C language: Input a string str from the keyboard, and count the number of 26 letters from lowercase letters a to z in str (the number of 0 will not be displayed, and other characters will not be counted).

For beginners, when you see this type of string statistics question, you will first think of if selection statement.
Do a good range of conditions such as: ('a'<=str[i])&&(str[i]<='z')
like this It is done to classify the input string. But the subsequent statistics on the specific number of occurrences of each letter in the string are helpless. (For example, I decided at the beginning that if I choose to determine the 26 letters to be counted one by one, and later I will write the case where the number is 0, which is too complicated)
At this time, a specific algorithm is required.
Drawing lessons from the code of other programmers, noticed that the subsequent letters can be expressed through the corresponding expressions.

Code display:

#include<stdio.h>
#include<string.h>
#define N 100//宏定义便于定义字符数组的宽度
int main()
{
    
    
    int cnt[26]={
    
    0},len,i,j;
	char str[N];//宏定义

	printf("请输入一个字符串！\n");
	gets(str);
     
	len=strlen(str);//strlen剔除'\0'
	for(i=0;i<len;i++)
	{
    
    
		for(j=0;j<26;j++)//26个字母
		{
    
    
			if(str[i]=='a'+j)//此处就是一个表达式，'z'与'a'相差26，在'a'上特定加上数字即可表示相应字母
			{
    
    
				cnt[j]++;//这样解决了特定字母出现个数特定数组下标统计。
				break;//内循环一个判定完之后暂停。等待外循环的下一个字符。
			}
		}
	}
printf("统计结果为：\n");
	for(i=0;i<26;i++)
	{
    
    
		if(cnt[i]!=0)
		{
    
    
			printf("%c:%d\n",i+'a',cnt[i]);//此处的i+'a'较str[i]更好，若是str[i]还得一一判定是否为字母，且i是记录了输入字符串中有哪些字母，在'a'之上相加个数即可表示。
		}
	}


	return 0;
}

The code analysis is detailed in the code.