void ReverseNumber(void){
int x = 0;
fprintf(stderr, "开始解答：逆序数字\n");
printf("请输入一个待逆序的整数\n");
scanf("%d", &x);
int digits;
if ( x > 0) digits = floor(log10(x)) + 1;
else if (x < 0) digits = floor(log10(-x)) + 1;
else digits = 0;
// fprintf(stderr, "digits=%d", digits);
int const kDIVIDER = 10;
int o = 0;
int i = digits;
int n = 0;
/* Reverse function */
while (i != 0) {
n = x % kDIVIDER;
x = x / kDIVIDER;
o = o + n * power10(i - 1);
// fprintf(stderr, "Last Digit=%d, Truncated Value=%d, Output=%d, step=%d\n", n, x, o, i);
i --;
}
printf("%d\n", o);
}
int myatoi(const char *str) //将字符str转换为整数
{
const char *p = NULL;
int neg; //标志位，0代表负数，1代表正数
if (str[0] == '-') //str为负整数，置neg标志位为0
{
neg = 0;
p = ++str; //将字符串str字符串中的'-'去除，从str的第二位开始转化
}
else //否则置neg标志位为1
{
neg = 1;
p = str;
}
int i_len = len(p); //调到字符串长度
int i_temp = i_len;
int value = 0;
int i = 0;
for (; i < i_len; i++) //从str[0]开始遍历字符串
{
value += (power10(i_temp - 1) * char2i(p[i]));
i_temp--;
}
if (neg == 0) //str为负数 ，将转化后的字符串前
{
return 0 - value;
}
else
{
return value;
}
}
//{4,3,2,1} -> 4321
int digstoval(int * digits)
{
int i;
int sum=0;
for(i=0; i<4; i++)
sum += digits[i] * power10(i);
return sum;
}
bool to_rational(const exprt &expr, rationalt &rational_value)
{
if(expr.id()!=ID_constant) return true;
const std::string &value=expr.get_string(ID_value);
std::string no1, no2;
char mode=0;
for(std::string::const_iterator it=value.begin();
it!=value.end();
++it)
{
const char ch=*it;
if(isdigit(ch))
{
if(mode==0)
no1+=ch;
else
no2+=ch;
}
else if(ch=='/' || ch=='.')
{
if(mode==0)
mode=ch;
else
return true;
}
else
return true;
}
switch(mode)
{
case 0:
rational_value=string2integer(no1);
break;
case '.':
rational_value=string2integer(no1);
rational_value+=rationalt(string2integer(no2))/power10(no2.size());
break;
case '/':
rational_value=string2integer(no1);
rational_value/=string2integer(no2);
break;
default:
return true;
}
return false;
}
bool to_rational(const exprt &expr, rationalt &rational_value)
{
if(expr.id()!="constant") return true;
const std::string &value=expr.get_string("value");
std::string no1, no2;
char mode=0;
for(unsigned i=0; i<value.size(); i++)
{
char ch=value[i];
if(isdigit(ch))
{
if(mode==0)
no1+=ch;
else
no2+=ch;
}
else if(ch=='/' || ch=='.')
{
if(mode==0)
mode=ch;
else
return true;
}
else
return true;
}
switch(mode)
{
case 0:
rational_value=string2integer(no1);
break;
case '.':
rational_value=string2integer(no1);
rational_value+=rationalt(string2integer(no2))/power10(no2.size());
break;
case '/':
rational_value=string2integer(no1);
rational_value/=string2integer(no2);
break;
default:
return true;
}
return false;
}
void number_to_str(float number, char *str,int afterdecimal){
int n = (int)number, f = (int)((number - n) * power10(afterdecimal)), i= 0;
if (number < 0)
str[i++] = '-';
// changing -ves to +ves
if (n < 0) n = -n;
if (f < 0) f = -f;
str[i++] = n % 10 + '0';
while (n /= 10)
str[i++] = n % 10 + '0';
rev(str, number < 0 ? 1 : 0, i - 1);
if (afterdecimal) {
str[i++] = '.';
str[i++] = f % 10 + '0';
while (f /= 10)
str[i++] = f % 10 + '0';
rev(str, i - afterdecimal, i - 1);
}
