[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] = '\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);
}

转自/article/1523906.html