macOS switch to catch non int

[KhuntienNang]

I'm new in objective-c (or programming in general)

so, I was trying out switch statement, say:

		switch (option){
			case 1:
				statement
				break;
			case 2:
				statement
				break;
			case 3:
				statement
				break;
			default:
				statement
				break;

option is an int type.

the default catches other int besides 123, but it doesnt catch other character (non int)

is there a way to add on the default so that it catches something other than non interger?

 

[lloyddean]

but it doesnt catch other character (non int)

is there a way to add on the default so that it catches something other than non interger?

Since you did say 'character' I'll throw out

int option = '?';

switch ( option )
{
    case 'a':
        ...
        break;
    case 'A':
        ...
        break;
    case '?':
        ...
        break;
}

 

[Bill McEnaney]

Can you your option variable be a character variable instead of an integer variable? That way, if each option number were a digit between, say, '1' and '6', the default clause would catch any character that was outside that range.

 

[mrbash]

There are two things you could do. One is to use atoi (stdlib), the other is to use int (iostream).

If you are fairly sure that you are only getting numbers, then atoi is your function.
char letter = '1';
int number = atoi(&letter);
number ->1
If you have a letter say, 'a', you would get 0.

If you want the ascii equivalent of the letter 'a', you would use:

char letter = 'a';
cout << int(letter) << endl;

would print 97.

 

[chown33]

char letter = '1';
int number = atoi(&letter);

You're taking some potentially serious risks with your "not quite a nul-terminated string" there.

 

[mrbash]

You're taking some potentially serious risks with your "not quite a nul-terminated string" there.

This should be better:

char letter = '1';
char string[2];
string[0] = letter;
string[1] = 0;
int number = atoi(string);

 

[Bill McEnaney]

Did you mean to put the integer zero into the string? Why not put the '0' character there instead?

This should be better:

char letter = '1';
char string[2];
string[0] = letter;
string[1] = 0;
int number = atoi(string);

 

[notjustjay]

Did you mean to put the integer zero into the string? Why not put the '0' character there instead?

The integer 0 is the equivalent of the NULL character. Every C-style string must consist of a number of characters (your string) followed by the NULL terminator. If your string does not include the terminator, very bad things can happen. (At best, you get garbage results -- at worst, you get either crashes or one of those security vulnerabilities that allows a hacker to inject his/her own code.)

 

[chown33]

This should be better:

char letter = '1';
char string[2];
string[0] = letter;
string[1] = 0;
int number = atoi(string);

C's quoted-string notation is legal as initializers for char arrays. So here's the same thing, more conciserishly:

char string[] = "1";
int number = atoi(string);

 

[chown33]

Did you mean to put the integer zero into the string? Why not put the '0' character there instead?

Because '0' doesn't have an integer zero value.

#include <stdio.h>
main( int argc, char** argv )
{  printf( "%d %d\n", '0', 0 );  }

http://en.wikipedia.org/wiki/C_string

 

[Bill McEnaney]

I know that, but I'm used to writing in programming languages where that assignment would have made the computer complain about a type mismatch.

Because '0' doesn't have an integer zero value.

#include <stdio.h>
main( int argc, char** argv )
{  printf( "%d %d\n", '0', 0 );  }

http://en.wikipedia.org/wiki/C_string

 

[Sander]

In that case, you'd probably have used '\0'.
Arguably, the languages you speak of make this more error-prone by this insistance, exactly as shown in your example. You want to null-terminate a string, so you assign the value 0 to its last character; the compiler doesn't like that, so you change it to '0' after which the compiler is happy (but you aren't!).
If the language has qualms about integer-to-character conversions (which is a good thing in itself), then the 0 should probably be an exception (just like it is for pointers).

This manual string composition looks rather brittle to me, anyway.

 

[Bill McEnaney]

I can ignore these details when I write in a strongly-typed language. I don't need to choose between zero and '\0' because that language won't let me put an integer zero into a string. I don't know about you. But I'd rather use a language that forbids the strange, potentially dangerous "tricks" that C allows. Mrbash was right: Unfortunately, I wrote a dangerous C-function where I assumed that an array had already been allocated. I just hadn't considered his point because it hadn't occured to me when I wrote that function. But many C programmers knowingly write dangerous code. So I wish someone would explain why those programmers do that.

In that case, you'd probably have used '\0'.
Arguably, the languages you speak of make this more error-prone by this insistance, exactly as shown in your example. You want to null-terminate a string, so you assign the value 0 to its last character; the compiler doesn't like that, so you change it to '0' after which the compiler is happy (but you aren't!).
If the language has qualms about integer-to-character conversions (which is a good thing in itself), then the 0 should probably be an exception (just like it is for pointers).

This manual string composition looks rather brittle to me, anyway.