Thread: Another noob question

hello once agin,

I am having a real hard time getting the (!negation) to sink home.
given:

Code:

ptr = Strprint;
if (!(*ptr)(str))
printf("Done!\n");

return 0;

}

Here is my problem (and I have read the chapter about it, and looked it up on google) all this true is false and false is true stuff. why are the vaules returned swaped out like that? in the if (!(*ptr)(str)) that means if it is true it moves to the next line? and if it is false, it just returns a vaule?

Once agin sorry for such basic questions here.

Code:

#include <stdio.h>

int StrPrint(char *str);

main()
{
       char str[24] = "Pointing to a function.";
       int (*ptr)(char *str);

       ptr = StrPrint;
       if (!(*ptr)(str))
             printf("Done!\n");
     
       return 0;
}

int StrPrint(char *str)
{
        printf("%s\n", str);
        return 0;
}

Cleaned it up a little:

Code:

#include <stdio.h>

// Change function to accept constant char *s, as you're not changing the      
// passed-in value in the function. Added a second function for demonstration purposes.
int StrPrint(const char *str);
int AsteriskPrint(const char *str);

// Make a nice typedef for our functions, so it's a bit more                    
// intuitive to read and write                                                  
typedef int (*MyPrintFunc)(const char *str);

int main(void)
{
    char str[24] = "Pointing to a function.";

    // Create a pointer to one of our functions                                 
    MyPrintFunc printFunc = StrPrint;

    // Then use it                                                              
    if (!printFunc(str))
        printf("Done!\n");

    // Reassign function pointer to second function                             
    printFunc = AsteriskPrint;

    // And use it                                                               
    if (!printFunc(str))
        printf("Done!\n");

    return 0;
}

int StrPrint(const char *str)
{
    // Return the number of characters printed, instead of just 0.              
    return printf("%s\n", str);
}

int AsteriskPrint(const char *str)
{
    return printf("***%s***\n", str);
}

An excellent page on function pointers

Originally Posted by Gikimish

Thank you very much for the ex.'s, and I am sorry I didnt make it very plain in the first post as to my quary.

I understand function pointers completly, my quary is the !negation.

I am having trouble with how, if the operand evalutes to nonzero vaule it is logically false, and when the operator yields 1 its logically true, why is this? pretty much what I am saying is I am haveing a problem looking at any:

In C the if statement only completes on boolean True... that is to say: if (this is true) do this.

When you use negation in a conditional statement you flip true to false so the statement becomes: if (this is not true) do this.

Try this for an extremely simple example:

Code:

int main(void)
     { int x;
         // this is expected   
        x = 9;
         if ( x < 10)
           puts("X is less than 10")    

        // this is negated
        x = 11;
       if (! (x < 10))
          puts("X is not less than 10")

}

In the first statement the if statement resolves to true -- x is less than 10-- so it prints out.
In the second statement if resolves to false --x is not less than 10-- it should not print but false negated is true... so it prints.

Does that help?

Originally Posted by Gikimish

Very much Thank you very very much!!!

But it gave me another question,

Why do this at all, I mean what power does the !negation operator give in the larger picture.
I mean that would be very simple with:

In the demonstration example it's pretty useless and you are correct, else would take care of the second part. But consider a larger program where you are using settings and flags to instigate certain behaviours... For example you have a setting for "Print out results"... your test could be a simple negation...

Code:

int main(void)
   { int PrinterResults = 0;  // no printer

      // lots of intervening code

      if (! PrinterResults)
        puts("the results are...");
}

... used to display the result on the screen if you don't want to waste paper.

Other times you may want to test the condition of a status flag... if (!systemerror)... but there's no reason to have an else statement or perhaps the only condition you care about at that moment is the false one.

As you write bigger and more complex programs you'll find a ton of uses for this little gem.

Thank you very very much after a long road I finally understand it and the power it gives.

Originally Posted by Gikimish

Thank you very very much after a long road I finally understand it and the power it gives.

Glad I could help...