str[i] = 0;
}
/*
* 作用:调整指数部分
* 参数:
* n:要处理的ln
* inc_power:指数增量 >0 增加指数,把整数部分结尾的0去掉 <0 减少指数,增加整数部分结尾的0
* 返回值:
* 成功:ln
* 失败:返回NULL
*/
ln ln_adjustpower(ln n,int inc_power)
{
int a;
int res,carry;
int zeronum;
cell p;
//验证参数
if(ln_checknull(n) !=0)
{
fprintf(stderr,"[%s %d] %s error,reason: ln_checknull fail\n",__FILE__,__LINE__,__FUNCTION__);
return NULL;
}
//如果n是0,让它随便设置指数,反正都一样
if(ln_cmp_int(n,0)==0)
{
//设置新的指数
n->power+=inc_power;
return n;
}
if(inc_power==0) //指数不变
return n;
if(inc_power<0) //减少指数
{
//设置新的指数
n->power+=inc_power;
//转正
inc_power*=-1;
//先乘上剩余部分的0
a=power10(inc_power%DIGIT_NUM);
p=n->lsd;
res=0;
carry=0;
while(1)
{
res=a*p->num+carry;
p->num=res%UNIT;
carry=res/UNIT;
if(p==n->msd)
{
if(carry>0) //存在进位
{
if(n->lsd->lcell ==n->msd) //节点不够
{
if(ln_addcell(n,INIT_SIZE) ==NULL) //分配失败
{
fprintf(stderr,"[%s %d] %s error,reason: ln_addcell fail\n",__FILE__,__LINE__,__FUNCTION__);
return NULL;
}
}
p->hcell->num=carry;
n->msd=p->hcell;
}
break;
}
p=p->hcell;
}
inc_power-=inc_power%DIGIT_NUM;
while(inc_power>0)
{
if(n->lsd->lcell !=n->msd) //还有多余的节点，利用它
{
n->lsd=n->lsd->lcell;
n->lsd->num=0;
inc_power-=DIGIT_NUM;
}
else //一次性分配剩下的节点
{
if(ln_addcell(n,inc_power/DIGIT_NUM)==NULL)
{
fprintf(stderr,"[%s %d] %s error,reason: ln_addcell fail\n",__FILE__,__LINE__,__FUNCTION__);
return NULL;
}
n->lsd=n->msd->hcell;
break;
}
}
}
else
{
//获取结尾0的个数
zeronum=ln_endingzeronum(n);
if(zeronum==-1)
{
fprintf(stderr,"[%s %d] %s error,reason: ln_endingzeronum fail\n",__FILE__,__LINE__,__FUNCTION__);
return NULL;
}
//不能去掉这么多0
if(inc_power>zeronum)
{
fprintf(stderr,"[%s %d] %s error,reason: inc_power too large (%d)\n",__FILE__,__LINE__,__FUNCTION__,inc_power);
return NULL;
}
//设置新的指数
n->power+=inc_power;
//先处理结尾的0节点
p=n->lsd;
while(p->num==0 && inc_power>=DIGIT_NUM)
{
inc_power-=DIGIT_NUM;
p=p->hcell;
}
n->lsd=p;
//除以剩下的0
if(inc_power==0)
return n;
a=power10(inc_power);
p=n->msd;
res=0;
carry=0;
while(1)
{
res=p->num+carry*UNIT;
p->num=res/a;
carry=res%a;
if(p==n->lsd)
break;
p=p->lcell;
}
}
return n;
}
/*
* 作用:把ln取精度
* 副作用:把ln前置0和后置0去掉
* 参数:
* n:要处理的ln
* precision:所需精度(保留的小数位数)
* mode:指定截断或者四舍五入
* 返回值:
* 成功:返回ln
* 失败:NULL
*/
ln ln_fix(ln n,int precision,divide_mode mode)
{
int pointnum;
int power;
cell p;
//验证参数
if(ln_checknull(n)!=0)
{
fprintf(stderr,"[%s %d] %s error,reason: ln_checknull fail\n",__FILE__,__LINE__,__FUNCTION__);
return NULL;
}
//精度参数有误
if(precision<0)
{
fprintf(stderr,"[%s %d] %s error,reason: precision error\n",__FILE__,__LINE__,__FUNCTION__);
return NULL;
}
//去除前置0
ln_stripleadingzero(n);
//去除后置0
ln_adjustpower(n,ln_endingzeronum(n));
//获取最低节点的小数点位数
pointnum=ln_pointnum(n,n->lsd);
//大于pointnum,直接返回
if(precision >=pointnum)
return n;
//只存在一个节点
if(n->lsd==n->msd)
{
p=n->lsd;
power=power10(pointnum-precision);
if(mode==trunc_res) //截断
p->num-=p->num%power;
else
{
if(p->num%power>=5*power/10)
p->num=(p->num/power+1)*power;
else
p->num-=p->num%power;
}
}
else
{
p=n->lsd;
while(pointnum>precision)
{
p=p->hcell;
pointnum-=DIGIT_NUM;
}
if(pointnum==precision) //现在p所指的节点精度就是precision
{
if(mode==round_res && p->lcell->num>=UNIT/2) //四舍五入
p->num++;
}
else
{
p=p->lcell;
