Why result of
include <stdio.h>
int main()
{
unsigned short int i = 0xff ;
unsigned short int j;
j= i<<2;
printf("%x n%x\n", i, j);
return 0;
}
is j = 3fc ?
if both i and j are short int - so they are 2bytes values, so j shouldnt =fc ??
thx in advance for explanations.
~
~
Shifting 0xff left two bits looks like this:
0000 0000 1111 1111
0 0 f f
0000 0011 1111 1100 -- Shifted left 2 bits.
0 3 f c
So 0x00ff << 2 = 0x03fc. Everything looks as it should be .
No, 0x3fc is correct. Note that a byte is two hex digits, so a (16-bit) short has a total of 4 hex digits.
0xff << 2 == 0xff * 4 == 0x3fc == 1020
Even if they are 2-bytes, they are allowed to hold this small value.
3FC requires only 12 bits to store so it can be stored in 2 bytes.
C++ makes no guarantee as to the number of bytes in an unsigned short int. It in fact makes almost no guarantee about the size of any type other than char which is guaranteed to be 1 byte.
In this case though it's irrelevant because 3fc can be successfully stored in only 2 bytes.
Maybe this is what you actually tried to write? (Remember an hex digit is 4 bits only, ie, half a byte)
#include <stdio.h>
int main()
{
unsigned short int i = 0xffff;
unsigned short int j;
j = i<<8;
printf("%x %x\n", i, j);
return 0;
}
This outputs ffff ff00
Related
x is an int,
I should be able to get the correct result when 0 is not involved. In attempts to account for the 0 case, I added "& x", which I believe now should return that a number is positive iff x > 0 and x is not 0 (because in c, any number other than 0 evaluates to true, correct?)
But, when running my tests, it says it has failed to evaulate 0x7fffffff as positive and I am not sure why!
Here is my code:
int mask = 0x1;
x = x >> 31;
int lsb = mask & x;
return ( (lsb) ^ 0x1) & (x) )
Edit: I have solved the problem by changing the code to the one below! Feedback is still very much appreciated, or any problems you may spot.
int mask = 0x1;
int lsb = (x >> 31) & mask;
int result = !(lsb ^ 0x1);
return !(result | !x);
If you know the representation is 2's complement, then you can do:
#include <stdio.h>
#define IS_NEG(a) (!!((1 << 31) & (a)))
int main(void)
{
int n;
while(1) {
scanf("%d", &n);
printf("negative: %d\n", IS_NEG(n));
}
return 0;
}
Explanation:
(1 << 31) will take the number 1 and shift it 31 times to the left, thus giving you 1000 0000 0000 0000 0000 0000 0000 0000. If you don't want to use the shift, you could use 0x80000000 too.
& (a) does a bitwise test with that big binary number. Since an AND operation only returns TRUE when both operands are TRUE, it follows that only if your number is negative (in 2's complement representation) that this will return TRUE.
!!(...) This double negation accounts for the fact that when you do that bitwise AND, the returned value by the expression will be (1 << 31) if the number is really negative. So we invert it (giving us zero), than invert it again (giving us 1). Therefore, this ensures that we get a ZERO or a ONE as a final result.
IS_NEG will return 0 on positive numbers AND 0, and returns 1 on all negative numbers.
Since the MSB will be a one when the number is negative, just test that bit. Note that this will only work for 32 bit integers (so you have to check that with a sizeof(int). The example returns 1 if a number is negative, but should be no problem reworking it to return 1 for positive numbers.
Let me know if this doesn't solve the problem. As I understand, you just want to test if any given int is positive/negative.
Edit: From the comments, I made a program to help you see what's going on.
