[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 = 0, i = 0;
    int sign = 1;
    
    for( i=0; isspace(s[i]); i++ );
    
    sign = (s[i] == '-')? -1:1;
    
    if( s[i] == '+' || s[i] == '-' )
        i++;
    
    for( ;isdigit(s[i]); i++ )
    {
        num = 10*num + (s[i]-'0');
    }
    
    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 = 0;
    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 = 0;
    char tmp;
    
    while( s[size] != NULL )
        size++;
    size--;
    
    int i=0;
    while( i <= size>>1 )
    {
        tmp = s[i], s[i] = s[size-i], s[size-i] = tmp;
        i++;
    }
}

void ItoA( int n, char *s )
{
    int sign = 1;
    
    if( n < 0 )
    {
        sign = -1;        
        n = -n;
    }
    
    int i=0; 
    do
    {
        s[i++] = n%10 + '0';
    }    while((n/=10) > 0);
    
    if( sign == -1 )
        s[i++] = '-';
    s[i] = '';
    Reverse( s );
}

4. atof( char * s )

把字符串转化成double类型

double AtoF( char *s )
{
    int sign = 1;
    int i = 0;
    for( i=0; isspace(s[i]); i++ );
    
    sign = (s[i] == '-')? -1:1;
    
    if( s[i] == '+' || s[i] == '-' )
        i++;
        
    double num = 0.0;
    double pow = 1.0;
    //整数 
    for( ;isdigit(s[i]); i++ )
        num = num*10 + (s[i]-'0');
        
    for( i++; isdigit(s[i]); i++ )
    {
        num = num*10 + (s[i]-'0');
        pow *= 10;
    }
    
    return sign * (num/pow);
}

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