blocks|key|1149437|text|char*+c+=+new+char[length]();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1149438|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>char* c = new char[length]();
</code></pre>

blocks|key|1364309|text|两种方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1364310|char+*c+=+new+char[length];
std::fill(c,+c+%2B+length,+INITIAL_VALUE);
//+just+this+once,+since+it's+char,+you+could+use+memset|code-block|syntax|javascript|1364311|或者：|1364312|std::vector<char>+c(length,+INITIAL_VALUE);|1364313|在我的第二种方法中，默认的第二个参数已经是0，所以在您的例子中它是不必要的：|1364314|std::vector<char>+c(length);|1364315|[编辑:去投票给弗雷德的答案，char*+c+=+new+char[length]();]|offset|length|style|CODE|1364316|entityMap^0|0|0|0|0|0|0|F|T|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@$Q|13|R|14|S|T]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

Two ways:

<pre><code>char *c = new char[length];
std::fill(c, c + length, INITIAL_VALUE);
// just this once, since it's char, you could use memset
</code></pre>

Or:

<pre><code>std::vector&lt;char&gt; c(length, INITIAL_VALUE);
</code></pre>

In my second way, the default second parameter is 0 already, so in your case it's unnecessary:

<pre><code>std::vector&lt;char&gt; c(length);
</code></pre>

[Edit: go vote for Fred's answer, <code>char* c = new char[length]();</code>]

blocks|key|1364073|text|也许可以使用std::fill_n()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1364074|char*+c+=+new+char[length];
std::fill_n(c,length,0);|code-block|syntax|javascript|1364075|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/algorithm/fill_n/^0|6|D|6|D|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Maybe use <a href="http://www.cplusplus.com/reference/algorithm/fill_n/" rel="noreferrer"><code>std::fill_n()</code></a>?

<pre><code>char* c = new char[length];
std::fill_n(c,length,0);
</code></pre>

blocks|key|1149224|text|C%2B%2B没有特定的特性来实现这一点。但是，如果您使用std::vector而不是数组(您可能应该这样做)，那么您可以指定一个值来初始化向量。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1149225|std::vector+<char>+v(+100,+42+);|code-block|syntax|javascript|1149226|创建一个大小为100的向量，所有值初始化为42。|1149227|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|$]]

C++ has no specific feature to do that. However, if you use a std::vector instead of an array (as you probably should do) then you can specify a value to initialise the vector with.

<pre><code>std::vector &lt;char&gt; v( 100, 42 );
</code></pre>

creates a vector of size 100 with all values initialised to 42.

blocks|key|1364510|text|既然有了c%2B%2B11，我们就可以使用list+initialization|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1364511|char*+c+=+new+char[length]{};|code-block|syntax|javascript|1364512|对于聚合类型，则会执行aggregate+initialization，其效果与char+c[2]+=+{};相同。|style|CODE|1364513|entityMap|0|LINK|mutability|MUTABLE|url|http://en.cppreference.com/w/cpp/language/list_initialization|1|http://en.cppreference.com/w/cpp/language/aggregate_initialization^0|H|J|0|0|0|14|F|B|O|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|11|8|@$A|12|B|13|K|L]]|9|@$A|14|B|15|1|16]]|C|$]]|$1|M|3|-4|5|6|7|17|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

Since c++11 we could use <a href="http://en.cppreference.com/w/cpp/language/list_initialization" rel="noreferrer">list initialization</a>:

<pre><code>char* c = new char[length]{};
</code></pre>

For an aggregate type, then <a href="http://en.cppreference.com/w/cpp/language/aggregate_initialization" rel="noreferrer">aggregate initialization</a> will be performed, which has the same effect like <code>char c[2] = {};</code>.

blocks|key|1364186|text|你不可能在一行代码中轻松做到这一点。您可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1364187|char*+c+=+new+char[length];
memset(c,+0,+length);|code-block|syntax|javascript|1364188|或者，您可以重载new运算符：|1364189|void+*operator+new(size_t+size,+bool+nullify)
{
++++void+*buf+=+malloc(size);

++++if+(!buf)+{
+++++++++++++//+Handle+this
++++}

++++memset(buf,+'\0',+size);

