blocks|key|1200396|text|这里有一个简单的方法：(ip+>>+8)、(ip+>>+16)和(ip+>>+24)将第二、第三和第四个字节移动到低位字节，而&+0xFF在每一步中隔离最低有效字节。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1200397|void+print_ip(unsigned+int+ip)
{
++++unsigned+char+bytes[4];
++++bytes[0]+=+ip+&+0xFF;
++++bytes[1]+=+(ip+>>+8)+&+0xFF;
++++bytes[2]+=+(ip+>>+16)+&+0xFF;
++++bytes[3]+=+(ip+>>+24)+&+0xFF;+++
++++printf("%25d.%25d.%25d.%25d\n",+bytes[3],+bytes[2],+bytes[1],+bytes[0]);++++++++
}|code-block|syntax|javascript|1200398|在第一步中有一个隐含的bytes[0]+=+(ip+>>+0)+&+0xFF;。|1200399|使用snprintf()将其打印为字符串。|1200400|entityMap^0|B|9|L|A|W|A|1R|6|0|0|B|S|0|2|A|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]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|P|$]]

Here's a simple method to do it: The <code>(ip &gt;&gt; 8)</code>, <code>(ip &gt;&gt; 16)</code> and <code>(ip &gt;&gt; 24)</code> moves the 2nd, 3rd and 4th bytes into the lower order byte, while the <code>&amp; 0xFF</code> isolates the least significant byte at each step.

<pre><code>void print_ip(unsigned int ip)
{
 unsigned char bytes[4];
 bytes[0] = ip &amp; 0xFF;
 bytes[1] = (ip &gt;&gt; 8) &amp; 0xFF;
 bytes[2] = (ip &gt;&gt; 16) &amp; 0xFF;
 bytes[3] = (ip &gt;&gt; 24) &amp; 0xFF; 
 printf("%d.%d.%d.%d\n", bytes[3], bytes[2], bytes[1], bytes[0]); 
}
</code></pre>

There is an implied <code>bytes[0] = (ip &gt;&gt; 0) &amp; 0xFF;</code> at the first step.

Use <code>snprintf()</code> to print it to a string.

blocks|key|1415514|text|另一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1415515|union+IP+{
++++unsigned+int+ip;
++++struct+{
++++++unsigned+char+d;
++++++unsigned+char+c;
++++++unsigned+char+b;
++++++unsigned+char+a;
++++}+ip2;
};

...
char++ips[20];
IP+ip;
ip.ip+=+0xAABBCCDD;

sprintf(ips,+"%25x.%25x.%25x.%25x",+ip.ip2.a,+ip.ip2.b,+ip.ip2.c,+ip.ip2.d);
printf("%25s\n",+ips);|code-block|syntax|javascript|1415516|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|$]]

Another approach:

<pre><code>union IP {
 unsigned int ip;
 struct {
 unsigned char d;
 unsigned char c;
 unsigned char b;
 unsigned char a;
 } ip2;
};

...
char ips[20];
IP ip;
ip.ip = 0xAABBCCDD;

sprintf(ips, "%x.%x.%x.%x", ip.ip2.a, ip.ip2.b, ip.ip2.c, ip.ip2.d);
printf("%s\n", ips);
</code></pre>

blocks|key|1200327|text|提示:将32位整数分解为4个8位整数，并打印出来。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1200328|类似的内容(未编译，YMMV)：|1200329|int+i+=+0xDEADBEEF;+//+some+32-bit+integer
printf("%25i.%25i.%25i.%25i",
++++++++++(i+>>+24)+&+0xFF,
++++++++++(i+>>+16)+&+0xFF,
++++++++++(i+>>+8)+&+0xFF,
++++++++++i+&+0xFF);|code-block|syntax|javascript|1200330|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|$]]

Hint: break up the 32-bit integer to 4 8-bit integers, and print them out.

