This question requires extracting all numeric characters ('0'...'9') in a string and converting it to an integer output.

I encountered a question today, I think it is very meaningful, so write it down and record it

topic:

This question requires extracting all numeric characters ('0'...'9') in a string and converting it to an integer output.

I didn't read the topic carefully at first, and then my first thought was to find the numeric characters in the string and print them out directly. The code is as follows

#include <stdio.h>
int main() 
{
    char arr[81];
    int len = 0;
    //输入字符串，并记录下len
    while (1) 
    {
        scanf("%c", &arr[len]);
        if (arr[len] == '\n') 
        {
            break;
        }
        len++;
    }
    //我当时就是直接找出数字字符，然后直接打印出来
    for(int i = 0;i < len;++i)
    {
        if(arr[i] >= '0' && arr[i] <= '9')
        {
            printf("%c", arr[i]);
        }
    }
    return 0;
}

When I submitted it later, I found out that it needs to be converted into an integer output.

The improved code is as follows

#include<stdio.h> 
int main()
{
	char arr[80];
	char b[80];
	int n = 0;
	int len = 0;
	while (1)
	{
		scanf("%c", &arr[len]);
		if (arr[len] == '\n')
		{
			break;
		}
		len++;
	}
	将数字备份到字符串b中 
	for (int i = 0; i < len; ++i) 
	{
		if (arr[i] >= '0' && arr[i] <= '9') 
		{
			b[n] = arr[i];
			n++;//记录b的个数
		}
	}
	int sum = 0;
	//最大的坑
	//本题要求提取一个字符串中的所有数字字符（'0'……'9'），将其转换为一个整数输出。
	for (int i = 0; i < n; ++i)
	{
		//输出最好按题目要求
		if (i == 0) 
		{
			sum = b[i] - '0';
		}
		//字符串转数字要-'0' 
		else 
		{
			sum = sum * 10 + b[i] - '0';
		}
	}
	printf("%d", sum);
}

Finally, make a small summary.

In fact, this question is not that difficult, it is very basic. However, due to my sloppyness, I did not write according to the requirements of the topic, which caused me to spend a certain amount of time.

So, veterans, you still have to read the questions carefully, and don’t make such low-level mistakes like me.