#include <stdio.h>
#define IS_NEG(a) (!!(0x80000000 & (a)))
char buf[65];
/* converts an integer #n to binary represention of #bits bits */
char *bin(int n, unsigned int bits)
{
char *s = buf;
for(bits = (1 << (bits - 1)); bits > 0; bits = bits >> 1)
/* look! double negation again! Why this? :) */
*s++ = !!(n & bits) + 48;
*s = 0;
return buf;
}
int main(void)
{
/* R will be our partial result through-out the loop */
int r, n;
while(1) {
/* get the number */
scanf("%d", &n);
/* this is the inner part of the macro
* after this, we could say IS_NEG "becomes"
* (!!(r))
*/
r = n & 0x80000000;
printf("n & 0x80000000: 0x%x\n", r);
printf(" n = %s\n", bin(n, 32));
printf(" r = %s\n", bin(r, 32));
/* now we print what R is, so you see that the bitwise AND will
* return 0x80000000 on negative numbers. It will also print
* the NEGATION of R...
*
* After the printf(), we just assign the negated value to R.
*/
printf("r = 0x%x, !r = 0x%x\n", r, !r);
r = !r;
printf(" r = %s\n", bin(r, 32));
/* After this, IS_NEG "becomes" (!(r)) */
/* In the MACRO, this would be the second negation. */
printf("r = 0x%x, !r = 0x%x\n", r, !r);
r = !r;
printf(" r = %s\n", bin(r, 32));
/* Now, if R is 0, it means the number is either ZERO or
* POSITIVE.
*
* If R is 1, then the number is negative
*/
}
return 0;
}
https://graphics.stanford.edu/~seander/bithacks.html#CopyIntegerSign
Technically, an int could be a different size on different machines; use of an C99 int32_t from inttypes.h may help with portability. It might not even be encoded in the format you expect, Are there any non-twos-complement implementations of C?.
The really portable easy way, is of course,
static int is_positive(const int a) {
return a > 0;
}
The compiler will probably do a better job optimising it.
Edit: From comments, I came up with this; I tried to make it agnostic of the int-size. It is very much the same style as your own, checking whether the number is negative or zero, and inverting.
#include <stdio.h> /* printf */
#include <limits.h> /* INT_ */
#include <assert.h> /* assert */
/** Assumes a is a 2's-compliment number
and ~INT_MAX = 0b100..00, (checks if negative.) */
static int is_positive(const int a) {
unsigned b = a;
return !((b & ~INT_MAX) | !b);
}
static int is_really_positive(const int a) {
return a > 0;
}
static void test(const int a) {
printf("Number %d. Is positive %d.\n", a, is_positive(a));
assert(is_positive(a) == is_really_positive(a));
}
int main(void) {
test(INT_MIN);
test(-2);
test(-1);
test(0);
test(1);
test(2);
test(INT_MAX);
return 0;
}
Also related, https://stackoverflow.com/a/3532331/2472827.
Given the allowed operators from your comment, ! ~ & ^ | + << >>, edit: with the later constraint of no casts only the second alternative fits:
static int is_positive(unsigned x)
{
return ~x+1 >> (CHAR_BIT*sizeof x-1);
}
Here's the deal: conversion to unsigned is very carefully specified in C: if the signed value is unrepresentable in the unsigned type, one plus the maximum value representable in the unsigned type is added (or subtracted, no idea what prompted them to include this possibility) to the incoming value until the result is representable.
So the result depends only on the incoming value, not its representation. -1 is converted to UINT_MAX, no matter what. This is correct, since the universe itself runs on twos-complement notation . That it also makes the conversion a simple no-op reinterpretation on most CPUs is just a bonus.
You can get a 32-bit wide zero-or-nonzero test using bitwise or and shifts as follows:
int t;
t = x | (x>>16);
t = t | (t >> 8);
t = t | (t >> 4);
t = t | (t >> 2)
t = (t | (t>>1)) & 1;
This sets t to the "OR" of the low 32 bits of x and will 0 if and only if the low 32 bits are all zero. If the int type is 32 bits or less, this will be equivalent to (x != 0). You can combine that with your sign bit test:
return t & (~x >> 31);
To check whether given number is positive or negative. As you mentioned x is an int & I assume its 32-bit long signed int. for e.g
int x = 0x7fffffff;
How above x represented in binary
x => 0111 1111 | 1111 1111 | 1111 1111 | 1111 1111
| |
MSB LSB
Now to check given number is positive or negative using bitwise opaertor, just find out the status of last(MSB or 31st(longint) or 15th(short int)) bit status whether it's 0 or 1, if last bit is found as 0 means given number is positive otherwise negative.