++++return+buf;
}|1364190|然后您将能够执行以下操作：|1364191|char*+c+=+new(true)+char[length];|1364192|而|1364193|char*+c+=+new+char[length];|1364194|将保持原有的行为。(请注意，如果您希望所有new都将它们创建的内容清零，可以使用上面相同的方法，但去掉bool+nullify部分)。|offset|length|style|CODE|1364195|请注意，如果您选择第二条路径，则应该重载标准的new操作符(不带bool的操作符)和delete操作符。这是因为您在这里使用的是malloc()，而标准说明malloc()+%2B+delete操作是未定义的。所以你必须重载delete才能使用free()，而普通的新手才能使用malloc()。|1364196|在实践中，尽管所有的实现都在内部使用malloc()/free()本身，所以即使你不这样做，你也不会遇到任何问题(除了语言律师对你大喊大叫之外)。|1364197|entityMap^0|0|0|0|0|0|0|0|0|L|3|1F|C|0|1S|8|26|8|2H|6|32|6|3C|6|3T|8|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|16|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|18|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1C|8|@$U|1D|V|1E|W|X]|$U|1F|V|1G|W|X]]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1H|8|@$U|1I|V|1J|W|X]|$U|1K|V|1L|W|X]|$U|1M|V|1N|W|X]|$U|1O|V|1P|W|X]|$U|1Q|V|1R|W|X]|$U|1S|V|1T|W|X]]|9|@]|A|$]]|$1|10|3|11|5|6|7|1U|8|@]|9|@]|A|$]]|$1|12|3|-4|5|6|7|1V|8|@]|9|@]|A|$]]]|13|$]]

You can't do it in one line easily. You can do:

<pre><code>char* c = new char[length];
memset(c, 0, length);
</code></pre>

Or, you can overload the new operator:

<pre><code>void *operator new(size_t size, bool nullify)
{
 void *buf = malloc(size);

 if (!buf) {
 // Handle this
 }

 memset(buf, '\0', size);

 return buf;
}
</code></pre>

Then you will be able to do:

<pre><code>char* c = new(true) char[length];
</code></pre>

while

<pre><code>char* c = new char[length];
</code></pre>

will maintain the old behavior. (Note, if you want all <code>new</code>s to zero out what they create, you can do it by using the same above but taking out the <code>bool nullify</code> part).

Do note that if you choose the second path you should overload the standard new operator (the one without the bool) and the delete operator too. This is because here you're using <code>malloc()</code>, and the standard says that <code>malloc()</code> + <code>delete</code> operations are undefined. So you have to overload <code>delete</code> to use <code>free()</code>, and the normal new to use <code>malloc()</code>.

In practice though all implementations use malloc()/free() themselves internally, so even if you don't do it most likely you won't run into any problems (except language lawyers yelling at you)

blocks|key|1149116|text|没有内部手段，AFAIK。使用这个:+memset(c，0，length)；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1149117|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|$]]

No internal means, AFAIK. Use this:
 memset(c, 0, length);

blocks|key|1364381|text|而许多发帖者的隐含评论=>不使用数组，而是使用向量。数组的所有优点，没有任何缺点。PLus你会得到很多其他的好东西|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1364382|如果你不了解STL，请阅读Josuttis的C%2B%2B标准库和meyers的有效STL。|1364383|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

and the implicit comment by many posters => Dont use arrays, use vectors. All of the benefits of arrays with none of the downsides. PLus you get lots of other goodies

If you dont know STL, read Josuttis The C++ standard library and meyers effective STL

blocks|key|1364456|text|你必须“手动”初始化它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1364457|char*+c+=+new+char[length];
for(int+i+=+0;i<length;i%2B%2B)
++++c[i]='\0';|code-block|syntax|javascript|1364458|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|$]]

you have to initialize it "by hand" :

<pre><code>char* c = new char[length];
for(int i = 0;i&lt;length;i++)
 c[i]='\0';
</code></pre>

How do I achieve the dynamic equivalent of this static array initialisation:

<pre><code>char c[2] = {}; // Sets all members to '\0';
</code></pre>

In other words, create a dynamic array with all values initialised to the termination character:

<pre><code>char* c = new char[length]; // how do i amend this? 
</code></pre>

How do you initialise a dynamic array in C++?

如何实现这种静态数组初始化的动态等价物：char c[2] = {};  // Sets all members to '\0';换句话说，创建一个动态数组，将所有值初始化为终止字符：char* c = new char[length]; // how do i amend this? 

问如何在C++中初始化动态数组？
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