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] = '\0';
	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);
}