C编程：将Hex Int转换为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.

Answer 1

itoa is non-standard. Stay away.

One possibility is to use sprintf and the proper format specifier for hexa ie 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.

Answer 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;

Answer 3

Using the sprintf() function like this -- sprintf(charBuffer, "%x", i); -- I think will work very well.

Answer 4

Using the sprintf() function to convert an integer to hexadecimal should accomplish your task.

Here is an example:

int i = 0x11111111;

char szHexPrintBuf[10];

int ret_code = 0;

ret_code = sprintf(szHexPrintBuf, "%x", i);

if(0 > ret_code)
{ 
   something-bad-happend();
}