Extract string (C language)

Extract string (C language)

Write a blog New Year's Eve (><) Happy New Year! Safe!
Directly on the picture:

When you look at this kind of topic, you know-it's so easy! A lot of pits! This damn detail!

• Pit 1: xxx.42xxx outputs 0.42
• Pit 2:
  xxx123
  456xxx outputs 123456
  -xxx123
  .456xx outputs 123.456
• Detail 1: How to extract numbers?
• Detail 2: How to sort floating-point numbers?

Gan, I stopped writing xxx, and under pressure, I still racked my brains and came out (Crying Haw,
don’t talk nonsense, go to the train of thought:

1. Preparation

• The end of multi-line input is judged by EOF, ie ctrl+z
• Open a two-dimensional array and store the extracted number string: m[j][k]
• Use the atof function to convert the string to a floating point number and store it in the double array num[35]. (Note that atof returns double, I thought it was a float and then I made an error)
• Use qsort to sort, notice:
  void qsort(void *base, size_t nitems, size_t size, int (*compar)(const void *, const void *)) The
  compar function returns an int type, so change its variant:
  return *(double *)a < *(double *)b ? 1 : -1;

2. The main part

• Use the isdigit function to determine whether it is a number and start to extract
• If the last of the line is a number, set a flag:
  f = 1;
  hand it over to the next line, and judge whether the beginning of the next line is a number, then continue to add (k) to the end of the original extracted number, and withdraw the increment of the number of numbers (j) :
  k += 1; j -= 1;
• If you encounter a number, add 0 at the beginning, then switch to the extraction of numbers, and set a flag as well:
  f1 = 1;

3. Code part

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
#include <math.h>
int j=0, k=0, f=0, f1=0, sum=0;
char m[35][25] = {
    
    "0"};        //m[j][k]数字字符串数组，j为数字的个数，k为数字的长度 
double num[35] = {
    
    0};
int cmp(const void *a, const void *b)    //降序排序 
{
    
    
    return *(double *)a < *(double *)b ? 1 : -1;
}

double *find(char str[])
{
    
    
	int n, a=0, i=0;
	n = strlen(str);
	for(i = 0; i < n; i++)
	{
    
    	
		if(i == 0)         //判断上一行是否为数字  是f=1 
		{
    
    
			if(f && (isdigit(str[i]) || str[i] == '.'))   //该行开头为数字或小数点 
			{
    
    
				k += 1;         //接在上一行提取的数字后面 
				j -= 1;         //不要递增，还是上一个提取的数 
				if(str[i] == '.')
				{
    
    
					m[j][k] = str[i];
					k++;
				}	         //此时f未归0，为了避免与.数字的情况冲突 
			}
			else       //如果该行开头不是数字或小数点，归0 
			{
    
    
				k = 0;
				f = 0;
			}
		}
		if(!(isdigit(str[i-1])) && str[i] == '.' && f == 0)    //出现.数字的情况，接下来处理为0.数字 
			f1 = 1;
		if(isdigit(str[i]) || f1 == 1)    //如果是数字或者是.数字 
		{
    
    
			f = 0;     //将可能未归0的f归0 
			if(f1 == 1)
			{
    
    
				m[j][k] = '0';   //加0 
				k++;
				f1 = 0;
			}
			m[j][k] = str[i];   //存第一个数字 
			i++;
			if(i == n)   //如果是一行的最后一位 
			{
    
    
				f = 1;
				j++;      //默认存完一串数字就递增 
			}
			for(a = i; a < n; a++)  //存接下来的数字 
			{
    
    
				if(isdigit(str[a]) || (str[a] == '.' && isdigit(str[a+1])))  //第二个括号判断条件舍去了数字.的情况 
				{
    
    
					k++;
					m[j][k] = str[a];     //存 
					if(a == n-1)     //直到一行最后一位 
					{
    
    
						j++;
						f = 1;
					}	
				}
				else     //存结束了 
				{
    
    
					k = 0;
					j++;
					i = a-1;
					break;
				}
			}
			
		}
		if(a == n)      //如果a增到结尾，就让i跳出循环，避免重复计算 
			i = n-1;
	}
	for(i = 0; i < j; i++)        //字符串转为浮点数 
		num[i] = atof(m[i]);
	return num;
}

int main(void)
{
    
    
	char str[1000] = {
    
    "0"};
	while(scanf("%s",str)!=EOF)
		find(str);
	qsort(num, j, sizeof(num[0]), cmp);
	printf("%.2f",num[0]);
	for(int i = 1; i < j; i++)
		printf(" %.2f",num[i]);
	printf("\n"); 
	return 0;
}