[转载]C函数的实现（strcpy,atoi,atof,itoa,reverse）

在笔试面试中经常会遇到让你实现C语言中的一些函数比如strcpy，atoi等

1. atoi

把字符串s转换成数字

 int Atoi( char *s )
 {
     int num = , i = ;
     int sign = ;

     for( i=; isspace(s[i]); i++ );

     sign = (s[i] == '-')? -:;

     if( s[i] == '+' || s[i] == '-' )
         i++;

     for( ;isdigit(s[i]); i++ )
     {
         num = *num + (s[i]-'');
     }

     return sign*num;
 }

2. strcpy(char *src, char *dst)

把src复制到dst中

 int Strcpy(char* src, char *dst)
 {
 //    register char *tmp;
     int i = ;
     while( src[i]!= NULL )
     {
         dst[i] = src[i++];
     }
 }

3.itoa( int n, char *s )

把数字转化成字符串

 void Reverse( char *s )
 {
     int size = ;
     char tmp;

     while( s[size] != NULL )
         size++;
     size--;

     int i=;
     while( i <= size>> )
     {
         tmp = s[i], s[i] = s[size-i], s[size-i] = tmp;
         i++;
     }
 }

 void ItoA( int n, char *s )
 {
     int sign = ;

     if( n <  )
     {
         sign = -;
         n = -n;
     }

     int i=;
     do
     {
         s[i++] = n% + '';
     }    while((n/=) > );

     if( sign == - )
         s[i++] = '-';
     s[i] = '\0';
     Reverse( s );
 }

4. atof( char * s )

把字符串转化成double类型

 double AtoF( char *s )
 {
     int sign = ;
     int i = ;
     for( i=; isspace(s[i]); i++ );

     sign = (s[i] == '-')? -:;

     if( s[i] == '+' || s[i] == '-' )
         i++;

     double num = 0.0;
     double pow = 1.0;
     //整数
     for( ;isdigit(s[i]); i++ )
         num = num* + (s[i]-'');

     for( i++; isdigit(s[i]); i++ )
     {
         num = num* + (s[i]-'');
         pow *= ;
     }

     return sign * (num/pow);
 }

转自http://blog.csdn.net/matrix_laboratory/article/details/12028431