Yes, but is is also more difficult to understand than just checking for source[j] != '\0' or even using the concise loop that I showed you.Originally Posted by peripatein
Looks fine, but of course, you should be printing outside of my_strcat.
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Were you referring to my latest attempt, namely while (source[j] != '\0') destination[i++] = source[j++]; ? Or rather to the inversion of order?
Moreover, is there no need to "manually" add '\0' to destination at the end? Is that taken care of by the loop?
Last edited by peripatein; 05-13-2013 at 04:56 AM.
You forget to terminate "destination". Your loop ends when source[j] == '\0' thus you are not copying '\0' to "destination".Code:void my_strcat (char destination[], const char source[]) { int i = 0, j = 0; while (destination[i] != '\0') i++; destination[i++] = ' '; while (source[j] != '\0') destination[i++] = source[j++]; printf("%s\n", destination); }
Bye, Andreas
In that case, I could just stick to my original code, writing thus:
right?Code:void my_strcat (char destination[], const char source[]) { int i = 0, j = 0; while (destination[i] != '\0') i++; destination[i++] = ' '; while ((j != 0 && source[j - 1] != '\0') || !j) destination[i++] = source[j++]; }
You cannot do that source[j - 1] is invalid when j is 0.
Kurt
EDIT: I see it now, could work but it's a really strange solution
Last edited by ZuK; 05-13-2013 at 06:05 AM.
The main real goal in programming is to write maintainable code.
You failed in that goal on you last post of code on this line.
Much more complex than will be easy to maintain.Code:while ((j != 0 && source[j - 1] != '\0') || !j) destination[i++] = source[j++];
Tim S.
Last edited by stahta01; 05-13-2013 at 06:05 AM.
"...a computer is a stupid machine with the ability to do incredibly smart things, while computer programmers are smart people with the ability to do incredibly stupid things. They are,in short, a perfect match.." Bill Bryson
But it doesn't access that location once I include the condition j != 0 !! Why is that not maintainable?
What functionality does the " || !j " add to your code?
Tim S.
"...a computer is a stupid machine with the ability to do incredibly smart things, while computer programmers are smart people with the ability to do incredibly stupid things. They are,in short, a perfect match.." Bill Bryson
Without that the loop wouldn't run at all. I said it "strange solution" just to save one line of code.
Kurt
Last edited by ZuK; 05-13-2013 at 06:12 AM.
I personally find it more sophisticated. Yes, more cumbersome and complicated perhaps, strange you might add, yet this way the string (viz. destination) is added the enclosing '\0' by the while loop itself.
I could have likewise written it thus:
Code:while (!j || source[j - 1] != '\0') destination[i++] = source[j++];
Don't write code to be unnecessarily clever; write code to be clear. Be considerate to the person who will maintain your code when you're gone. Even if you're just learning, get into this habit.
My comments correspond to what I quoted, i.e., the first comment for the first snippet; the second comment for the second snippet.
You should add the null character at the end, and in fact I implied that in my earlier posts. My mistake on saying that your second code snippet is fine though, but in my defense I was too busy checking your first code snippet for correctness and did not notice that you missed that part out.
It is obfuscated, not sophisticated. Even the method that I showed you in post #14 is somewhat obfuscated, but at least it is well known. With it, you can simplify your function to:
Code:void my_strcat(char destination[], const char source[]) { int i = 0, j = 0; while (destination[i] != '\0') i++; destination[i++] = ' '; while (destination[i++] = source[j++]); }
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