I have this number in hexadecimal:
FFFFFFFFFFFF
and I need to save it, so I used double
double a=0xffffffffffff;
but I need to print it and I don't know how to. Each time I use %f, %d, %x, it doesn't print the value of it; I just need to print ffffffffffff. My code:
int main()
{
double a=0xffffffffffff;
printf("%x\n",a);
printf("%d\n",a);
printf("%X\n",a);
printf("%f\n",a);
return 0;
}
The only true value is %f; that returns the decimal value of the hexadecimal — it returns this:
ffffffe0
-32
FFFFFFE0
281474976710655.000000
with this I need to change from that hexadecimal to string, to compare it, because I have FFFFFFFFFFFF in string too and I need to compare both. If I can't printf it, neither will sprintf will work.
That's an integer, and a long one. Don't use double to store such a value, that's for floating-point.
Just use:
unsigned long long temp = 0xffffffffffffull;
You have 12 hexadecimal digits, so your number needs at least 12 * 4 = 48 bits. Most platforms should have an unsigned long long of 64 bits, which should be fine.
If your compiler is supported enough to support C99, you can do:
#include <stdint.h>
and then use the uint_least64_t type as suggested in a comment. In Linux I guess you're using GCC so you should be fine, you might need to specify that you intend your code to be compiled as C99 (with -std=c99).
To print it, use:
printf("temp=%llx\n", temp);
Also note that the value itself is not hexadecimal, but you can print it as hexadecimal. THe value is just a value, the base matters only when converting to/from text, i.e. an external representation of the number. Internally on a binary computer, the number is stored in binary.
While you store the value in a double, there's no sane way to get hexadecimal output.
Your code passes a double to printf() and repeatedly tries to format that as a plain int; that is not going to work well. If you used GCC with warnings enabled, you should have got warnings about mismatches between format string and value (and if you use GCC without warnings enabled, repent your evil/lazy ways and turn the compiler warnings on — use -Wall at least, and probably -Wextra too, and fix any resulting warnings).
The value is 12 F's, so it is a long long value (or unsigned long long); store it and treat it as such:
int main()
{
unsigned long long a = 0xFFFFFFFFFFFF;
printf("%llx\n", a);
printf("%lld\n", a);
printf("%llX\n", a);
printf("0x%llX\n", a);
return 0;
}
An old question, but I think the OP was trying to get to a hex string representation of a double and back again.
If so, here's some code to do that:
union {
double dbl;
uint8_t i8[sizeof(double) + 1];
} buf;
printf("sizeof(dbl) = %ld\n", sizeof(double));
memset(buf.i8, 0x00, sizeof(buf.i8));
buf.dbl = 3.14159265359;
printf("%f\n", buf.dbl);
for(int i = 0; i < (int) sizeof(buf.i8) - 1; ++i) {
printf("%02X", buf.i8[i]);
}
printf("\n");
The resulting output is:
sizeof(dbl) = 8
3.141593
EA2E4454FB210940
To read in the 16 character string and get the double back:
const char* str = "EA2E4454FB210940";
int j = 0;
for (int i = 0; i < (int) strlen(str); i += 2) {
uint8_t val = 0;
for (int k = 0; k < 2; ++k) {
char ch = str[i + k];
val *= 16;
if (ch >= '0' && ch <= '9') {
val += ch - '0';
} else if (ch >= 'A' && ch <= 'F') {
val += ch - 'A' + 10;
}
}
buf.i8[j++] = val;
}
for(int i = 0; i < (int) sizeof(buf.i8) - 1; ++i) {
printf("%02X", buf.i8[i]);
}
printf("\n");
printf("%f\n", buf.dbl);
The output is:
EA2E4454FB210940
3.141593
Note that this is on OSX, different platforms may not have the same binary representation for a double. Also, not sure what happens with NaN and other oddball values.
In another context, his question is valid. Say you want to print the output of the pow() function in hexadecimal - I just encountered one now.
printf( "pow(a, b) is : %f\n", pow(a, b) ); //Will give decimal value
printf( "pow(a, b) is : %X\n", pow(a, b) ); //Not allowed
So do
unsigned long long int temp;
temp = pow(a, b);
printf( "pow(a, b) is : %X\n", temp); //Will give hexadecimal value
Related
I got large HEX string in result into int i could be more than 10 ^ 30, and I converted in hex. I need sum (3 hex string) and remove last 12 numbers.
hex example "000000000000000000000000bd4c61f945644cf099d41ab8a0ab2ac5d2533835", "000000000000000000000000000000000000000000000000f32f5908b7f3c000", "00000000000000000000000000000000000000000000000000e969cd49be4000". And I need to sum them and get result into int. Thank you
I "made" a little two functions and they work but i think could be better, and they dont convert to normal integer number
// convert hex to unsigned char decimal
unsigned char div10(unsigned char *hex, unsigned size)
{
unsigned rem = 0;
for(int i = 0; i < size; i++)
{
unsigned n = rem * 256 + hex[i];
hex[i] = n / 10;
rem = n % 10;
}
return rem;
}
unsigned char hex_to_dec_summer(char *local){
unsigned char result[32]={0};
unsigned char output[18]={};
char input[64];
strcpy(input, local);
unsigned char hexnr[sizeof(input)/2]={};
for (int i=0; i<sizeof(input)/2; i++) {
sscanf(&input[i*2], "%02xd", &hexnr[i]);
}
unsigned char hexzero[32] = {0};
unsigned i = 0;
while(memcmp(hexnr, hexzero, sizeof(hexnr)) != 0 && i < sizeof(result))
{
result[sizeof(result) - i - 1] = div10(hexnr, sizeof(hexnr));
i++;
}
printf("\n");
for(unsigned j = 0; j < sizeof output; j++)
{
output[j]=result[j];
printf("%d", output[j]);
}
output[18]='\0';
}
I know how its make in python3 -> int(hex_number, 16)/(10**12) - like that but i need it in c
The reason this sort of thing works so easily in Python is that, unusually, Python supports arbitrary-precision integers natively.
Most languages, including C, use fixed sizes for their native types. To perform arbitrary-precision arithmetic, you generally need a separate library, such as GMP.
Here is a basic example of using GMP to solve your problem:
#include <stdio.h>
#include <gmp.h>
char *inputs[] = {
"000000000000000000000000bd4c61f945644cf099d41ab8a0ab2ac5d2533835",
"000000000000000000000000000000000000000000000000f32f5908b7f3c000",
"00000000000000000000000000000000000000000000000000e969cd49be4000"
};
int main()
{
char outstr[100];
mpz_t x; mpz_init(x);
mpz_t y; mpz_init(y);
mpz_t sum; mpz_init(sum);
mpz_t ten; mpz_init_set_si(ten, 10);
mpz_t fac; mpz_init(fac);
mpz_pow_ui(fac, ten, 12); /* fac = 10**12 */
int i;
for(i = 0; i < 3; i++) {
mpz_set_str(x, inputs[i], 16);
mpz_tdiv_q(y, x, fac);
mpz_add(sum, sum, y); /* sum += x / fac */
}
printf("%s\n", mpz_get_str(outstr, 10, sum));
}
The code is a bit verbose, because arbitrary-precision integers (that is, variables of type mpz_t) have nontrivial memory allocation requirements, and everything you do with them requires explicit function calls. (Working with extended types like this would be considerably more convenient in a language with good support for object-oriented programming, like C++.)
To compile this, you'll need to have GMP installed. On my machine, I used
cc testprog.c -lgmp
When run, this program prints
1080702647035076263416932216315997551
Or, if I changed 10 to 16 in the last line, it would print d022c1183a2720991b1fea332a6d6f.
It will make a slight difference whether you divide by 1012 and then sum, or sum and then divide. To sum and then divide, you could get rid of the line mpz_tdiv_q(y, x, fac) inside the loop, change mpz_add(sum, sum, y) to mpz_add(sum, sum, x), and add the line
mpz_tdiv_q(sum, sum, fac);
outside the loop, just before printing.
It's fairly straight forward to add up the (in this case hex) digits of two strings.
This doesn't try to be "optimal", but it does give a sum (as a string of hex digits). vals[0] acts as the accumulator.
When OP clarifies what is meant by "I need sum (3 hex string) and remove last 12 numbers", this answer could be extended.
If more speed is needed, the accumulator could be allocated and used as an array of uint8_t's (saving converting back to ASCII hex until a final total is available.) Also the LUT to convert ASCII hex to '0-F' could be 'binary' (not requiring the subtraction of ASCII character values.)
Anyway...
#include <stdio.h>
char *vals[] = {
"000000000000000000000000bd4c61f945644cf099d41ab8a0ab2ac5d2533835",
"000000000000000000000000000000000000000000000000f32f5908b7f3c000",
"00000000000000000000000000000000000000000000000000e969cd49be4000",
};
char *frmHex =
"................................................0000000000......"
".777777..........................WWWWWW.........................";
char *tohex = "0123456789ABCDEF";
void addTo( char *p0, char *p1 ) {
printf( " %s\n+ %s\n", p0, p1 );
char *px = p0 + strlen( p0 ) - 1;
char *py = p1 + strlen( p1 ) - 1;
for( int carry = 0; px >= p0 && py >= p1; px--, py-- ) {
int val = *px - frmHex[ *px ] + *py - frmHex[ *py ] + carry;
carry = val / 0x10; *px = tohex[ val % 0x10 ];
}
printf( "= %s\n\n", p0 );
}
int main() {
addTo( vals[ 0 ], vals[ 1 ] );
addTo( vals[ 0 ], vals[ 2 ] );
return 0;
}
Output
000000000000000000000000bd4c61f945644cf099d41ab8a0ab2ac5d2533835
+ 000000000000000000000000000000000000000000000000f32f5908b7f3c000
= 000000000000000000000000BD4C61F945644CF099D41AB993DA83CE8A46F835
000000000000000000000000BD4C61F945644CF099D41AB993DA83CE8A46F835
+ 00000000000000000000000000000000000000000000000000e969cd49be4000
= 000000000000000000000000BD4C61F945644CF099D41AB994C3ED9BD4053835
If this were to progress (and use binary accumulators), 'compaction' after summing would quickly lead into integer division (that could be done simply with shifting and repeated subtraction.) Anyway...
I've made function that prints memory address to the stdout but it dosent work the sameway as printf("%p", ...). For example the printf gives me this: 0x7fff58a2d2bb but my function prints something like this: 58a2d2bb. I know what 0x mean but cant get what 7fff mean. Can someone explain me what does that part mean and how can I add it to my code:
char *ft_itoa_base(uintmax_t num, uintmax_t base){
int i;
uintmax_t val_cp;
uintmax_t rem;
char *str;
val_cp = num;
i = 1;
while((val_cp /= base) >= 1)
i++;
str = ft_strnew(i); // Basically what it does (char*)malloc(sizeof(char) * (i + 1))
str[i] = '\0';
while(i-- > 0)
{
rem = num % base;
str[i] = (rem > 9)? (rem-10) + 'a' : rem + '0';
num /= base;
}
return (str);}
char a = 'x';
void* p0 = &a;
uintmax_t i = (uintmax_t)p0;
ft_putstr(ft_itoa_base(i, 16));
ft_putchar('\n');
printf("PrintF: %p", p0);
What does 0x7fff mean?
The missing "0x7fff" is part of the correct hexadecimal address lost in errant code of ft_itoa_base().
OP's original ft_itoa_base() was not fully uintmax_t ready given one of the variables was int instead of uintmax_t.
That caused the output to print a truncated address of "58a2d2bb" rather than the correct "7fff58a2d2bb".
The posted corrected ft_itoa_base() has minor weaknesses.
// char* ft_itoa_base(uintmax_t num, uintmax_t base) {
// No need for base to be so wide with uintmax_t
char* ft_itoa_base(uintmax_t num, int /* or unsigned */ base) {
int i;
uintmax_t val_cp;
// uintmax_t rem;
// No need for rem to be so wide with uintmax_t
int /* or unsigned */ rem;
char *str;
...
0x7fff58a2d2bb is 140,734,680,453,819 expressed in hexadecimal (or base 16) format where the digit a represents 10, b represents 11, and up to f which represents 15.
Base 16 or hexadecimal format is preferred over base 10 for memory addresses because 16 is a power of 2 which makes it handy for viewing as bit masks, and 10 is not which makes it difficult to view in terms of bit masks.
I have written a C program which uses two different algorithms to convert a string constant representing a numeric value to its integer value. For some reasons, the first algorithm, atoi(), doesn't execute properly on large values, while the second algorithm, atoi_imp(), works fine. Is this an optimization issue or some other error? The problem is that the first function makes the program's process to terminate with an error.
#include <stdio.h>
#include <string.h>
unsigned long long int atoi(const char[]);
unsigned long long int atoi_imp(const char[]);
int main(void) {
printf("%llu\n", atoi("9417820179"));
printf("%llu\n", atoi_imp("9417820179"));
return 0;
}
unsigned long long int atoi(const char str[]) {
unsigned long long int i, j, power, num = 0;
for (i = strlen(str) - 1; i >= 0; --i) {
power = 1;
for (j = 0; j < strlen(str) - i - 1; ++j) {
power *= 10;
}
num += (str[i] - '0') * power;
}
return num;
}
unsigned long long int atoi_imp(const char str[]) {
unsigned long long int i, num = 0;
for (i = 0; str[i] >= '0' && str[i] <= '9'; ++i) {
num = num * 10 + (str[i] - '0');
}
return num;
}
atoi is part of C standard library, with signature int atoi(const char *);.
You are declaring that a function with that name exists, but give it different return type. Note that in C, function name is the only thing that matters, and the toolchain can only trust what you tell in the source code. If you lie to the compiler, like here, all bets are off.
You should select different name for your own implementation to avoid issues.
As researched by #pmg, C standard (link to C99.7.1.3) says, using names from C standard library for your own global symbols (functions or global variables) is explicitly Undefined Behavior. Beware of nasal demons!
Ok there is at least one problem with your function atoi.
You are looping down on an unsigned value and check if its bigger equal zero, which should be an underflow.
The most easy fix is index shifting i.e.:
unsigned long long int my_atoi(const char str[]) {
unsigned long long int i, j, power, num = 0;
for (i = strlen(str); i != 0; --i) {
power = 1;
for (j = 0; j < strlen(str) - i; ++j) {
power *= 10;
}
num += (str[i-1] - '0') * power;
}
return num;
}
Too late, but may help. I did for base 10, in case you change the base you need to take care about how to compute the digit 0, in *p-'0'.
I would use the Horner's rule to compute the value.
#include <stdio.h>
void main(void)
{
char *a = "5363", *p = a;
int unsigned base = 10;
long unsigned x = 0;
while(*p) {
x*=base;
x+=(*p-'0');
p++;
}
printf("%lu\n", x);
}
Your function has an infinite loop: as i is unsigned, i >= 0 is always true.
It can be improved in different ways:
you should compute the length of str just once. strlen() is not cheap, it must scan the string until it finds the null terminator. The compiler is not always capable of optimizing away redundant calls for the same argument.
power could be computed incrementally, avoiding the need for a nested loop.
you should not use the name atoi as it is a standard function in the C library. Unless you implement its specification exactly and correctly, you should use a different name.
Here is a corrected and improved version:
unsigned long long int atoi_power(const char str[]) {
size_t i, len = strlen(str);
unsigned long long int power = 1, num = 0;
for (i = len; i-- > 0; ) {
num += (str[i] - '0') * power;
power *= 10;
}
return num;
}
Modified this way, the function should have a similar performance as the atoi_imp version. Note however that they do not implement the same semantics. atoi_pow must be given a string of digits, whereas atoi_imp can have trailing characters.
As a matter of fact neither atoi_imp nor atoi_pow implement the specification of atoi extended to handle larger unsigned integers:
atoi ignored any leading white space characters,
atoi accepts an optional sign, either '+' or '-'.
atoi consumes all following decimal digits, the behavior on overflow is undefined.
atoi ignores and trailing characters that are not decimal digits.
Given these semantics, the natural implementation or atoi is that of atoi_imp with extra tests. Note that even strtoull(), which you could use to implement your function handles white space and an optional sign, although the conversion of negative values may give surprising results.
I have a problem to convert integer type's double rage number to binary as the below,
void intToBin(int digit) {
int b;
int k = 0;
char *bits;
int i;
bits= (char *) malloc(sizeof(char));
while (digit) {
b = digit % 2;
digit = digit / 2;
bits[k] = b;
k++;
}
for ( i = k - 1; i >= 0; i--) {
printf("%d", bits[i]);
}
}
But as you can see the that function's arguments input is integer.
I came across the error when I tried with intToBin(10329216702565230)
because 10329216702565230 is over integer range.
How can I extend what that have integer type's double rage number to binary ?
update
I've updated the below code
void intToBin(uint64_t digit) {
int b;
int k = 0;
char *bits;
int i;
bits = malloc(sizeof digit * 64);
while (digit) {
b = digit % 2;
digit = digit / 2;
bits[k] = b;
k++;
}
for ( i = k - 1; i >= 0; i--) {
printf("%d", bits[i]);
}
}
But I didn't get it what should I do to get the 2's complement ?
m
dmnngn
Solution is to use type which supports that range of numbers.
Use unsigned long long or uint64_t(assuming you are passing non negative integers, otherwise use long long or int64_t). Then you call the function like this Edited to add int64_t to uint64_t from the comment posted. unsigned long long is 64 bits atleast - can even be wider. With OP's comment of getting 64 bits output - better to use (u)int64_t
intToBin(10329216702565230U)
In case you want to use negative numbers use long long.Call it like this
intToBin(10329216702565230L).
You didn't allocate enough memory - you were accessing memory that you haven't allocated, resulting in Undefined behavior. You have allocated 1 char first and then you didn't allocate. You can solve this by reallocating - reallocate memory inside the loop (reallocate 1 char at a time inside loop). And then use it. Instead of calling realloc multiple times why don't you allocate memory for 64 chars and then use it to store the result. And in the end, the left over space can be freed with another realloc call.
You don't need to cast the return value of malloc (void* to char* conversion is done implicitly).
You didn't check the return value of malloc. malloc may return NULL and in that case you have to handle that separately. For example:-
#define NBITS 64
...
...
bits = malloc(NBITS);
if( bits == NULL ){
perror("malloc failed");
exit(EXIT_FAILURE);
}
Note: The 64 magic number is coming introduced with the thought that unsigned long long is 64 bits atleast. So while converting we will be using that in case the number of bits exceeds 64 we will reallocate. A better choice is to use what chux said - sizeof digit * CHAR_BIT.
Also
bits[k] = b+'0';
We are putting the ascii value and then you can print it like this
printf("%c", bits[i]);
You forgot to free the allocated memory. Without freeing it (free(bits)), you have memory leak.
Davic C. Rankins comment
void intToBin(int digit)
{
int b;
int k = 0;
char *bits;
int i;
bits= (char *) malloc(sizeof(char));
while (digit) {
b = digit % 2;
digit = digit / 2;
bits[k] = b;
k++;
}
for ( i = k - 1; i >= 0; i--) {
printf("%d", bits[i]);
}
}
The answer is simple,
Replace int with int64_t to use 64 bits instead of 32.
Please try it and let us know
Replace int with int64_t to use 64 bits instead of 32.
I want do the two's complement of a float data.
unsigned long Temperature ;
Temperature = (~(unsigned long)(564.48))+1;
But the problem is that the cast loses information, 564 instead of 564.48.
Can i do the two's complement without a loss of information?
That is a very weird thing to do; floating-point numbers are not stored as 2s complement, so it doesn't make a lot of sense.
Anyway, you can perhaps use the good old union trick:
union {
float real;
unsigned long integer;
} tmp = { 564.48 };
tmp.integer = ~tmp.integer + 1;
printf("I got %f\n", tmp.real);
When I tried it (on ideone) it printed:
I got -0.007412
Note that this relies on unspecified behavior, so it's possible it might break if your compiler does not implement the access in the most straight-forward manner. This is distinct form undefined behavior (which would make the code invalid), but still not optimal. Someone did tell me that newer standards make it clearer, but I've not found an exact reference so ... consider yourself warned.
You can't use ~ over floats (it must be an integer type):
#include <stdio.h>
void print_binary(size_t const size, void const * const ptr)
{
unsigned char *b = (unsigned char *) ptr;
unsigned char byte;
int i, j;
for (i = size - 1; i >= 0; i--) {
for (j = 7; j >= 0; j--) {
byte = b[i] & (1 << j);
byte >>= j;
printf("%u", byte);
}
}
printf("\n");
}
int main(void)
{
float f = 564.48f;
char *p = (char *)&f;
size_t i;
print_binary(sizeof(f), &f);
for (i = 0; i < sizeof(float); i++) {
p[i] = ~p[i];
}
print_binary(sizeof(f), &f);
f += 1.f;
return 0;
}
Output:
01000100000011010001111010111000
10111011111100101110000101000111
Of course print_binary is there for test the result, remove it, and (as pointed out by #barakmanos) print_binary assumes little endian, the rest of the code is not affected by endiannes:
#include <stdio.h>
int main(void)
{
float f = 564.48f;
char *p = (char *)&f;
size_t i;
for (i = 0; i < sizeof(float); i++) {
p[i] = ~p[i];
}
f += 1.f;
return 0;
}
Casting a floating-point value to an integer value changes the "bit contents" of that value.
In order to perform two's complement on the "bit contents" of a floating-point value:
float f = 564.48f;
unsigned long Temperature = ~*(unsigned long*)&f+1;
Make sure that sizeof(long) == sizeof(float), or use double instead of float.