C Programming: Convert Hex Int to Char*

Question

My question is how I would go about converting something like:

    int i = 0x11111111;

to a character pointer? I tried using the itoa() function but it gave me a floating-point exception.

Solution 1

itoa is non-standard. Stay away.

One possibility is to use sprintf and the proper format specifier for hexa i.e. x and do:

char str[ BIG_ENOUGH + 1 ];
sprintf(str,"%x",value);

However, the problem with this computing the size of the value array. You have to do with some guesses and FAQ 12.21 is a good starting point.

The number of characters required to represent a number in any base b can be approximated by the following formula:

⌈log_b(n + 1)⌉

Add a couple more to hold the 0x, if need be, and then your BIG_ENOUGH is ready.

Solution 2

char buffer[20];

Then:

sprintf(buffer, "%x", i);

Or:

itoa(i, buffer, 16);

Character pointer to buffer can be buffer itself (but it is const) or other variable:

char *p = buffer;

Comments

[Ruben]

For int (4 bytes) sufficient size is easily determinated: 9 for hex, 11 for decimal, 33 for binary.

[dirkgently]

@Ruben: Without the 0x prefix for hex, yes.

[Ruben]

Of cource, without prefix - only for number itself and \0 at the end of string.

[Ben Voigt]

@octopusgrabbus: Why did you add a cast? It provides no benefit at all, and is potentially harmful (disabling useful compiler warnings that would catch type errors during compile instead of crashing at runtime).

[octopusgrabbus]

@BenVoigt Thanks for catching this. I don't program enough in C anymore and just plain forgot that rule.

[octopusgrabbus]

The edit you approved @BenVoigt was rejected. I just submitted again.

[Robaticus]

I rolled back this answer. @octopusgrabbus - if you want to answer a question, please do so as an answer. Please don't edit another answer to add your own code.