power=power10(DIGIT_NUM+pointnum-precision);
if(mode==trunc_res) //截断
{
p->num-=p->num%power;
}
else
{
if(p->num%power>=5*power/10)
p->num=(p->num/power+1)*power;
else
p->num-=p->num%power;
}
}
//p后面有效节点被舍去,因此重置为0
if(p!=n->lsd)
{
do
{
p=p->lcell;
p->num=0;
}
while(p!=n->lsd);
}
}
//有可能四舍五入导致溢出,需要做调整
p=n->lsd;
while(1)
{
if(p->num>=UNIT)
{
p->num-=UNIT;
if(p==n->msd)
{
if(ln_addcell(n,INIT_SIZE) !=NULL)
{
fprintf(stderr,"[%s %d] %s error,reason: ln_addcell fail\n",__FILE__,__LINE__,__FUNCTION__);
return NULL;
}
n->msd=p->hcell;
n->msd->num=1;
break;
}
else
{
p=p->hcell;
p->num+=1;
}
}
else
break;
}
return n;
}
//double *
//ft_numparse(char **s, bool whole)
double *
ft_numparse(char **s, int whole)
{
double mant = 0.0;
int sign = 1, exsign = 1, p;
double expo = 0.0;
static double num;
char *string = *s;
/* See if the number begins with + or -. */
if (*string == '+') {
string++;
} else if (*string == '-') {
string++;
sign = -1;
}
/* We don't want to recognise "P" as 0P, or .P as 0.0P... */
if ((!isdigit(*string) && *string != '.') ||
((*string == '.') && !isdigit(string[1])))
return (NULL);
/* Now accumulate a number. Note ascii dependencies here... */
while (isdigit(*string))
mant = mant * 10.0 + (*string++ - '0');
/* Now maybe a decimal point. */
if (*string == '.') {
string++;
p = 1;
while (isdigit(*string))
mant += (*string++ - '0') / power10(p++);
}
/* Now look for the scale factor or the exponent (can't have both). */
switch (*string) {
case 'e':
case 'E':
/* Parse another number. */
string++;
if (*string == '+') {
exsign = 1;
string++;
} else if (*string == '-') {
exsign = -1;
string++;
}
while (isdigit(*string))
expo = expo * 10.0 + (*string++ - '0');
if (*string == '.') {
string++;
p = 1;
while (isdigit(*string))
expo += (*string++ - '0') / power10(p++);
}
expo *= exsign;
break;
case 't':
case 'T':
expo = 12.0;
string++;
break;
case 'g':
case 'G':
expo = 9.0;
string++;
break;
case 'k':
case 'K':
expo = 3.0;
string++;
break;
case 'u':
case 'U':
expo = -6.0;
string++;
break;
case 'n':
case 'N':
expo = -9.0;
string++;
break;
case 'p':
case 'P':
expo = -12.0;
string++;
break;
case 'f':
case 'F':
expo = -15.0;
string++;
break;
case 'm':
case 'M':
/* Can be either m, mil, or meg. */
if (string[1] && string[2] &&
((string[1] == 'e') || (string[1] == 'E')) &&
((string[2] == 'g') || (string[2] == 'G')))
{
expo = 6.0;
string += 3;
} else if (string[1] && string[2] &&
((string[1] == 'i') || (string[1] == 'I')) &&
((string[2] == 'l') || (string[2] == 'L')))
{
expo = -6.0;
mant *= 25.4;
string += 3;
} else {
expo = -3.0;
string++;
}
break;
}
if (whole && *string != '\0') {
return (NULL);
} else if (ft_strictnumparse && *string && isdigit(string[-1])) {
if (*string == '_')
while (isalpha(*string) || (*string == '_'))
string++;
else
return (NULL);
} else {
while (isalpha(*string) || (*string == '_'))
string++;
}
*s = string;
num = sign * mant * pow(10.0, expo);
if (ft_parsedb)
fprintf(cp_err, "numparse: got %e, left = %s\n", num, *s);
return (&num);
}
void testPower10()
{
assertIntEqual(power10(0), 1);
assertIntEqual(power10(2), 100);
assertIntEqual(power10(4), 10000);
}