macOS Please diagnose my pointer problem

[orangezorki]

Dear all,

I'm writing a program for a project that revolves around the need to dynamically allocate a big chunk of memory. But for whatever reason, and despite much searching, reading the relevant chapters in the books I have, I am still unable to do something as easy as setting the value of a char in the memory allocated. For example:

int main(int argc, const char * argv[])
{

    unsigned char* ptr;
    
    ptr = calloc(1000, sizeof(char));
    
    *ptr  = 0;
    *(ptr+1) = 1;
    *(ptr + 2) = 2;
    *(ptr + 3) = 3;
    
    printf("%c" , *(ptr + 2) );
    
    return 0;
}

This prints an upside down question mark - not what I expected!

Any ideas on what stupid thing I'm doing?

Many thanks,

David

 

[subsonix]

Use %d instead of %c. %c prints the variable as a character mapping to ascii, ascii values in that region are not printable characters:

     The decimal set:

       0 nul    1 soh    2 stx    3 etx    4 eot    5 enq    6 ack    7 bel
       8 bs     9 ht    10 nl    11 vt    12 np    13 cr    14 so    15 si

Edit:

If you actually intend to use characters, you need to enclose the values in single quotes:

    *ptr  = '0';
    *(ptr+1) = '1';
    *(ptr + 2) = '2';
    *(ptr + 3) = '3';
    
    printf("%c" , *(ptr + 2) );

 

[chown33]

First, what did you expect?

You've stored the binary value 2 at *(ptr+2). You're printing this value with the %c format, which will output it as a single character. Look up the ASCII character code for that value.

If you wanted it to print as the numeric string "2", then don't use %c.


Second, you can subscript the pointer, like this:

    ptr[0]  = 0;
    ptr[1] = 1;
    ptr[2] = 2;
    ptr[3] = 3;

This is a universal rule in C: pointers can always be subscripted. It's not limited to just calloc'ed memory.

 

[orangezorki]

Thank you so much - I said it would be a stupid mistake!

I do want to store just a value in the char, and had tried ptr [2] = 2; but did it the long way to try and work out the problem.

David