retreiving variable from within loop.

This is a discussion on retreiving variable from within loop. within the C Programming forums, part of the General Programming Boards category; Hmm, the program compiles when I fix this error: Code: char key[16] = "aeiluozcgraolifi"; //27 initializer-string for array of chars ...

  1. #16
    The larch
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    May 2006
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    Hmm, the program compiles when I fix this error:
    Code:
    char key[16] = "aeiluozcgraolifi";
    //27 initializer-string for array of chars is too long
    Other than that there may be better ways to get the size of the file than getting 150000 characters and then counting how many there actually is.

    And then
    Code:
    for(x = 0; x < count; x++){
                encrypt[x] = file[x] ^ key[x];
                }
    The length of key was only 16. On my particular run the program didn't crash, but since there apparently were 0-bites beyond key, only the beginning of the text got encrypted.
    I would use x%KEYSIZE to get the index for key[] in the loop.
    Last edited by anon; 08-04-2006 at 05:39 PM.

  2. #17
    Registered User
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    Alright, Ill try these suggestions, thanks.

  3. #18
    Just Lurking Dave_Sinkula's Avatar
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    So I gather you're experimenting with XOR encryption? (Sometimes it's easier -- and better -- to state the problem, and/or the "big picture", rather than perceived issues with assumed solutions.) Anyways, a number of your latest posts seems to point me toward that.

    And it's been kinda slow, and I've never written one that I can remember, so I decided to play along at home. Here is something to compare with after you've finished or as you continue developing your code. I'm sure others will be along soon to nitpick it as well.
    Code:
    #include <stdio.h>
    /*
     * argv[1] - name of input file to encrypt/decrypt
     * argv[2] - name of output file
     * argv[3] - crypt key
     */
    int main(int argc, char *const *argv)
    {
       if ( argc == 4 )
       {
          FILE *input  = fopen(argv[1], "rb");
          FILE *output = fopen(argv[2], "wb");
          if ( input != NULL && output != NULL )
          {
             unsigned char buffer[BUFSIZ];
             size_t count, i, j = 0;
             do {
                count = fread(buffer, sizeof *buffer, sizeof buffer, input);
                for ( i = 0; i < count; ++i )
                {
                   buffer[i] ^= argv[3][j++];
                   if ( argv[3][j] == '\0' )
                   {
                      j = 0; /* restart at the beginning of the key */
                   }
                }
                fwrite(buffer, sizeof *buffer, count, output);
             } while ( count == sizeof buffer );
             fclose(input);
             fclose(output);
          }
       }
       return 0;
    }
    Doing something like this...
    H:\Forums>xorcrypt main.c file.bin key
    H:\Forums>xorcrypt file.bin file.txt key
    You should find that main.c matches file.txt.
    7. It is easier to write an incorrect program than understand a correct one.
    40. There are two ways to write error-free programs; only the third one works.*

  4. #19
    Registered User
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    Code:
    for(x = 0; x < count; x++){
                encrypt[x] = file[x] ^ key[x];
                }
                printf("%s", encrypt);
    KABLAM.

    Surely one of the XOR operations will produce zero, by chance. encrypt[x] is non-zero up to a certain x. You will only see the first x characters printed. After that, well, there goes. printf() will die there and then because it has encountered a zero character.

    Listen everyone, it gets better...

    And what if no zero character is produced? Then printf() will print all 150,000 bytes of encrypt and perhaps more, until it does reach a zero.


    Toodles.
    ~
    Last edited by jafet; 08-05-2006 at 09:28 PM.
    Code:
    #include <stdio.h>
    
    void J(char*a){int f,i=0,c='1';for(;a[i]!='0';++i)if(i==81){
    puts(a);return;}for(;c<='9';++c){for(f=0;f<9;++f)if(a[i-i%27+i%9
    /3*3+f/3*9+f%3]==c||a[i%9+f*9]==c||a[i-i%9+f]==c)goto e;a[i]=c;J(a);a[i]
    ='0';e:;}}int main(int c,char**v){int t=0;if(c>1){for(;v[1][
    t];++t);if(t==81){J(v[1]);return 0;}}puts("sudoku [0-9]{81}");return 1;}

  5. #20
    Registered User
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    so I should use puts()?

  6. #21
    ATH0 quzah's Avatar
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    Oct 2001
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    Quote Originally Posted by jafet
    Code:
    for(x = 0; x < count; x++){
                encrypt[x] = file[x] ^ key[x];
                }
                printf("%s", encrypt);
    KABLAM.

    Surely one of the XOR operations will produce zero, by chance. encrypt[x] is non-zero up to a certain x. You will only see the first x characters printed. After that, well, there goes. printf() will die there and then because it has encountered a zero character.
    No it won't:
    Code:
    #include<stdio.h>
    int main( void )
    {
        char foo[] = { 0 };
        printf("%s", foo );
        return 0;
    }
    It will simply print nothing. A string is defined as zero or more characters, terminated by a null character. If all you have is the null, you have what is considered an empty string. It's still a legal string. Thus, any Standard C function which handles strings, will still be able to handle it.
    Quote Originally Posted by jafet
    And what if no zero character is produced? Then printf() will print all 150,000 bytes of encrypt and perhaps more, until it does reach a zero.
    So? As long as you're not running off the end of your allocated space, who cares how much it prints? It's still legal, assuming you are in fact staying in bounds of your allocated space.


    Quzah.
    Last edited by quzah; 08-06-2006 at 09:47 AM.
    Hope is the first step on the road to disappointment.

  7. #22
    Registered User
    Join Date
    Mar 2006
    Posts
    725
    Yep, so printf only prints part of the string... that's prolly why the OP thought the rest of the data had "vanished"...
    Code:
    #include <stdio.h>
    
    void J(char*a){int f,i=0,c='1';for(;a[i]!='0';++i)if(i==81){
    puts(a);return;}for(;c<='9';++c){for(f=0;f<9;++f)if(a[i-i%27+i%9
    /3*3+f/3*9+f%3]==c||a[i%9+f*9]==c||a[i-i%9+f]==c)goto e;a[i]=c;J(a);a[i]
    ='0';e:;}}int main(int c,char**v){int t=0;if(c>1){for(;v[1][
    t];++t);if(t==81){J(v[1]);return 0;}}puts("sudoku [0-9]{81}");return 1;}

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