字符串拷贝与进制转换的简单应用

源码中需要将整数分成前后两个部分，可以用strncpy()函数来实现

#include<stdio.h> 
#include<string.h> 

int main() 
{ 
     char str1[10]={0},str2[10]={0};
     char temp[10]={0};
     char *string="abcdefghi"; 
     int i =12345678;
     sprintf(temp,"%d",i);
     strncpy(str1,temp,4);
     strncpy(str2,temp+4,4); 
     printf("str1 is %s,str2 is %s\n",str1,str2); 
	
     strncpy(str1,string,3);
     str1[3]='\0';
     strncpy(str2,string+4,5); 
     str2[5]='\0';
     printf("str1 is %s,str2 is %s\n",str1,str2); 
     return 0; 
} 

运行结果：
str1 is 1234,str2 is 5678
str1 is abc,str2 is efghi

项目应用场景：将一个十进制数转换成十六进制，然后将十六进制分成前后两部分，再分别以十进制数显示出来

#include<stdio.h> 
#include<string.h> 
#include<stdlib.h>

int main() 
{ 
     char str1[10]={0},str2[10]={0};
     char temp[10]={0},str[10]={0};
     char *stopstring;
     int front=0,rear=0;
	 int next_lockid = 18913516;
	 ltoa(next_lockid,temp,16);  
     sprintf(str,"%08s",temp);
     printf("temp is %s,str is %s\n",temp,str);
     strncpy(str1,str,4);
     strncpy(str2,str+4,4); 
     front = strtol(str1,&stopstring,16);
     rear = strtol(str2,&stopstring,16);
     printf("str1 is %s,str2 is %s\n",str1,str2); 
     printf("front is %d,rear is %d\n",front,rear); 
	
     return 0; 
} 

运行结果：
temp is 12098ec,str is 012098ec
str1 is 0120,str2 is 98ec
front is 288, rear is 39148

        通常在<stdlib.h>头文件中包含函数atoi()，该函数是标准的C语言函数，其功能是：把字符串转换成32位以下的任意进制数字。  
   
        而与其相反的itoa()函数则非标准C语言扩展函数。由于它不是标准C语言函数，所以不能在所有的编译器中使用。但是，大多数的编译器（如Windows上)在<stdlib.h>头文件中包含这个函数，而在linux系统中则不支持。      
        由于项目是在linux下编程，所以itoa()函数不能直接用，需要自己实现itoa()函数，下面是实现的itoa()函数源码：

  

#include<stdio.h> 
#include<string.h> 
#include<stdlib.h>

char* _itoa(int value, char* string, int radix)
{
    char tmp[33];
    char* tp = tmp;
    int i;
    unsigned v;
    int sign;
    char* sp;
 
    if (radix > 36 || radix <= 1)
    {
 
        return 0;
    }
 
    sign = (radix == 10 && value < 0);
    if (sign)
        v = -value;
    else
        v = (unsigned)value;
    while (v || tp == tmp)
    {
        i = v % radix;
        v = v / radix;
    if (i < 10)
        *tp++ = i+'0';
    else
        *tp++ = i + 'a' - 10;
    }
 
    if (string == 0)
    string = (char*)malloc((tp-tmp)+sign+1);
    sp = string;
 
    if (sign)
        *sp++ = '-';
    while (tp > tmp)
        *sp++ = *--tp;
        *sp = 0;
    return string;
} 

    另附上实现的atol()函数，以备不时之需

#include "stdio.h"
#include "conio.h"
#include <ctype.h>
#include <stdlib.h>

long  atol(const char *nptr)
{
    int c; /* current char */
    long total; /* current total */
    int sign; /* if ''-'', then negative, otherwise positive */
 
/* skip whitespace */
    while ( isspace((int)(unsigned char)*nptr) )
        ++nptr;
 
    c = (int)(unsigned char)*nptr++;
    sign = c; /* save sign indication */
    if (c == '-' || c == '+')
        c = (int)(unsigned char)*nptr++; /* skip sign */
 
    total = 0;
 
    while (isdigit(c)) {
        total = 10 * total + (c - '0'); /* accumulate digit */
        c = (int)(unsigned char)*nptr++; /* get next char */
    }
 
    if (sign == '-')
        return -total;
    else
        return total; /* return result, negated if necessary */
}

     atoi()函数可以通过调用atol()函数获取

int atoi(const char *nptr)
{
    return (int)atol(nptr);
}