C: Sorting a char array

The way I am trying to go about it is to find the lowest value in the array, concatenate it onto the end of an output string, then change that value to null so it won't be compared in the next loop iteration. I have had little success.


    char input[40], output[40], lowest, buffer[100];
    int i=0, j=0, position;

    printf("Enter a string (1-40 characters): ");
    fgets(input, 40, stdin);
   
    /*converts to lower case and deletes special characters*/
    for(j=0;j<40;j++)
    {  
        if(input[j]<91 && input[j]>64)
            input[j]+=32;
        if(input[j]<97 || input[j]>122)
            input[j]='*';
    }

    j=0;

    for(j=0;j<40;j++)
    { 
        lowest=input[j];
        position=j;
        for(i=0;i<40;i++)
        {
            if(lowest!=NULL);
            {
                if(input[i] < lowest)
                {
                    position=i;
                    lowest=input[i];
                }
            }
        }
        if(lowest!=NULL);
        {
            sprintf(buffer, "%c", cvalue);
            strcat(output, buffer);
            memset(buffer, 0, sizeof(buffer));
            input[cposition]='*';
            cvalue=NULL;
        }
    }

    printf("Output: %s", output);

So far nothing is printed.

That's supposed to be NULL, not '*' on line 13.



    printf("Output: %s", &output[0]); ?

hmm... what about an in-place recursive binary sort? ^^

>>4
^____________________________________________________________________________________^;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;

8 7 6 5 4 3 2 1

>4 3 2 1 8 7 6 5 (4<>8 / 7<>3 /...) step 4 x4

>2 1 4 3 6 5 8 7 (2<>4 / 3<>1 /...) step 2 x6

>1 2 3 4 5 6 7 8  (2<>1 / 2<|>4 /...) step 1 x7

I managed to sort it. Now I'm having trouble removing special characters from the string.

'C' is not < 91, dipshit. use casts

>>8
Actually that would make a lot of sense.
Finished btw. Not quickly enough to get some sleep though.