CODEWITHSUNDEEP
cscanf
RobotsNeverSleep
RobotsNeverSleep

Reputation: 79

scanf characters into an int without conversion

How can one use scanf to scan in an integer amount of characters and simply stuff them into an unsigned int without conversion?

Take an example, I have the following input characters (I have put them in hex for visibility):

5A 5F 03 00 FF FF 3D 2A

I want the first 4 (because 4 char's fit in an int). In base 10 (decimal) this is equal to 221018 (big-endian). Great! That's what I want in my int. This seems to work as expected:

scanf("%s", &my_integer);

Somehow it seems to get the endianness right, placing the first character in the LSB of the int (why?). As you would expect however this produces a compiler warning as the pointer must be to a character array (man 3 scanf).

An alternate approach without using scanf():

for (int i = 0; i < 4; i++)
{
    my_integer |= (getchar() << i * 8);
}

Note that I don't intend to do any conversion here, I simple wish to use the pointer type to specify how many characters to read. The same is true if &my_integer was a long, I would read and store eight characters.

Simple really.

It appears my idea behind the use of scanf isn't correct and there must be a better approach. How would you do it?

N.B. I'm aware type sizes are architecture dependent.

Upvotes: 2

Views: 490

Answers (1)

Jens
Jens

Reputation: 72639

So you want to read 4 bytes from stdin and use them as they are as the representation of a 32-bit big-endian value:

int my_integer;

if (fread (&my_integer, sizeof my_integer, 1, stdin) != 1) {
    /* Some problem... */
}

Upvotes: 3

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