Something along the lines of this (not compiled, YMMV):

<pre><code>int i = 0xDEADBEEF; // some 32-bit integer
printf("%i.%i.%i.%i",
 (i &gt;&gt; 24) &amp; 0xFF,
 (i &gt;&gt; 16) &amp; 0xFF,
 (i &gt;&gt; 8) &amp; 0xFF,
 i &amp; 0xFF);
</code></pre>

blocks|key|1200472|text|如果传递一个字符串缓冲区来填充，并且我知道缓冲区足够大(即至少16个字符)，我就会这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1200473|sprintf(buffer,+"%25d.%25d.%25d.%25d",
++(ip+>>+24)+&+0xFF,
++(ip+>>+16)+&+0xFF,
++(ip+>>++8)+&+0xFF,
++(ip++++++)+&+0xFF);|code-block|syntax|javascript|1200474|这会比先创建一个字节数组稍微快一点，而且我认为它更具可读性。我通常使用snprintf，但是IP地址的长度不能超过16个字符，包括终止null。|1200475|或者，如果要求我提供一个返回char*的函数|1200476|char*+IPAddressToString(int+ip)
{
++char[]+result+=+new+char[16];

++sprintf(result,+"%25d.%25d.%25d.%25d",
++++(ip+>>+24)+&+0xFF,
++++(ip+>>+16)+&+0xFF,
++++(ip+>>++8)+&+0xFF,
++++(ip++++++)+&+0xFF);

++return+result;
}|1200477|entityMap^0|0|0|0|0|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

This is what I would do if passed a string buffer to fill and I knew the buffer was big enough (ie at least 16 characters long):

<pre><code>sprintf(buffer, "%d.%d.%d.%d",
 (ip &gt;&gt; 24) &amp; 0xFF,
 (ip &gt;&gt; 16) &amp; 0xFF,
 (ip &gt;&gt; 8) &amp; 0xFF,
 (ip ) &amp; 0xFF);
</code></pre>

This would be slightly faster than creating a byte array first, and I think it is more readable. I would normally use snprintf, but IP addresses can't be more than 16 characters long including the terminating null.

Alternatively if I was asked for a function returning a char*:

<pre><code>char* IPAddressToString(int ip)
{
 char[] result = new char[16];

 sprintf(result, "%d.%d.%d.%d",
 (ip &gt;&gt; 24) &amp; 0xFF,
 (ip &gt;&gt; 16) &amp; 0xFF,
 (ip &gt;&gt; 8) &amp; 0xFF,
 (ip ) &amp; 0xFF);

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

blocks|key|1415590|text|#include+"stdio.h"

void+print_ip(int+ip)+{
+++unsigned+char+bytes[4];
+++int+i;
+++for(i=0;+i<4;+i%2B%2B)+{
++++++bytes[i]+=+(ip+>>+i*8)+&+0xFF;
+++}
+++printf("%25d.%25d.%25d.%25d\n",+bytes[3],+bytes[2],+bytes[1],+bytes[0]);
}

int+main()+{
+++int+ip+=+0xDEADBEEF;
+++print_ip(ip);+++
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1415591|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>#include "stdio.h"

void print_ip(int ip) {
 unsigned char bytes[4];
 int i;
 for(i=0; i&lt;4; i++) {
 bytes[i] = (ip &gt;&gt; i*8) &amp; 0xFF;
 }
 printf("%d.%d.%d.%d\n", bytes[3], bytes[2], bytes[1], bytes[0]);
}

int main() {
 int ip = 0xDEADBEEF;
 print_ip(ip); 
}
</code></pre>

blocks|key|1200566|text|我的另一种减法解决方案:)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1200567|void+convert(+unsigned+int+addr+)
{++++
unsigned+int+num[OCTET],+
++++++++++++next_addr[OCTET];

int+bits+=+8;
unsigned+int+shift_bits;
int+i;

next_addr[0]+=+addr;+
shift_bits+-=+bits;++
num[0]+=+next_addr[0]+>>+shift_bits;

for+(+i+=+0;+i+<+OCTET-1;+i+%2B%2B+)
{+++++++
++++next_addr[i+%2B+1]+=+next_addr[i]+-+(+num[i]+<<+shift_bits+);+//+next+subaddr
++++shift_bits+-=+bits;+//+next+shift
++++num[i+%2B+1]+=+next_addr[i+%2B+1]+>>+shift_bits;+//+octet
}

printf(+"%25d.%25d.%25d.%25d\n",+num[0],+num[1],+num[2],+num[3]+);
}|code-block|syntax|javascript|1200568|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|$]]

