写了个ipv4地址字符串转换为4个整型数的函数, 大佬们帮我看一下有没有啥问题, 跪谢
正确条件是: 仅包含数字, 空格, 点,
允许数字和点之间的空格,
点与点之间为空或为空格, 则默认为0, 如 '...' 视为'0.0.0.0',
点数量为3,
数值不大于255.
其他情况均为非法地址.
代码如下:
程序代码：
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int isIpv4Correct(const char*, int*);
int main(int argc, char const *argv[])
{
char ipv4str[50] = {};
int ipv4int[4] = {};
strcpy(ipv4str, "...");
puts(ipv4str);
if (isIpv4Correct(ipv4str, ipv4int)) {
puts("Correct!");
for (int i = 0; i < 4; i++) {
printf("%d ", ipv4int[i]);
}
} else {
puts("Error!");
}
return 0;
}
int isIpv4Correct(const char *ipv4str, int *ipv4int) {
int i = 0, ip = 0;
int res = 0;
int dotCount = 0;
int inNum = 0;
char cur = ipv4str[i];
while (cur) {
if (!isdigit(cur) && cur != ' ' && cur != '.') {
return 0;
} else if (isdigit(cur)) {
inNum = 1;
res = res*10 + cur-'0';
if (res > 255) return 0;
} else if (cur == ' ') {
if (inNum) {
return 0;
} else {
while (cur == ' ') {
cur = ipv4str[++i];
}
continue;
}
} else {
++dotCount;
inNum = 0;
ipv4int[ip++] = res;
res = 0;
}
cur = ipv4str[++i];
}
if (dotCount != 3) return 0;
ipv4int[ip] = res;
return 1;
}
