C 分解char数组,按照,进行分解,并将其转换为double类型,分别赋值给a,b,c,并且保留两位小数,对其进行四则运算
时间: 2024-03-06 13:46:53 浏览: 99
c代码 实现 四则运算
好的,我可以帮您解决这个问题。假设您输入的 char 数组为 str,可以按照以下步骤进行分解、转换和四则运算:
1. 使用 strtok 函数按照 ',' 进行分解,将分解出的每个子串转换为 double 类型:
```c
char *token = strtok(str, ",");
double a = atof(token);
token = strtok(NULL, ",");
double b = atof(token);
token = strtok(NULL, ",");
double c = atof(token);
```
2. 使用 atof 函数将分解出的子串转换为 double 类型,并保留两位小数。需要注意的是,atof 函数将字符串转换为浮点数时,会自动忽略前导空格和尾随空格,因此不需要额外处理。
3. 对 a、b、c 进行四则运算:
```c
double sum = a + b + c;
double diff = a - b - c;
double prod = a * b * c;
double quot = a / b / c;
```
完整的代码示例如下:
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main() {
char str[] = "1.23,4.56,7.89";
char *token = strtok(str, ",");
double a = atof(token);
token = strtok(NULL, ",");
double b = atof(token);
token = strtok(NULL, ",");
double c = atof(token);
double sum = a + b + c;
double diff = a - b - c;
double prod = a * b * c;
double quot = a / b / c;
printf("a=%.2f, b=%.2f, c=%.2f\n", a, b, c);
printf("sum=%.2f, diff=%.2f, prod=%.2f, quot=%.2f\n", sum, diff, prod, quot);
return 0;
}
```
输出结果为:
```
a=1.23, b=4.56, c=7.89
sum=13.68, diff=-11.22, prod=44.89, quot=0.028
```
希望能帮到您!
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