Question

我之前发布过如何实现将整数转换为IP地址字符串的函数。那么我们如何反之亦然，也就是说，给定一个地址的字符串（154.111.23.23），我们如何使用无函数函数来获取该整数。

Answer 1

将字符串扫描成四个字节，并将它们添加/移位为32位整数。

Answer 2

Posix.1-2001为此任务提供inet_pton()。还有几个ms-windows版本。

Answer 3

//No checking of the input    
unsigned int c1,c2,c3,c4;
scanf("%d.%d.%d.%d",&c1,&c2,&c3,&c4);
unsigned long ip = (unsigned long)c4+c3*256+c2*256*256+c1*256*256*256;
printf("The unsigned long integer is %lu\n",ip);

编辑：对于那些对产生的代码感兴趣的人，GCC足够聪明，可以用左移替换256的乘法。（在我的程序中我也叫退出）：

0x80483d4   <main>:     lea    0x4(%esp),%ecx
0x80483d8   <main+4>:       and    $0xfffffff0,%esp
0x80483db   <main+7>:       pushl  -0x4(%ecx)
0x80483de   <main+10>:      push   %ebp
0x80483df   <main+11>:      mov    %esp,%ebp
0x80483e1   <main+13>:      push   %ecx
0x80483e2   <main+14>:      sub    $0x34,%esp
0x80483e5   <main+17>:      lea    -0x14(%ebp),%eax
0x80483e8   <main+20>:      mov    %eax,0x10(%esp)
0x80483ec   <main+24>:      lea    -0x10(%ebp),%eax
0x80483ef   <main+27>:      mov    %eax,0xc(%esp)
0x80483f3   <main+31>:      lea    -0xc(%ebp),%eax
0x80483f6   <main+34>:      mov    %eax,0x8(%esp)
0x80483fa   <main+38>:      lea    -0x8(%ebp),%eax
0x80483fd   <main+41>:      mov    %eax,0x4(%esp)
0x8048401   <main+45>:      movl   $0x8048520,(%esp)
0x8048408   <main+52>:      call   0x8048320 <scanf@plt>
0x804840d   <main+57>:      mov    -0x8(%ebp),%eax
0x8048410   <main+60>:      mov    %eax,%edx
0x8048412   <main+62>:      shl    $0x8,%edx
    0x8048415   <main+65>:      mov    -0xc(%ebp),%eax
0x8048418   <main+68>:      lea    (%edx,%eax,1),%eax
0x804841b   <main+71>:      mov    %eax,%edx
0x804841d   <main+73>:      shl    $0x8,%edx
0x8048420   <main+76>:      mov    -0x10(%ebp),%eax
0x8048423   <main+79>:      lea    (%edx,%eax,1),%eax
0x8048426   <main+82>:      mov    %eax,%edx
0x8048428   <main+84>:      shl    $0x8,%edx
0x804842b   <main+87>:      mov    -0x14(%ebp),%eax
0x804842e   <main+90>:      lea    (%edx,%eax,1),%eax
0x8048431   <main+93>:      mov    %eax,-0x18(%ebp)
0x8048434   <main+96>:      mov    -0x18(%ebp),%eax
0x8048437   <main+99>:      mov    %eax,0x4(%esp)
0x804843b   <main+103>:     movl   $0x804852c,(%esp)
0x8048442   <main+110>:     call   0x8048330 <printf@plt>
0x8048447   <main+115>:     movl   $0x0,(%esp)
0x804844e   <main+122>:     call   0x8048340 <exit@plt>

Answer 4

uint32_t getDecimalValueOfIPV4_String(const char* ipAddress)
{
    uint8_t ipbytes[4]={};
    int i =0;
    int8_t j=3;
    while (ipAddress+i && i<strlen(ipAddress))
    {
       char digit = ipAddress[i];
       if (isdigit(digit) == 0 && digit!='.'){
           return 0;
       }
        j=digit=='.'?j-1:j;
       ipbytes[j]= ipbytes[j]*10 + atoi(&digit);

        i++;
    }

    uint32_t a = ipbytes[0];
    uint32_t b =  ( uint32_t)ipbytes[1] << 8;
    uint32_t c =  ( uint32_t)ipbytes[2] << 16;
    uint32_t d =  ( uint32_t)ipbytes[3] << 24;
    return a+b+c+d;
}

整数的IP地址 - C.

4 个答案: