Thread: printing char **array;

  1. #1
    Registered User
    Join Date
    Feb 2008
    Posts
    93

    printing char **array;

    Code:
    #include <stdio.h>
    
    int main(void)
    {
    
    char **array;
    
     while ( **array != '\0' ){
        printf("%s",*array);
        printf("\n");
        **array++;
        if (**array == '\0')
        break;
     }
    
    return 0;
    
    }


    This is printing the words in the array but it won't end. It creates an infinite loop. I always thought a string ended in a null character. Its like it can't find the null character. Anybody know why this would create an infinite loop?

  2. #2
    Deathray Engineer MacGyver's Avatar
    Join Date
    Mar 2007
    Posts
    3,210
    Well, this isn't real code.... please try to come up with a small, but real example that actually demonstrates the problem.

    One issue I will say right away is that you're using a char **. There is no requirement that a char ** have a terminating '\0' since it is not a string in and of itself. Also.... **array++; is probably not really what you want.

  3. #3
    Registered User
    Join Date
    Feb 2008
    Posts
    93
    I think I figured it out. Thanks anyway.

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