C language counts the number of characters in a string

Objective:

Enter a line of characters and count the number of various characters in it.

Specific code:

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define M 1024
void main() {
	char str[M];
	fgets(str, M, stdin);
	int space = 0;
	int letter = 0;
	int num = 0;
	int other = 0;
	for (int i = 0; i < (int)strlen(str); ++i) {
		if (str[i] == &#39; &#39;) {
			space += 1;
		}
		else if (str[i] > 64 && str[i] < 91 || str[i]>96 && str[i] < 123)  {
			letter += 1;
		}
		else if (str[i] > 47 && str[i] < 58)  {
			num += 1;
		}
		else {
			if (str[i] != &#39;\n&#39;) {//因为fgets()函数会在末尾自动加上\n，影响判断结果，需要判断是否为换行符
				other += 1;
			}
		}
	}
	printf("空格的个数为：%d\n", space);
	printf("英文字母的个数为：%d\n", letter);
	printf("数字的个数为：%d\n", num);
	printf("其他字符的个数为：%d\n", other);
	system("pause");
}

Note: The fgets() function will read the carriage return character \n that we type on the keyboard at the end of the string (before \0).

The running results are as follows:

Recommended tutorial: c language tutorial

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