Not so. sizeof(char) != sizeof(int). What you mean is that character literals like 'a' are actually int values, not char ones. And because char can be signed or unsigned by default, most character handling functions like tolower() take ints, not chars.
dwk
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Not literally, of course. They have different sizes, but they're just integers both of them, and they're treated as such:
Code:int a = 'a'; char a2 = 'a'; int b = 70; char b2 = 70;
I have tried rewriting it several times, and still the same problem occurs. Here is my latest attempt:
But it still exhibits the same behaviour. I am truly stumped. Anyone have any ideas or can suggest a better way to achieve the desired interface using a different method?Code:exitfunction() { char exitquestion; printf("\n\nDo you wanna quit?"); exitquestion = getchar(); if(exitquestion == 'N' || exitquestion == 'n') { main(); } else if(exitquestion == 'Y' || exitquestion == 'y') { exit (0); } else if(exitquestion != 'Y' && exitquestion != 'y' && exitquestion != 'N' && exitquestion != 'n') { printf("\n\nOnly [Y]es Or [N]o are acceptable inputs!"); exitfunction(); } }
Cheers.
What is this "recursion instead of loop" sickness that has hit us right now.
You DO NOT call main() again to repeat something within main. You do not call your own function to repeat it. In both cases, some form of loop construction should be used, such as while, for, do-while - even goto is better than calling recursively [but it's the least good solution, in my opinion].
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Mats
Compilers can produce warnings - make the compiler programmers happy: Use them!
Please don't PM me for help - and no, I don't do help over instant messengers.
I tried writing it as a while loop but I couldn't work out how to make it work like the above if statements should.
ie. Do you want to quit? Y/ N
N then go back to main, repeat the program again.
Y then exit.
Neither then tell user that only Y/N can be input and ask the 'Do you want to quit?' question again.
I could not think of any other way to achieve this.
So, show us the code where you are doing that. What you show above is WRONG, WRONG, WRONG!Originally Posted by jim.rattlehead
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Mats
Compilers can produce warnings - make the compiler programmers happy: Use them!
Please don't PM me for help - and no, I don't do help over instant messengers.
When do you bother to ask for help if you're not listening to us? We say "use a loop," you do "use lots of IFs."
What part do you not understand? And you have yet to actually give a return type to exitfunction. Use void exitfunction if you don't plan on returning anything from it.
I just couldn't work out how to use a loop for this situation.
I have this:
But I wouldn't know how to add the exit if 'y' and go back to main if 'n' without using if statements.Code:void exitquestion() { char exitquestion; printf("\n\nDo you wanna quit?"); exitquestion = getchar(); while (exitquestion != 'Y' && exitquestion != 'y' && exitquestion != 'N' && exitquestion != 'n') { printf("\n\nOnly [Y]es Or [N]o are acceptable inputs!"); printf("\n\nDo you wanna quit?"); exitquestion = getchar(); } }
You should have posted that before because it looks very correct.
The preferred method for doing this is to let main handle it. exitquestion should return a bool or int, with true or 1 if the user wants to quit, and 0 or false if the user wants to continue.
Inside main, have a loop that loops if exitquestion returns false or 0, otherwise it terminates the loop, cleans up and exits.
