Thread: Copying strings, all or nothing?

My goal is to copy only the elements of string2 that are equal to string 1 into a new string. I tested this idea with an array of integers and it worked, but didn't work for the strings. My question is: Am I missing something that would make this idea possible?

Code:

#include<stdio.h>
main()
{
    int scan1;
    char arr1[40] ;
    char arr2[40] ;
    char arr3[40] = {'_',.....,'_'}; /*for sake of brevity with post*/

    printf("line1:");
    fgets(arr1,40,stdin);

    printf("line2:");
    fgets(arr2,40,stdin);

    scan1=0;

    /**If arr1 & arr2 are equal, copy parts of arr2 that
    equal arr1 into arr3, if not equal leave those elements of arr3
    alone**/

    while(arr1[scan1]!= '\0'){
        arr3[scan1]=(arr2[scan1]==arr1[scan1]) ? arr2[scan1] : arr3[scan1];
        ++scan1;
    }

    arr3[scan1]='\0'; /**adding null after copy**/

    printf("%s", arr3);

    return 0;
    }

Expected input/output:

string1:"HELLO"
string2:"L"
string3:"__LL_"

Actual input output:

string1:"HELLO"
string2:"L"
string3:"_____"

or

string1:"HELLO"
string2:"HELLO"
string3"HELLO"

String 3 wants it all, or nothing at all.

Thank you for your time, Alpo.

Originally Posted by Alpo

My goal is to copy only the elements of string2 that are equal to string 1 into a new string.

Your goal does not make sense: an element of string 2 would be a char, therefore, no element of string 2 can be equal to string 1.

Perhaps you want to copy the elements of string 1 that can be found in string 2 into a new string? If so, your code needs to do more than just compare elements pairwise: you would need to loop over string 2 to check whether the current element of string 1 can be found in it.

I would recommend rephrasing the part inside your while loop into something less verbose. Try explaining what you're doing there in English step by step, make sure its correct, then rewrite it back to C.

Basically if the any char in string 2 is equal to a char in string 1, copy only that char from string 2 into string 3, but else wise leave string 3 chars alone. It works on this int array:

Code:

 #include<stdio.h>
 main()
 {
    int array []={3,1,3,7,5,6,3,7,8,1};
     int array2[]={3,1,3,7,5,6,3,7,7,1};
     int array3[10]={0};
    int scan1, go;
     scan1=0;
    while( scan1 < 10 ){
         array3[scan1]=(array[scan1]==array2[scan1]) ? array2[scan1] : array3[scan1];
        ++scan1;
     }
    for (go=0; go<=9; ++go){
        printf("%d", array3[go]);
    }
    return 0;
}

array 3=0,...,0 (before copy)

array=3,1,3,7,5,6,3,7,8,1

array2=3,1,3,7,5,6,3,7,7,1 (changed 9th element)

array3=3,1,3,7,5,6,3,7,0,1 (after copy)

Originally Posted by Alpo

It works on this int array:

Here's proof that it does not work:

Code:

#include<stdio.h>

int main(void)
{
    int array [] = {3, 1, 3, 7, 5, 6, 3, 7, 8, 1};
    int array2[] = {3};
    int array3[10] = {0};
    int scan1, go;
    scan1 = 0;
    while (scan1 < 10 ) {
        array3[scan1]=(array[scan1]==array2[scan1]) ? array2[scan1] : array3[scan1];
        ++scan1;
    }

    for (go = 0; go <= 9; ++go) {
        printf("%d", array3[go]);
    }
    return 0;
}

I get the output:

Code:

3000000000

instead of the expected:

Code:

3030003000

EDIT:
Actually, this is confusing. Look at what you wrote: "if the any char in string 2 is equal to a char in string 1, copy only that char from string 2 into string 3". Now, look at what you wrote:

Expected input/output:

string1:"HELLO"
string2:"L"
string3:"__LL_"

There is only one char in string2, hence only that char can possibly be copied, so your expected output must be wrong... unless you're mixing up string1 and string2. You need to be absolutely clear.

@ LaserLight, the first output was expected, only elements of array2 that match array are suppose to be copied, therefor if only a 3 matches, only a 3 is copied, the rest of array 3 should equal what it did before any copying took place.

Therefor

array = 1,2,3,4
array2= 1,3,2,4

array 3 should output= 1,0,0,4

I just don't understand why it doesn't work on char strings?

Good point Laserlight, the L's should have been in the same place. Let me go check it. Thanks.

Originally Posted by Alpo

@ LaserLight, the first output was expected, only elements of array2 that match array are suppose to be copied, therefor if only a 3 matches, only a 3 is copied, the rest of array 3 should equal what it did before any copying took place.

Let's consider:

Code:

int array [] = {3, 1, 3, 7, 5, 6, 3, 7, 8, 1};
int array2[] = {6};

What is the expected output?

What I am trying to do is to produce an example input for the integer array version that corresponds to your example input for the string version that you say is wrong, in order to demonstrate to you that your algorithm is wrong (or right, as the case may be).

SUCCESS!!! OMG, thank you laserlight.

The expected output would be all 0's on array3 in your example. Since there are no matchs.

Char array1:HELLO
Char array2:**LL*

Char array3(output):__LL_

I just wasn't matching the correct positions. Seriously thanks.

YES!!! Thanks for your hints Matticus. It finds the elements anywhere now, and I added a tolower function to make it easier.

Code:

#include<stdio.h>
char lower(char tolow[], int size)
{
    int go;
    for (go=0; go<size; ++go){
        tolow[go]=(tolow[go]>='A' && tolow[go]<='Z') ? tolow[go]  +=32 : tolow[go];
    }
 return tolow[go];
}
main()
{
    int scan1, scan2;
    char arr1[40] ;
    char arr2[40] ;
    char arr3[40] = {'_','_',...,'_'};
    printf("line1:");
    fgets(arr1,40,stdin);
    lower(arr1, 40);
    printf("line2:");
    fgets(arr2,40,stdin);
    scan1=0;
    scan2=0;
    /**If arr1 & arr2 are equal, copy parts of arr2 that
    equal arr1 into arr3, if not equal leave those elements of arr3
    alone**/
    while(arr1[scan1]!= '\0'  && arr2[scan2] !='\0'){
        arr3[scan1]=(arr2[scan2]==arr1[scan1]) ? arr2[scan2] : arr3[scan1];
        scan2=(arr2[scan2]!=arr1[scan2])? scan2++ : scan2;
        ++scan1;
    }
    arr3[scan1]='\0'; /**adding null after copy**/

        printf("%s", arr3);
 
    return 0;
    }

Originally Posted by Alpo

... I might need time to think on it.

That's the most encouraging thing I've heard from a new poster in a long time! :-)

Usually we have to tell people to do that.