blocks|key|292377|text|简而言之，不是。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|292378|长答案是，memset设置字节并对字符起作用，因为它们是单字节，但整数不是。|offset|length|style|CODE|292379|entityMap^0|0|5|6|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

Short answer, NO. 

Long answer, <code>memset</code> sets bytes and works for characters because they are single bytes, but integers are not.

blocks|key|292410|text|memset的第三个参数是字节大小。因此您应该设置arr[15]总字节大小|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|292411|memset(arr,+1,+sizeof(arr));|code-block|syntax|javascript|292412|但是，您可能希望将值1设置为arr中的整个元素。那么你最好在循环中设置。|292413|for+(i+=+0;+i+<+sizeof(arr)/sizeof(arr[0]);+i%2B%2B)+{
++++arr[i]+=+1;
}|292414|因为memset()在每个字节中设置1。所以这不是你所期望的。|292415|entityMap^0|P|7|0|0|0|0|2|8|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

The third argument of memset is byte size. So you should set total byte size of <code>arr[15]</code>

<pre><code>memset(arr, 1, sizeof(arr));
</code></pre>

However probably, you should want to set value 1 to whole elements in arr. Then you've better to set in the loop.

<pre><code>for (i = 0; i &lt; sizeof(arr)/sizeof(arr[0]); i++) {
 arr[i] = 1;
}
</code></pre>

Because <code>memset()</code> set 1 in each bytes. So it's not your expected.

blocks|key|292452|text|不，除非期望的目标值是0，否则您不能为此目的而移植使用memset。memset将目标内存区域视为字节数组，而不是int数组。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|292453|一个相当流行的用重复模式填充内存区域的方法实际上是基于memcpy的。它在很大程度上依赖于memcpy以正向拷贝数据的预期|292454|int+arr[15];

arr[0]+=+1;
memcpy(&arr[1],+&arr[0],+sizeof+arr+-+sizeof+*arr);|code-block|syntax|javascript|292455|当然，这是一个相当丑陋的技巧，因为当源和目标内存区域重叠时，标准memcpy的行为是未定义的。不过，您可以编写自己的memcpy版本，确保它以正向复制数据，并以上述方式使用。但这并不是真正值得的。只需使用一个简单的循环将数组的元素设置为所需的值。|292456|entityMap^0|B|1|R|6|Y|6|1L|3|0|R|6|19|6|0|0|W|6|1M|6|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|P|$]]

No, you can't [portably] use <code>memset</code> for that purpose, unless the desired target value is <code>0</code>. <code>memset</code> treats the target memory region as an array of bytes, not an array of <code>int</code>s.

A fairly popular hack for filling a memory region with a repetitive pattern is actually based on <code>memcpy</code>. It critically relies on the expectation that <code>memcpy</code> copies data in forward direction

<pre><code>int arr[15];

arr[0] = 1;
memcpy(&amp;arr[1], &amp;arr[0], sizeof arr - sizeof *arr);
</code></pre>

This is, of course, a pretty ugly hack, since the behavior of standard <code>memcpy</code> is undefined when the source and destination memory regions overlap. You can write your own version of <code>memcpy</code> though, making sure it copies data in forward direction, and use in the above fashion. But it is not really worth it. Just use a simple cycle to set the elements of your array to the desired value.

blocks|key|292493|text|Memset为具有1个字节的数据类型设置值，但整数具有4个字节或更多字节，因此它将不起作用，您将得到无用的值。它主要用于处理char和string类型。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|292494|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Memset sets values for data types having 1 byte but integers have 4 bytes or more , so it won't work and you'll get garbage values.
It's mostly used when you are working with char and string types.

blocks|key|75282|text|在Linux、OSX和其他类UNIX操作系统上，wchar_t是32位的，您可以使用wmemset()而不是memset()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|75283|#include<wchar.h>
...
int+arr[15];
wmemset(+arr,+1,+6+);|code-block|syntax|javascript|75284|请注意，MS-Windows上的wchar_t是16位的，所以这个技巧可能不起作用。|75285|entityMap^0|O|7|16|9|1I|8|0|0|G|7|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

On Linux, OSX and other UNIX like operating systems where <code>wchar_t</code> is 32 bits and you can use <code>wmemset()</code> instead of <code>memset()</code>.

