如何将字符串从C传递到swift？

Question

I am using OpenSSL as a library in swift. Since it's in C, I need to pass some data from C language back to swift. The data was an array, but after some research I have learned that I cannot pass an array from C to swift. One way I can think of it is to change the array into a string, an then pass the string back to swift. In CI wrote:

char teststring()
{
    char c[]={"H","E","L","L","O"};
    printf(c);
    return c;
}

In Swift I wrote:

teststring()

The result is that it only prints H . What did I do wrong with it? It there a easier way to pass an array from C back to Swift?

Answer 1

You might want to change

char c[]={"H","E","L","L","O"};

to

char c[]={'H','E','L','L','O','\0'};

Or

char c[]="HELLO"; 

By char c[]={"H","E","L","L","O"}; an array of strings is getting declared, rather than an array of characters. There is a difference between a string and a char.

A string literal enclosed within a double quotes represents a set of characters. So, "H" seems to represent a single character to you, but internally it contains an additional character with 'H' and the very necessary '\\0' which is terminating character in C. That was possibly the reason that you were getting a H printed on your console as output.

This is because as per C11 standard - Section: 6.4.5 String literals which states,

A character string literal is a sequence of zero or more multibyte characters enclosed in double-quotes, as in "xyz". A UTF-8 string literal is the same, except prefixed by u8. A wide string literal is the same, except prefixed by the letter L, u, or U.

Additionally, section: 6.4.4.4 Character constants states that,

An integer character constant is a sequence of one or more multibyte characters enclosed in single-quotes, as in 'x'. A wide character constant is the same, except prefixed by the letter L, u, or U. With a few exceptions detailed later, the elements of the sequence are any members of the source character set; they are mapped in an implementation-defined manner to members of the execution character set.

Also,

char teststring()

means that teststring is supposed to return a char , and you are returning c which is now a string. So either change it to return a pointer to char ie

char* teststring()

Or, return a char like return c[0];

Answer 2

there's a couple things...

First, it'd be a lot easier to format your string this way:

char* c="HELLO";

But, if you want to do it your way still, you need to null-terminate your string, otherwise the computer will not know where the string ends and new memory begins:

char c[]={"H","E","L","L","O"};

becomes:

char c[]={'H','E','L','L','O',0};

See the single quotes vs the double quotes? The double quotes indicate a whole string (null terminated with zero)... the single quotes indicate a single char.

So when you did it with double quotes, in characters in memory for the whole thing looked something like this:

H0E0L0L0O0

With single quotes, the characters in memory looks like this:

HELLO0

The printf function will print until it hits a zero and then quit. That's why you were getting just the 'H' printing out.

Answer 3

Yes you have to declare array of char s.

So the it will be

Case-1

char c[]={'H','E','L','L','O','\0');

Here you have created something like array of array of chars. Coorect way of doing it would be.

Case-2

char c[][2]={"H","E","L","L","O"};

But in this case you have to pass c[0] or c[1] in printf .

Also note that the first thing can alternatively done using

Case-3

   char c[]="HELLO";

This is an array of 6 characters, \\0 included.

And once again you could have also done

Case-4

   char *c[]={"H","E","L","L","O"};

Here you have to pass the c[0] or c[1] which will print the string "H" or "E" .

Also note that

char c[]={'H','E','L','L','O'};

is a valid initialization but this is not null terminated. So you can't pass it where a null terminated char array is expected. (like strlen() etc).

Now the case-2, here the strings are modifiable. So you can change the character in them where as the strings in case-4 are not. Those are string literals which are non-modifiable.

Simply put if you wanted to have a string which contains those as characters then Case-1 and case-3 are the way for you.

Answer 4

c is an array and when you return c you return first element. You need to return entire array. Check

printf("%s", c);
return *c;

And remember to change signature of your function