Removing Punctuations in a string in C using Switch case

问题

Can someone please help me and tell what is wrong with my code. I made the solution using switch case and replace the punctuations with empty string.

#include<stdio.h>
#include<string.h>
int main() 
{
    char st[50];
    int i;
    printf("ENter the string:\n");       
    gets(st);
    for(i=0;i<strlen(st);i++)
    {
        switch(st[i])
        {
            case '!':
            case '"':
            case '#':
            case '$':
            case '%':
            case '&':strcpy(st[i]," ");
                     break;
        }
        printf("String is:\n");
        puts(st);
    }
    return 0;
}

回答1:

1. strcpy(st[i]," ") is wrong use st[i]=' '; (strcpy is for copy strings , is case of single char direct assignment is the process ) .
2. gets(st) is now removed from C .It causes buffer overflows . Use fgets(). Read more on gets() and fgets()

Here can replace gets() using fgets() by :-

fgets(st,50,stdin);

Modified code :-

#include <stdio.h>
#include <string.h>
int main()
{
    char st[50];
    int i;
    printf("ENter the string:\n");
    fgets(st, 50, stdin);
    for (i = 0; i < strlen(st); i++)
    {
        switch (st[i])
        {
        case '!':
        case '"':
        case '#':
        case '$':
        case '%':
        case '&':
            st[i] = ' ';
            break;
        }
        printf("String is:\n");
        puts(st);
    }
    return 0;
}

Recommended :- move puts() outside for-loop.

Output :-

ENter the string:
!hello#%worl$
String is:
 hello#%worl$

String is:
 hello#%worl$

String is:
 hello#%worl$

String is:
 hello#%worl$

String is:
 hello#%worl$

String is:
 hello#%worl$

String is:
 hello %worl$

String is:
 hello  worl$

String is:
 hello  worl$

String is:
 hello  worl$

String is:
 hello  worl$

String is:
 hello  worl$

String is:
 hello  worl 

String is:
 hello  worl 

回答2:

" " is a space. I don't think this is your intended behavior. Use another buffer to make a copy. For example,

 #include<stdio.h>
        #include<string.h>
        int main() 
        {
            char sta[50];
            char stb[50];
            int i,j;
            printf("ENter the string:\n");       
            gets(sta);
            for(i=0,j=0;i<strlen(sta);i++)
            {
                switch(st[sta])
                {
                    case '!':
                    case '"':
                    case '#':
                    case '$':
                    case '%':
                    case '&': break;
                    default: 
                             stb[j++]=sta[i];
                             break;
                }
                stb[j] = (char)0; // C str termination...
                printf("String is:\n");
                puts(stb);
            }
            return 0;
        }

回答3:

The following proposed code:

1. cleanly compiles
2. eliminates the 'magic' numbers
3. properly replaces the listed punctuation marks with a space
4. only prints the result once
5. uses the 'valid' function fgets() rather than the (currently) nonexistent gets() function
6. avoid comparing signed and unsigned values
7. limits the scope of the variable i

and now, the proposed code:

#include<stdio.h>
#include<string.h>

#define MAX_STR_LEN 50

int main( void ) 
{
    char st[ MAX_STR_LEN + 1 ];

    printf("ENter the string, max 50 characters\n");       
    fgets( st, sizeof( st ), stdin );

    for( size_t i=0;i<strlen(st);i++)
    {
        switch(st[i])
        {
            case '!':
            case '"':
            case '#':
            case '$':
            case '%':
            case '&':
                st[i] = ' ';
                break;

            default:
                break;
        }
    }

    printf( "Modified String is:\n %s\n", st );
    return 0;
}

来源：https://stackoverflow.com/questions/51600333/removing-punctuations-in-a-string-in-c-using-switch-case