<pre><code>#include&lt;wchar.h&gt;
...
int arr[15];
wmemset( arr, 1, 6 );
</code></pre>

Note that <code>wchar_t</code> on MS-Windows is 16 bits so this trick may not work.

blocks|key|75299|text|我尝试了下面的程序，似乎您可以使用memset()来初始化您的数组，其中只有-1和0|type|unstyled|depth|inlineStyleRanges|entityRanges|data|75300|#include<stdio.h>
#include<string.h>

void+printArray(int+arr[],+int+len)
{
++++++++int+i=0;
++++for(i=0;+i<len;+i%2B%2B)
++++{
++++++++printf("%25d+",+arr[i]);
++++}
++++puts("");
}

int+main()
{
++++int+arrLen+=+15;
++++int+totalNoOfElementsToBeInitialized+=+6;

++++int+arr[arrLen];
++++printArray(arr,+arrLen);
++++memset(arr,+-1,+totalNoOfElementsToBeInitialized*sizeof(arr[0]));
++++printArray(arr,+arrLen);
++++memset(arr,+0,+totalNoOfElementsToBeInitialized*sizeof(arr[0]));
++++printArray(arr,+arrLen);
++++memset(arr,+1,+totalNoOfElementsToBeInitialized*sizeof(arr[0]));
++++printArray(arr,+arrLen);
++++memset(arr,+2,+totalNoOfElementsToBeInitialized*sizeof(arr[0]));
++++printArray(arr,+arrLen);
++++memset(arr,+-2,+totalNoOfElementsToBeInitialized*sizeof(arr[0]));
++++printArray(arr,+arrLen);
++++return+0;
}|code-block|syntax|javascript|75301|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|$]]

I tried the following program and it appears that you can initialize your array using memset() with -1 and 0 only 

<pre><code>#include&lt;stdio.h&gt;
#include&lt;string.h&gt;

void printArray(int arr[], int len)
{
 int i=0;
 for(i=0; i&lt;len; i++)
 {
 printf("%d ", arr[i]);
 }
 puts("");
}

int main()
{
 int arrLen = 15;
 int totalNoOfElementsToBeInitialized = 6;

 int arr[arrLen];
 printArray(arr, arrLen);
 memset(arr, -1, totalNoOfElementsToBeInitialized*sizeof(arr[0]));
 printArray(arr, arrLen);
 memset(arr, 0, totalNoOfElementsToBeInitialized*sizeof(arr[0]));
 printArray(arr, arrLen);
 memset(arr, 1, totalNoOfElementsToBeInitialized*sizeof(arr[0]));
 printArray(arr, arrLen);
 memset(arr, 2, totalNoOfElementsToBeInitialized*sizeof(arr[0]));
 printArray(arr, arrLen);
 memset(arr, -2, totalNoOfElementsToBeInitialized*sizeof(arr[0]));
 printArray(arr, arrLen);
 return 0;
}
</code></pre>

blocks|key|292557|text|实际上，使用memset_pattern4是可能的(它一次设置4位)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|292558|memset_pattern4(your_array,+your_number+,+sizeof(youre_array));|code-block|syntax|javascript|292559|entityMap^0|6|F|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Actually it is possible with <code>memset_pattern4</code> (it sets 4 bits at a time).
<pre><code>memset_pattern4(your_array, your_number , sizeof(youre_array));
</code></pre>

<pre><code>char str[] = "beautiful earth";
memset(str, '*', 6);
printf("%s", str);

Output:
******ful earth
</code></pre>

Like the above use of memset, can we initialize only a few integer array index values to 1 as given below?

<pre><code>int arr[15];
memset(arr, 1, 6);
</code></pre>

Using memset for integer array in C

Windows

Linux

char str[] = "beautiful earth";memset(str, '*', 6);printf("%s", str);Output:******ful earth就像上面memset的用法一样，我们可以像下面给出的那样，只将几个整数数组索引值初始化为1吗？int arr[15];memset(arr...

问在C++中使用memset处理整数数组
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