C函数的实现（strcpy,atoi,atof,itoa,reverse）

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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);}