My alternative solution with subtraction :)

<pre><code>void convert( unsigned int addr )
{ 
unsigned int num[OCTET], 
 next_addr[OCTET];

int bits = 8;
unsigned int shift_bits;
int i;

next_addr[0] = addr; 
shift_bits -= bits; 
num[0] = next_addr[0] &gt;&gt; shift_bits;

for ( i = 0; i &lt; OCTET-1; i ++ )
{ 
 next_addr[i + 1] = next_addr[i] - ( num[i] &lt;&lt; shift_bits ); // next subaddr
 shift_bits -= bits; // next shift
 num[i + 1] = next_addr[i + 1] &gt;&gt; shift_bits; // octet
}

printf( "%d.%d.%d.%d\n", num[0], num[1], num[2], num[3] );
}
</code></pre>

blocks|key|1415685|text|void+ul2chardec(char*pcIP,+unsigned+long+ulIPN){
int+i;+int+k=0;+char+c0,+c1;
for+(i+=+0;+i<4;+i%2B%2B){
++++c0+=+((((ulIPN+&+(0xff+<<+((3+-+i)+*+8)))+>>+((3+-+i)+*+8)))+/+100)+%2B+0x30;
++++if+(c0+!=+'0'){+*(pcIP+%2B+k)+=+c0;+k%2B%2B;+}
++++c1+=+(((((ulIPN+&+(0xff+<<+((3+-+i)+*+8)))+>>+((3+-+i)+*+8)))+%25+100)+/+10)+%2B+0x30;
++++if+(!(c1+=='0'+&&+c0=='0')){+*(pcIP+%2B+k)+=+c1;+k%2B%2B;+}
++++*(pcIP+%2Bk)+=+(((((ulIPN+&+(0xff+<<+((3+-+i)+*+8))))+>>+((3+-+i)+*+8)))+%25+10)+%2B+0x30;
++++k%2B%2B;
++++if+(i<3){+*(pcIP+%2B+k)+=+'.';+k%2B%2B;}
}
*(pcIP+%2B+k)+=+0;+//+pcIP+should+be+x10+bytes|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1415686|}|unstyled|1415687|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>void ul2chardec(char*pcIP, unsigned long ulIPN){
int i; int k=0; char c0, c1;
for (i = 0; i&lt;4; i++){
 c0 = ((((ulIPN &amp; (0xff &lt;&lt; ((3 - i) * 8))) &gt;&gt; ((3 - i) * 8))) / 100) + 0x30;
 if (c0 != '0'){ *(pcIP + k) = c0; k++; }
 c1 = (((((ulIPN &amp; (0xff &lt;&lt; ((3 - i) * 8))) &gt;&gt; ((3 - i) * 8))) % 100) / 10) + 0x30;
 if (!(c1 =='0' &amp;&amp; c0=='0')){ *(pcIP + k) = c1; k++; }
 *(pcIP +k) = (((((ulIPN &amp; (0xff &lt;&lt; ((3 - i) * 8)))) &gt;&gt; ((3 - i) * 8))) % 10) + 0x30;
 k++;
 if (i&lt;3){ *(pcIP + k) = '.'; k++;}
}
*(pcIP + k) = 0; // pcIP should be x10 bytes
</code></pre>

}

I'm preparing for a quiz, and I have a strong suspicion I may be tasked with implementing such a function. Basically, given an IP address in network notation, how can we get that from a 32 bit integer into a string in it's dotted decimal notation (something like 155.247.182.83)...? Obviously we can't be using any type of inet functions either...I'm stumped!

Integer to IP Address - C

我正在准备一个测验，我有一个强烈的怀疑，我的任务可能是实现这样一个功能。基本上，给定一个采用网络表示法的IP地址，我们如何将其从32位整数转换为点分十进制表示法(类似于155.247.182.83)的字符串？显然，我们也不能使用任何类型的inet函数……我被难住了！

问整数到IP地址-C
EN

回答 7

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问整数到IP地址-CEN

回答 7

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问整数到IP地址-C
EN