Removing spaces from string, adding words to array in C

Question

I am trying to remove all spaces from a string and add each word into a string array.

For instance, provided the string:

char command[120] = "apple   banana orange  dogfood"

an array, defined as:

char *items[20+1]; //Maximum 20 items in string + 0

would contain the elements:

{"apple", "banana", "orange" "dogfood"}

That is, all spaces have been removed.

My goal is to be able to make the call:

echo    1    3 2

and printing out

1 3 2

instead of

    1    3 2

CODE

So far, I have the following:

    char temp_array[120];
    //Populate array with words
    int i, j = 0, ant = 0;
    for(i = 0; i < strlen(command); i++) {
        if(isspace(command[i]) || command[i] == '\0') {
            if(ant <= 20) {
                temp_array[j] = '\0';
                memo = (char *) malloc(sizeof(temp_array)); //allocate memory
                strcpy(memo, temp_array);
                items[ant] = memo;
                memset(temp_array, 0, j);
                ant++;
                j= 0;
            }
        }
        else if(!isspace(command[i])) {
            temp_array[j] = command[i];
            j++;
        }
    }//Done populating array
    items[ant] = NULL;

Any help is highly appreciated!

char *items[20+1] is array of 21 string pointers. To define N elements of 20+1 chars, write char items[][20+1]. — i486
– i486, Commented Nov 4, 2015 at 9:23
@i486: The char array should be able to contain a maximum of 20 words (pluss 1), not a maximum of 20 characters. — yulai
– yulai, Commented Nov 4, 2015 at 9:25
Can you outline what part of your code does not seem to work? (I'm only assuming it doesn't.) Must you duplicate the strings, or is it enough to have pointers into command? This mainly depends on where command came from and what its lifetime is through your program. — Jongware
– Jongware, Commented Nov 4, 2015 at 9:37
There are library functions that will do this for you. Look up strtok() and strsep(). — r3mainer
– r3mainer, Commented Nov 4, 2015 at 9:48

tdao · Accepted Answer · 2016-01-03 04:15:53Z

4

This problem may be solved naturally using strtok.

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

int main()
{
    char str[] = "apple   banana orange  dogfood";
    char *items[20] = { NULL };
    char *pch;

    pch = strtok( str," \t\n" );
    int i = 0;
    while( NULL != pch )
    {
        items[i++] = pch;
        pch = strtok( NULL, " \t\n" );
    }

    for( i = 0; i < 20; i++ )
    {
        if( NULL != items[i] )
        {
            printf( "items[%d] = %s\n", i, items[i] );
        }
        else
        {
            break;
        }
    }
    return 0;
}

Output:

items[0] = apple
items[1] = banana
items[2] = orange
items[3] = dogfood

edited Jan 3, 2016 at 4:15

answered Nov 4, 2015 at 9:51

tdao

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1 Comment

David C. Rankin Over a year ago

Note: the pointers assigned to the items array are only valid so long as the memory containing str remains valid. You will need to provide adequate storage for, and copy, each string pointed to by pch to items (instead of simply assigning the pointer) if for instance you call strtok in a function and return a pointer to items.

John · Accepted Answer · 2015-11-04 09:32:58Z

0

Loop through each character. If it's a space, continue. If it's not, add the address of that character to the array, loop until the first occurring space and replace it with a null terminator. Repeat.

Note that this destroys your initial string (command in your case). Also remember it must be modifiable (cannot be a read-only string).

answered Nov 4, 2015 at 9:32

John

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Comments

machine_1 · Accepted Answer · 2015-11-04 10:56:41Z

Here is a function that identify words in a string and stores them in an array.

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

int GetWords(char *String,char **buffer);

int main(void)
{
    char command[120] = "   apple   banana orange  dogfood";

    char *items[20+1];

    int num_of_words = GetWords(command,items);

    for( int n = 0 ; n < num_of_words ; n++ )
    {
        printf("%s ",items[n]);
        free(items[n]);
    }
    puts("");
    return 0;
}

int GetWords(char *String,char **buffer)
{
    int x = -1 , y = 0 , z = 0 ;

    size_t len = strlen(String) , n ;

    for( n = 0 ; n < len ; n++ )
    {
        y++;

        if( String[n] == ' ' )
        {
            y = 0;
            z = 0;
        }

        if( y == 1 )
        {
            x++;
            buffer[x] = (char*)malloc(len+1);
        }

        if( y > 0 )
        {
            buffer[x][z] = String[n];
            z++;
            buffer[x][z] = '\0';
        }
    }
    //return number of words
    return (x+1);
}

Anurag Yadav · Accepted Answer · 2015-11-04 13:29:35Z

0

String Current = "hello java example";
String trimstr=Current.trim().replace(" ", "");

int count=trimstr.length();

String tryarray[]=new String[count];
for(int i =0;i<count;i++){
tryarray[i]=trimstr.substring(i, i+1);
System.out.println(trimstr.substring(i, i+1));
}
System.out.println("arrays is "+Arrays.toString(tryarray));

answered Nov 4, 2015 at 13:29

Anurag Yadav

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1 Comment

chux Over a year ago

What is String tryarray[]=new String[count]; in this C language post?

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