1 ．設(shè)計函數(shù) int atoi(char *s) 。
//atio函數(shù)實現(xiàn)，總感覺自己代碼太繁瑣了，by leetaolion@126.com
//atio轉(zhuǎn)換舉例 "123"->123
//atio轉(zhuǎn)換舉例 "00123"->123
//atio轉(zhuǎn)換舉例 " 0 0123"->123
//atio轉(zhuǎn)換舉例 "+123"->123
//atio轉(zhuǎn)換舉例 "-123"->-123
//atio轉(zhuǎn)換舉例 "(+/-)( /e)00123"->0
//atio轉(zhuǎn)換舉例 "e00123"->0

#include "iostream.h"
#include "ctype.h"

int my_atoi(char *s);

int my_atoi(char *s)
{
?int i=0,j;
?while (s[0]=='0'||s[0]==' ')//過濾首部0和空格
?{
??s++;
?};
?if ( !isdigit(s[0]) )?? //如果第一位不是數(shù)字，則可能是符號位
?{
?switch ( s[0] )
??{
??case '+':
??s++;
??while (s[0]>='0'&&s[0]<='9')
??{
???j=int(s[0]-'0');
???i=i*10+j;
???s++;
??}
??return i;
??break;

??case '-':
??s++;
??while (s[0]>='0'&&s[0]<='9')
??{
???j=int(s[0]-'0');
???i=i*10+j;
???s++;
??}
??return 0-i;
??break;
?
??default:
??return i;
??break;?
??}
?}

??while (s[0]>='0'&&s[0]<='9')
??{
???j=int(s[0]-'0');
???i=i*10+j;
???s++;
??}
??return i;

}

void main( void )
{
?char *s=NULL;
?int i;

?s = "?? +123456";
?i = my_atoi( s );
?cout<<i<<endl;

?s = "?? 0-0123456";
?i = my_atoi( s );
?cout<<i<<endl;

?s = "12345e6";
?i = my_atoi( s );
?cout<<i<<endl;
?
}

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