How to display a hexa decimal value with a space after 4 digits

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Hi All,

I am trying to display a hexa decimal value with length ranging from (40 digits to 60 digits)
as a result of SHA1HMAC algorithm. I have to display that hexa number in the format of 4 digits and after that a space (eg: 0x1234 5678 90AB CDEF ....)

I am using the below code but i am getting an error saying

"Run-Time Check Failure #2 - Stack around the variable 'formatString' was corrupted."

Please let me know how to handle this scenario and fix this. It would be great if I dynamically format because the length of the hexa decimal number may vary(from 40 digits to 60 digits)

Appreciate your help.

Thanks,
Sudhakar

code:

C++

#ifdef _MSC_VER 
inline unsigned __int64 strtoull(const char *nptr, char **endptr, int base)
{ return _strtoui64(nptr, endptr, base); }
#endif

GDString result = "123456789ABCDEF123456789ABCDEF123456789"
GDUINT64 value = strtoull(result.c_str(),NULL,16 )  ;

GDString resultString;
char formatString[50] = {0};
		
sprintf(formatString,"0x%04X %04X %04X %04X %04X %04X %04X %04X %04X %04X %04X",
			(unsigned)(value >> 160ULL) & 0xFFFF, 
			(unsigned)(value >> 144ULL) & 0xFFFF, 
			(unsigned)(value >> 128ULL) & 0xFFFF, 
			(unsigned)(value >> 112ULL) & 0xFFFF, 
			(unsigned)(value >> 96ULL) & 0xFFFF, 
			(unsigned)(value >> 80ULL) & 0xFFFF, 
			(unsigned)(value >> 64ULL) & 0xFFFF, 
			(unsigned)(value >> 48ULL) & 0xFFFF, 
			(unsigned)(value >> 32ULL) & 0xFFFF, 
			(unsigned)(value >> 16) & 0xFFFF, 
			(unsigned)value & 0xFFFF);

		resultString = formatString;

Posted 7-Mar-14 3:30am

Member 3975629

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Comments

nv3 7-Mar-14 16:51pm

How do you expect to 11 * 4 = 44 hex digits into a 64 bit uint value? There are no more than 16 hex digits in a 64 bit value. Besides that solution 2 is a good start for what you want to do.

merano 7-Mar-14 18:11pm

You are right, something is wrong with the shift. But your comment does not catch the facts.

He tries to shift 176 Bits, converted to hex-text into a 50 char output buffer.

Both datatypes are to small for this action. See Solution 3.

value: 11 * 2 * 8 Bit = 176 Bit

formatString: 2 + 44 + 10 + 1 = 57 chars

The shift is strange and the buffer to small.

nv3 7-Mar-14 18:28pm

Yes, formatString is too short to receive the 57 character result of the printf. Nevertheless, his variable "value" is of type _int64 and hence cannot hold 176 bits in the first place.

4 solutions

Solution 2

C++

static const char lookup[] = "0123456789ABCDEF";

// value to be printed.
long long work = value;
// text output
char text[32] = {0}; // should be 21 chars max.
size_t len = (sizeof text) - 1;
do
{
    // display least significant 16 bits.
    text[--len] = lookup[work & 0xF];
    // shift down for next nibble.
    work >>= 4;
    text[--len] = lookup[work & 0xF];
    // shift down for next nibble.
    work >>= 4;
    text[--len] = lookup[work & 0xF];
    // shift down for next nibble.
    work >>= 4;
    text[--len] = lookup[work & 0xF];
    // shift down for next nibble.
    work >>= 4;
    // any non-zero bits left?
    if (work == 0)
    {
        // no: hex prefix.
        text[--len] = 'x';
        text[--len] = '0';
    }
    else
    {
        text[--len] = ' ';
    }
}
while (work);
// output is at &text[len].
fprintf(stdout, "%s", text + len);

Posted 7-Mar-14 6:15am

bling

Updated 10-Mar-14 5:42am

v2

Solution 3

formatString at 50 chars is not big enough for the result in the snippet you posted, you'll need at least 57 if I'm not mistaken.
0x = 2
11 groups of 4 digits = 44
10 spaces = 10
null terminator = 1

Posted 7-Mar-14 6:23am

jeron1

Comments

merano 7-Mar-14 17:59pm

You need at least 57 chars for output buffer, but before that you need

11 * 2 * 8 Bit = 176 Bit datatype for value, but i think there are only 64 Bit.

The shift is strange.

jeron1 7-Mar-14 18:09pm

Yeah, I wasn't really looking at the logic or datatypes, merely the usage of formatString.

Member 3975629 10-Mar-14 4:23am

Thanks for your solution. it seems working on windows fine. but on Linux it is giving the below error with the same above code though i have used unsignd long long as the data type for value .Please let me know if you have any idea/suggetion on this.

error: right shift count >= width of type

And also please let me know how to avoid hard coding of here with shifts. I mean here i used 11 groups of 4 digits by using >> operator. but is there any way that we can dynamically get the format string with different length of value ? and also what could be the max value of a int64 value ? i mean how many max digits can it have ?

Appreciate your help.

Thanks,
Sudhakar

Solution 1

I'm not a C++ specialist, at all, so I won't be able to post any code, but, as far as I know, there is no pre-made format string that could display an hexadecimal value with every four-digits groups isolated.
Maybe you could break it down to several simple steps :
1) Build an hexadecimal string from the value
2) Create a custom method that will accept your hexadecimal string as input, and return it under the form of 4-digits groups. I could easily write that in C# or in VB, but not in C++; sorry :(

Hope this helps, anyway. Good luck!

Posted 7-Mar-14 4:47am

phil.o

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Jochen Arndt · Accepted Answer · 2014-03-10T02:30:00

There is no need to convert the string into numbers. Just print out the string portions inserting spaces at the required positions:

const char *sha1 = "123456789ABCDEF123456789ABCDEF123456789";

printf("In: %s\n", sha1);
printf("Out: 0x");
size_t len = strlen(sha1);
// Get number of digits for first part if not a multiple of 4.
size_t digits = len % 4;
if (digits == 0)
    digits = 4;
const char *src = sha1;
while (*src)
{
    if (src > sha1)
        printf(" ");
    for (size_t i = 0; i < digits; i++)
        printf("%c", *src++);
    digits = 4;
}
printf("\n");

If you need the output stored in a buffer use something similar:

size_t len = strlen(sha1);
char *buf = new char[len + 4 + len / 4];
const char *src = sha1;
char *dst = buf;
*dst++ = '0';
*dst++ = 'x';
// Get number of digits for first part if not a multiple of 4.
size_t digits = len % 4;
if (digits == 0)
    digits = 4;
while (*src)
{
    if (src > sha1)
        *dst++ = ' ';
    for (size_t i = 0; i < digits; i++)
        *dst++ = *src++;
    digits = 4;
}
*dst = '\0';
printf("In: %s\nOut: %s\n", sha1, buf);
delete [] buf