The C Programming Language 练习题4-2
2017-10-11 22:53
330 查看
题目
对atof函数进行扩充,使它可以处理形如123.45e-6的科学表示法,其中,浮点数后面可能会紧跟一个e或E以及一个指数(可能有正负号)。
题目分析
教材中atof函数已经处理到指数之前的位置,扩充只需判断后面是否有指数。
代码实现
对atof函数进行扩充,使它可以处理形如123.45e-6的科学表示法,其中,浮点数后面可能会紧跟一个e或E以及一个指数(可能有正负号)。
题目分析
教材中atof函数已经处理到指数之前的位置,扩充只需判断后面是否有指数。
代码实现
#include <ctype.h> #include <stdio.h> double atof(char s[]) { double val, power; int i, sign, eval, epower; for (i = 0; isspace(s[i]); i++) ; sign = (s[i] == '-') ? -1 : 1; if (s[i] == '+' || s[i] =='-') i++; for (val = 0.0; isdigit(s[i]); i++) val = 10.0 * val + (s[i] - '0'); if (s[i] == '.') i++; for (power = 1.0; isdigit(s[i]); i++) { val = 10.0 * val + (s[i] - '0'); power *= 10; } val = sign * val / power; /*从此处开始判断是否有指数部分*/ if (s[i] == 'e' || s[i] == 'E') i++; sign = (s[i] == '-') ? -1 : 1; if (s[i] == '-' || s[i] =='+') i++; for (eval = 0; isdigit(s[i]); i++) eval = 10 * eval + (s[i] - '0'); for (epower = 1; eval > 0; eval--) epower = 10 * epower; if (sign > 0) return val * epower; else return val / epower; } int main() { char s1[] = "123.45e-3"; printf("%f\n", atof(s1)); }
相关文章推荐
- The C Programming Language 练习题1-21
- The C Programming Language 练习题2-2
- The C Programming Language 练习题2-8
- <<The C Programming Language>>学习之路-练习题参考答案 1-4
- The C Programming Language 练习题1-22
- The C Programming Language 练习题2-6
- The C Programming Language 练习题3-1
- The C Programming Language 练习题4-5
- The C Programming Language 练习题2-1
- The C Programming Language 练习题4-3
- <<The C Programming Language>>学习之路-练习题参考答案 1-1
- <<The C Programming Language>>学习之路-练习题参考答案 1-2
- The C Programming Language 练习题2-3
- The C Programming Language 练习题2-5
- The C Programming Language 练习题2-9
- The C Programming Language 练习题3-2
- The C Programming Language 练习题3-3
- The C Programming Language 练习题2-4
- The C Programming Language 练习题2-10
- The C Programming Language 练习题3-5