Now How to check last bit(31st) status ? Shift last(31st) bit to 0th bit and perform bitwise AND & operation with 1.
x => 0111 1111 | 1111 1111 | 1111 1111 | 1111 1111
x>>31 => 0000 0000 | 0000 0000 | 0000 0000 | 0000 0000
---------------------------------------------
&
1 => 0000 0000 | 0000 0000 | 0000 0000 | 0000 0001
---------------------------------------------
0000 0000 | 0000 0000 | 0000 0000 | 0000 0000 => its binary of zero( 0 ) so its a positive number
Now how to program above
static inline int sign_bit_check(int x) {
return (x>>31) & 1;
}
And call the sign_bit_check() like
int main(void) {
int x = 0x7fffffff;
int ret = sign_bit_check(x);
if(ret) {
printf("Negative\n");
}
else {
printf("positive \n");
}
return 0;
}
I was thinking this world work, but it does not:
int a = -500;
a = a << 1;
a = (unsigned int)a >> 1;
//printf("%d",a) gives me "2147483148"
My thought was that the left-shift would remove the leftmost sign bit, so right-shifting it as an unsigned int would guarantee that it's a logical shift rather than arithmetic. Why is this incorrect?
Also:
int a = -500;
a = a << 1;
//printf("%d",a) gives me "-1000"
TL;DR: the easiest way is to use the abs function from <stdlib.h>. The rest of the answer involves the representation of negative numbers on a computer.
Negative integers are (almost always) represented in 2's complement form. (see note below)
The method of getting the negative of a number is:
Take the binary representation of the whole number (including leading zeroes for the data type, except the MSB which will serve as the sign bit).
Take the 1's complement of the above number.
Add 1 to the 1's complement.
Prefix a sign bit.
Using 500 as an example,
Take the binary representation of 500: _000 0001 1111 0100 (_ is a placeholder for the sign bit).
Take the 1's-complement / inverse of it: _111 1110 0000 1011
Add 1 to the 1's complement: _111 1110 0000 1011 + 1 = _111 1110 0000 1100. This is the same as 2147483148 that you obtained, when you replaced the sign-bit by zero.
Prefix 0 to show a positive number and 1 for a negative number: 1111 1110 0000 1100. (This will be different from 2147483148 above. The reason you got the above value is because you nuked the MSB).
Inverting the sign is a similar process. You get leading ones if you use 16-bit or 32-bit numbers leading to the large value that you see. The LSB should be the same in each case.
Note: there are machines with 1's complement representation, but they are a minority. The 2's complement is usually preferred because 0 has the same representation, i.e., -0 and 0 are represented as all-zeroes in the 2's complement notation.
Left-shifting negative integers invokes undefined behavior, so you can't do that. You could have used your code if you did a = (unsigned int)a << 1;. You'd get 500 = 0xFFFFFE0C, left-shifted 1 = 0xFFFFFC18.
a = (unsigned int)a >> 1; does indeed guarantee logical shift, so you get 0x7FFFFE0C. This is decimal 2147483148.
But this is needlessly complex. The best and most portable way to change the sign bit is simply a = -a. Any other code or method is questionable.
If you however insist on bit-twiddling, you could also do something like
(int32_t)a & ~(1u << 31)
This is portable to 32 bit systems, since (int32_t) guarantees two's complement, but 1u << 31 assumes 32 bit int type.
Demo:
#include <stdio.h>
#include <stdint.h>
int main (void)
{
int a = -500;
a = (unsigned int)a << 1;
a = (unsigned int)a >> 1;
printf("%.8X = %d\n", a, a);
_Static_assert(sizeof(int)>=4, "Int must be at least 32 bits.");
a = -500;
a = (int32_t)a & ~(1u << 31);
printf("%.8X = %d\n", a, a);
return 0;
}
As you put in the your "Also" section, after your first left shift of 1 bit, a DOES reflect -1000 as expected.
