#include<stdio.h>
#include<string.h>
#include <math.h>
long long convertDecimalToBinary(int n);
int main() {
int verify;
long long bip, dip;
char str1[100];
printf("Enter dotted decimal ip address :\n");
scanf("%s",str1);
verify = bin_verify(str1);
seperate(str1);
return 0;
}
int bin_verify(char str1[]) {
int i;
for(i = 0; i < strlen(str1); i++) {
if((str1[i] < 255) && (str1[i] > 0)) {
return 1;
}
}
}
// function to get first decimal no sepreted
int seperate(char str1[]) {
int s1, s2, s3, s4;
int i, j, k = 0, m, cnt = 0, cnt1 = 0, pos = 0;
char a[4], str2[100];
for(i = 0; i < strlen(str1); i++) {
pos = cnt;
if(str1[i] == '.') {
k = i;
pos = cnt;
for(j = 0; j < i; j++) {
a[j] = str1[k-cnt];
cnt = cnt - 1;
}
break;
}
else {
cnt++;
//goto one;
}
}
for(m = 0; m <= pos; m++) {
str1++;
}
s1 = atoi(a);
s1 = convertDecimalToBinary(s1);
printf("Binary Format of IP :\n");
printf("%d.",s1);
seperate2(str1);
return 0;
}
// function to get second decimal no sepreted
int seperate2(char str1[]) {
int s1, s2, s3, s4;
int i, j, k = 0, m, cnt = 0, cnt1 = 0, pos = 0;
char a[4], str2[100];
for(i = 0; i < strlen(str1); i++) {
pos = cnt;
if(str1[i] == '.') {
k = i;
pos = cnt;
for(j = 0; j < i; j++) {
a[j] = str1[k-cnt];
cnt = cnt - 1;
}
break;
}
else {
cnt++;
//goto one;
}
}
for(m = 0; m <= pos; m++) {
str1++;
}
s2 = atoi(a);
s2 = convertDecimalToBinary(s2);
printf("%d.",s2);
seperate3(str1);
return 0;
}
// function to get third decimal no sepreted
int seperate3(char str1[]) {
int s1, s2, s3, s4;
int i, j, k = 0, m, cnt = 0, cnt1 = 0, pos = 0;
char a[4], str2[100];
for(i = 0; i < strlen(str1); i++) {
pos = cnt;
if(str1[i] == '.') {
k = i;
pos = cnt;
for(j = 0; j < i; j++) {
a[j] = str1[k-cnt];
cnt = cnt - 1;
}
break;
}
else {
cnt++;
//goto one;
}
}
for(m = 0; m <= pos; m++) {
str1++;
}
s3 = atoi(a);
s3 = convertDecimalToBinary(s3);
printf("%d.",s3);
seperate4(str1);
return 0;
}
// function to get fourth decimal no sepreted
int seperate4(char str1[]) {
int s1, s2, s3, s4;
int i, j, k = 0, m, cnt = 0, cnt1 = 0, pos = 0;
char a[4], str2[100];
for(i = 0; i < strlen(str1); i++) {
pos = cnt;
if(str1[i] == '.') {
k = i;
pos = cnt;
for(j = 0; j < i; j++) {
a[j] = str1[k-cnt];
cnt = cnt - 1;
}
break;
}
else {
cnt++;
}
}
for(m = 0; m <= pos; m++) {
str1++;
}
s4 = atoi(a);
s4 = convertDecimalToBinary(s4);
printf("%d\n",s4);
return 0;
}
//to convert decimal to binary
long long convertDecimalToBinary(int n)
{
//printf("%d", n);
long long binaryNumber = 0;
int remainder, i = 1,step=0;
while (n!=0)
{
remainder = n%2;
// printf("Step %d: %d/2, Remainder = %d, Quotient = %d\n", step++, n, remainder, n/2);
n /= 2;
binaryNumber += remainder*i;
i *= 10;
}
return binaryNumber;
}
输出：
Enter dotted decimal ip address :
192.15.7.4
Binary Format of IP :
11000000.1111.111.0
我想将ip地址转换为二进制，但是，
它总是返回0作为最后一个十进制数的二进制。
为什么它没有执行seperate4()功能？
答案 0 :(得分：1)
你所做的错误已经在评论中列出，但你做得过于复杂，几乎就像Rube-Goldberg一样。它非常简单，无需任何复杂的技巧即可完成。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// ALL CHECKS OMMITTED!
char *int8_to_bin(int n, char *buf)
{
int i = 8;
// we need an unsigned integer for the bit-juggling
// because of two's complement numbers.
// Not really needed here because the input is positive
unsigned int N;
// check for proper size
if (n < 0 || n > 255) {
return NULL;
}
// is safe now
N = (unsigned int) n;
// we work backwards here, least significant bit first
// but we want it least significant bit last.
while (i--) {
// make a (character) digit out of an integer
buf[i] = (char) (N & 0x1) + '0';
// shift one bit to the right and
// drop the least significant bit by doing it
N >>= 1;
}
// return the pointer to the buffer we got (for easier use)
return buf;
}
int main(int argc, char **argv)
{
// keeps the intermediate integer
int addr;
// command-line argument, helper for strtol() and int8_to_bin()
char *s, *endptr, *cp;
// buffer for int8_to_bin() to work with
// initialize to all '\0';
char buf[9] = { '\0' };
// array holding the end-result
// four 8-character groups with three periods and one NUL
char binaddr[4 * 8 + 3 + 1];
// iterator
int i;
if (argc < 2) {
fprintf(stderr, "Usage: %s dotted_ipv4\n", argv[0]);
exit(EXIT_FAILURE);
}
// set a pointer pointing to the first argument as a shortcut
s = argv[1];
// the following can be done in a single loop, of course
// strtol() skips leading spaces and parses up to the first non-digit.
// endptr points to that point in the input where strtol() decided
// it to be the first non-digit
addr = (int) strtol(s, &endptr, 0);
// work on copy to check for NULL while keeping the content of buf
// (checking not done here)
cp = int8_to_bin(addr, buf);
// if(cp == NULL)...
// copy the result to the result-array
// cp has a NUL, is a proper C-string
strcpy(binaddr, cp);
// rinse and repeat three times
for (i = 1; i < 4; i++) {
// skip anything between number and period,
// (or use strchr() to do so)
while (*endptr != '.'){
endptr++;
}
// skip the period itself
endptr++;
// add a period to the output
strcat(binaddr, ".");
// get next number
addr = (int) strtol(endptr, &endptr, 0);
cp = int8_to_bin(addr, buf);
// if(cp == NULL)...
strcat(binaddr, cp);
}
printf("INPUT: %s\n", s);
printf("OUTPUT: %s\n", binaddr);
exit(EXIT_SUCCESS);
}
我们不需要复杂的解析算法，strtol()为我们做了，我们只需要自己找到下一个时期。所有输入和输出的大小是已知的和/或如果它们在其限制范围内，则可以很容易地检查 - 例如：不需要繁琐且容易出错的内存分配，我们可以使用固定大小的缓冲区和strcat()
有一个原则，不仅适用于海军，也适用于编程：KISS。