blocks|key|939605|text|你不需要一个完整的函数；一个简单的类型转换就足够了：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|939606|int+x;
static_cast<char*>(static_cast<void*>(&x));|code-block|syntax|javascript|939607|C%2B%2B中的任何对象都可以重新解释为字节数组。如果您希望将字节实际复制到单独的数组中，则可以使用std::copy|offset|length|style|CODE|939608|int+x;
char+bytes[sizeof+x];
std::copy(static_cast<const+char*>(static_cast<const+void*>(&x)),
++++++++++static_cast<const+char*>(static_cast<const+void*>(&x))+%2B+sizeof+x,
++++++++++bytes);|939609|这两种方法都没有考虑字节顺序，但是由于您可以将int重新解释为字节数组，因此自己执行任何必要的修改都很容易。|939610|entityMap^0|0|0|1B|9|0|0|N|3|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

You don't need a whole function for this; a simple cast will suffice:

<pre><code>int x;
static_cast&lt;char*&gt;(static_cast&lt;void*&gt;(&amp;x));
</code></pre>

Any object in C++ can be reinterpreted as an array of bytes. If you want to actually make a copy of the bytes into a separate array, you can use <code>std::copy</code>:

<pre><code>int x;
char bytes[sizeof x];
std::copy(static_cast&lt;const char*&gt;(static_cast&lt;const void*&gt;(&amp;x)),
 static_cast&lt;const char*&gt;(static_cast&lt;const void*&gt;(&amp;x)) + sizeof x,
 bytes);
</code></pre>

Neither of these methods takes byte ordering into account, but since you can reinterpret the <code>int</code> as an array of bytes, it is trivial to perform any necessary modifications yourself.

blocks|key|939629|text|return+((byte[0]<<24)%7C(byte[1]<<16)%7C(byte[2]<<8)%7C(byte[3]));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|939630|:D|unstyled|939631|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>return ((byte[0]&lt;&lt;24)|(byte[1]&lt;&lt;16)|(byte[2]&lt;&lt;8)|(byte[3]));
</code></pre>

:D

blocks|key|723576|text|您可以通过and和移位操作获得单个字节：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|723577|byte1+=++nint+&+0x000000ff
byte2+=+(nint+&+0x0000ff00)+>>+8
byte3+=+(nint+&+0x00ff0000)+>>+16
byte4+=+(nint+&+0xff000000)+>>+24|code-block|syntax|javascript|723578|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 can get individual bytes with anding and shifting operations:

<pre><code>byte1 = nint &amp; 0x000000ff
byte2 = (nint &amp; 0x0000ff00) &gt;&gt; 8
byte3 = (nint &amp; 0x00ff0000) &gt;&gt; 16
byte4 = (nint &amp; 0xff000000) &gt;&gt; 24
</code></pre>

blocks|key|723612|text|使用std::vector<unsigned+char>|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|723613|#include+<vector>
using+namespace+std;

vector<unsigned+char>+intToBytes(int+paramInt)
{
+++++vector<unsigned+char>+arrayOfByte(4);
+++++for+(int+i+=+0;+i+<+4;+i%2B%2B)
+++++++++arrayOfByte[3+-+i]+=+(paramInt+>>+(i+*+8));
+++++return+arrayOfByte;
}|code-block|syntax|javascript|723614|entityMap^0|2|Q|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|$]]

Using <code>std::vector&lt;unsigned char&gt;</code>:

<pre><code>#include &lt;vector&gt;
using namespace std;

vector&lt;unsigned char&gt; intToBytes(int paramInt)
{
 vector&lt;unsigned char&gt; arrayOfByte(4);
 for (int i = 0; i &lt; 4; i++)
 arrayOfByte[3 - i] = (paramInt &gt;&gt; (i * 8));
 return arrayOfByte;
}
</code></pre>

blocks|key|939696|text|std::vector<unsigned+char>+intToBytes(int+value)
{
++++std::vector<unsigned+char>+result;
++++result.push_back(value+>>+24);
++++result.push_back(value+>>+16);
++++result.push_back(value+>>++8);
++++result.push_back(value++++++);
++++return+result;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|939697|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>std::vector&lt;unsigned char&gt; intToBytes(int value)
{
 std::vector&lt;unsigned char&gt; result;
 result.push_back(value &gt;&gt; 24);
 result.push_back(value &gt;&gt; 16);
 result.push_back(value &gt;&gt; 8);
 result.push_back(value );
 return result;
}
</code></pre>

blocks|key|939788|text|我使用的另一种有用的方法是联合：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|939789|union+byteint
{
++++byte+b[sizeof+int];
++++int+i;
};
byteint+bi;
bi.i+=+1337;
for(int+i+=+0;+i<4;i%2B%2B)
++++destination[i]+=+bi.b[i];|code-block|syntax|javascript|939790|这将使得字节数组和整数将“重叠”(共享相同的内存)。所有类型都可以这样做，只要字节数组的大小与类型相同(否则其中一个字段不会受到另一个字段的影响)。当你必须在整数操作和字节操作/复制之间切换时，将它们作为一个对象也是很方便的。|939791|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|$]]

Another useful way of doing it that I use is unions:

<pre><code>union byteint
{
 byte b[sizeof int];
 int i;
};
byteint bi;
bi.i = 1337;
for(int i = 0; i&lt;4;i++)
 destination[i] = bi.b[i];
</code></pre>

This will make it so that the byte array and the integer will "overlap"( share the same memory ).
this can be done with all kinds of types, as long as the byte array is the same size as the type( else one of the fields will not be influenced by the other ). And having them as one object is also just convenient when you have to switch between integer manipulation and byte manipulation/copying.

blocks|key|939818|text|我知道这个问题已经有了答案，但我会给出我的解决方案。我使用的是模板函数和整数约束。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|939819|以下是我的解决方案：|939820|#include+<type_traits>
#include+<vector>

template+<typename+T,
++++++++++typename+std::enable_if<std::is_arithmetic<T>::value>::type*+=+nullptr>
std::vector<uint8_t>+splitValueToBytes(T+const&+value)
{
++++std::vector<uint8_t>+bytes;

++++for+(size_t+i+=+0;+i+<+sizeof(value);+i%2B%2B)
++++{
++++++++uint8_t+byte+=+value+>>+(i+*+8);
++++++++bytes.insert(bytes.begin(),+byte);
++++}

++++return+bytes;
}|code-block|syntax|javascript|939821|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I know this question already has answers but I will give my solution to this problem. I am using template function and integer constraint on it.

Here is my solution:

<pre><code>#include &lt;type_traits&gt;
#include &lt;vector&gt;

template &lt;typename T,
 typename std::enable_if&lt;std::is_arithmetic&lt;T&gt;::value&gt;::type* = nullptr&gt;
std::vector&lt;uint8_t&gt; splitValueToBytes(T const&amp; value)
{
 std::vector&lt;uint8_t&gt; bytes;

 for (size_t i = 0; i &lt; sizeof(value); i++)
 {
 uint8_t byte = value &gt;&gt; (i * 8);
 bytes.insert(bytes.begin(), byte);
 }

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

blocks|key|723805|text|我希望我的能帮上忙|type|unstyled|depth|inlineStyleRanges|entityRanges|data|723806|template+<typename+t_int>
std::array<uint8_t,+sizeof+(t_int)>+int2array(t_int+p_value)+{
++++static+const+uint8_t+_size_of+(static_cast<uint8_t>(sizeof+(t_int)));
++++typedef+std::array<uint8_t,+_size_of>+buffer;
++++static+const+std::array<uint8_t,+8>+_shifters+=+{8*0,+8*1,+8*2,+8*3,+8*4,+8*5,+8*6,+8*7};

++++buffer+_res;
++++for+(uint8_t+_i=0;+_i+<+_size_of;+%2B%2B_i)+{
++++++++_res[_i]+=+static_cast<uint8_t>((p_value+>>+_shifters[_i]));
++++}
++++return+_res;
}|code-block|syntax|javascript|723807|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 hope mine helps

<pre><code>template &lt;typename t_int&gt;
std::array&lt;uint8_t, sizeof (t_int)&gt; int2array(t_int p_value) {
 static const uint8_t _size_of (static_cast&lt;uint8_t&gt;(sizeof (t_int)));
 typedef std::array&lt;uint8_t, _size_of&gt; buffer;
 static const std::array&lt;uint8_t, 8&gt; _shifters = {8*0, 8*1, 8*2, 8*3, 8*4, 8*5, 8*6, 8*7};

 buffer _res;
 for (uint8_t _i=0; _i &lt; _size_of; ++_i) {
 _res[_i] = static_cast&lt;uint8_t&gt;((p_value &gt;&gt; _shifters[_i]));
 }
 return _res;
}
</code></pre>

I have this method in my java code which returns byte array for given int:

<pre><code>private static byte[] intToBytes(int paramInt)
{
 byte[] arrayOfByte = new byte[4];
 ByteBuffer localByteBuffer = ByteBuffer.allocate(4);
 localByteBuffer.putInt(paramInt);
 for (int i = 0; i &lt; 4; i++)
 arrayOfByte[(3 - i)] = localByteBuffer.array()[i];
 return arrayOfByte;
}
</code></pre>

Can someone give me tip how can i convert that method to C++?

C++ int to byte array

我在我的java代码中有这个方法，它返回给定int的字节数组：private static byte[] intToBytes(int paramInt){     byte[] arrayOfByte = new byte[4];     ByteBuffer localByteBuffer = ByteBuffer...

问C++整数到字节数组
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