
- 效果
- 普通数字
- 小数
- 科学记数法
- 源码
平台:
STC89C52
Keil uVision V5.29.0.0
PK51 Prof.Developers Kit Version:9.60.0.0
效果
这里以51单片机为例,对串口接收的字符串中的第三个数进行解码:
#ifndef STR2NUM_H_ #define STR2NUM_H_ #include "stdint.h" extern void str2int(char * str, char flag, int32_t no, int32_t *Output); extern void str2double(char * str, char flag, int32_t no, double *Output); #endif
#include "Str2Num.h"
void str2int(char * str, char flag, int32_t no, int32_t *Output)
{
int32_t id_end, count, output;
for (id_end = 0, count = 0; count != no; ++id_end)
{
if (str[id_end] == flag || str[id_end] == 'r' || str[id_end] == 'n' || str[id_end] == '')
++count;
}
id_end -= 2;
for (output = 0, count = 1; str[id_end] != flag && id_end >= 0; --id_end)
{
if (str[id_end] == '-')
output *= -1;
else if (str[id_end] == '+')
;
else
{
output += (str[id_end] - '0') * count;
count *= 10;
}
}
*Output = output;
}
void str2double(char * str, char flag, int32_t no, double *Output)
{
int32_t id_end, id_dot, id_e, count, id, pow;
double output;
for (id_end = 0, count = 0; count != no; ++id_end)
{
if (str[id_end] == flag || str[id_end] == 'r' || str[id_end] == 'n' || str[id_end] == '')
++count;
}
id_end -= 2; //寻找最后一个数字的下标
for (id_dot = id_end, id_e = -1; str[id_dot] != '.'; --id_dot)
{
if (str[id_dot] == 'e' || str[id_dot] == 'E')
id_e = id_dot; //标记e的下标
if (str[id_dot] == flag || str[id_dot] == 'r' || str[id_dot] == 'n' || str[id_dot] == '')
{
if(id_e > 0)
id_dot = id_e;
else
id_dot = id_end + 1;
break;
}
}
output = 0;
if (str[id_dot] == '.')
{
for (count = 10, id = id_dot + 1; str[id] != flag && str[id] != 'r' && str[id] != 'n' && str[id] != '' && str[id] != 'e' && str[id] != 'E'; ++id)
{
output += (str[id] - '0') / (double)count; //加入小数部分
count *= 10;
}
}
for (count = 1, id = id_dot - 1; str[id] != flag && id >= 0; --id)
{
if (str[id] == '-')
output *= -1;
else if (str[id] == '+')
;
else
{
output += (str[id] - '0') * count;
count *= 10; //加入整数部分
}
}
if(id_e > 0) //指数计算
{
for (pow = 0, count = 1; id_end != id_e; --id_end)
{
if (str[id_end] == '+')
;
else if (str[id_end] == '-')
pow *= -1;
else
{
pow += (str[id_end] - '0') * count;
count *= 10;
}
}
if (pow > 0)
while (pow--)
output *= 10;
else if(pow < 0)
while (pow++)
output /= 10;
}
*Output = output;
}
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