根据一些评论：以下是我所拥有的： char buf[50]; char *ptr = buf; strncpy(ptr, info.a, strlen(info.a)); ptr += strle...

问动态指针值分配
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