1-4 Exercise 3: String details

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Please write a program, the user will enter a line of characters, then the program will calculate number of digits, alphabets and another characters.

寫一個程式 輸入一個字串，計算字串裡的數字、英文字母和其他符號的次數。

Example input:

[email protected]#$%^&*()_+

Example output:

digit:10
alphabet:26
other:13

 

[Exercise] Coding (C) - asked in Chapter 13: Strings by (5.2k points)
Available when: 2018-01-04 18:00 - Due to: Unlimited
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53 Answers

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#include <stdio.h>
#include <string.h>
int main()
{
    char a[100];
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}
answered by (-140 points)
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prog.c: In function 'main':
prog.c:17:23: error: expected ')' before 'a'
             if('A'+k++a[i] || 'a'+k==a[i])
                       ^
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#include <stdio.h>
#include <ctype.h>
#include <string.h>

int main()
{
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prog.c: In function 'main':
prog.c:9:13: warning: format '%s' expects argument of type 'char *', but argument 2 has type 'char (*)[100]' [-Wformat=]
     scanf("%s", &str);
             ^
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Hidden content!
#include <stdio.h>
#include <string.h>
#include <ctype.h>

int main()
{
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#include <stdio.h>
#include <string.h>
int main()
{
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}
answered by (-191 points)
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Hidden content!
#include <stdio.h>
#include <string.h>
int main()
{
    char A[999];
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answered by (-196 points)
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prog.c: In function 'main':
prog.c:7:15: warning: implicit declaration of function 'strlen' [-Wimplicit-function-declaration]
     for(i=0;i<strlen(A);i++)
               ^~~~~~
prog.c:7:15: warning: incompatible implicit declaration of built-in function 'strlen'
prog.c:7:15: note: include '<string.h>' or provide a declaration of 'strlen'
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Hidden content!
#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int main()
{
    char a[999],letter;
    int alphat=0,digit=0,all=0,other=0;

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    return 0;

}
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#include<stdio.h>
#include<string.h>
int main()
{
    char a[1000];
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}
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#include<stdio.h>
#include<string.h>

int main()
{
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}
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Hidden content!
#include<stdio.h>
#include<string.h>

int main()
{
****       *** *** ****** a[100];
 * * **  **    * ** *** **   *   * ** *  * **
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*  * **   *   * *****  * *  = 0 ; i < len ; i++)
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*  ******  * ** *    * **** *   *   0;
}
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