The issue is in your cast to unsigned int. As explained above, the negative number is represented as 2's complement, meaning the sign is determined by the left most bit (most significant bit). When cast to an unsigned int, that value no longer represents sign but increases the maximum value your int can take.
Assuming 32 bit ints, the MSB used to represent -2^31 (= -2147483648) and now represents positive 2147483648 in an unsigned int, for an increase of 2* 2147483648 = 4294967296. Add this to your original value of -1000 and you get 4294966296. Right shift divides this by 2 and you arrive at 2147483148.
Hoping this may be helpful: (modified printing func from Print an int in binary representation using C)
void int2bin(int a, char *buffer, int buf_size) {
buffer += (buf_size - 1);
for (int i = buf_size-1; i >= 0; i--) {
*buffer-- = (a & 1) + '0';
a >>= 1;
}
}
int main() {
int test = -500;
int bufSize = sizeof(int)*8 + 1;
char buf[bufSize];
buf[bufSize-1] = '\0';
int2bin(test, buf, bufSize-1);
printf("%i (%u): %s\n", test, (unsigned int)test, buf);
//Prints: -500 (4294966796): 11111111111111111111111000001100
test = test << 1;
int2bin(test, buf, bufSize-1);
printf("%i (%u): %s\n", test, (unsigned int)test, buf);
//Prints: -1000 (4294966296): 11111111111111111111110000011000
test = 500;
int2bin(test, buf, bufSize-1);
printf("%i (%u): %s\n", test, (unsigned int)test, buf);
//Prints: 500 (500): 00000000000000000000000111110100
return 0;
}
I'm trying to convert a decimal number into binary (16 bits max). My function works perfectly for up to 8 bits but when I want to print numbers up to 16 bits, it stop printing characters.
I used "int" as my data type for 8 bits but since I want to store 16 bits I'm using an unsigned long int in every variable.
Here's the code:
/* Program to convert decimal to binary (16 bits) */
#include <stdio.h>
#include <string.h>
char *byte_to_binary_str(long unsigned int byte);
int main()
{
printf("%s",byte_to_binary_str(32768)); //1000000 0000000
return 0;
}
char *byte_to_binary_str(long unsigned int byte)
{
static char bit_string[17];
bit_string[0] = '\0';
long unsigned int mask;
for (mask = 2^15; mask > 0; mask >>= 1) {
/* Check if the mask bit is set */
strcat(bit_string, byte & mask ? "1" : "0");
}
return bit_string;
}
My output gives me:
0000
Process returned 0 (0x0) execution time : 0.063 s
Press any key to continue.
Anyone know why is this happening? Thanks in advance.
mask = 2^15;
does not set the value of mask to what you are expecting 2^15 is no 2 raised to the power 15. It is bitwise XOR of 2 and 15.
You need something that is 1000 0000 0000 0000 in binary. That number will be 0x8000 in hex. Hence, use:
mask = 0x8000;
You can also use something that makes sense in your algorithm.
mask = 1u << 15;
How can I get the most significative 1-bit index from an unsigned integer (uint16_t)?
Example:
uint16_t x = // 0000 0000 1111 0000 = 240
printf("ffs=%d", __builtin_ffs(allowed)); // ffs=4
There is a function (__builtin_ffs) that return the least significative 1-bit (LSB) from a unsigned integer.
I want something opposite, I want some function which returns 8 applied to above example.
Remark: I have tried building my own function but I have found some problems with datatype size, which depends by compiler.
From the GCC manual at http://gcc.gnu.org/onlinedocs/gcc-4.1.2/gcc/Other-Builtins.html:
Built-in Function: int __builtin_clz (unsigned int x)
Returns the number of leading 0-bits in x, starting at the most significant bit position. If x is 0, the result is undefined.
So, highest set bit:
#define ONE_BASED_INDEX_OF_HIGHEST_SET_BIT(x) \
(CHAR_BIT * sizeof 1 - __builtin_clz(x)) // 1-based index!!
beware of x == 0 or x<0 && sizeof(x)<sizeof 0 though.
if I am reading this right (am I? I'm a little rusty on this stuff) you can do this as follows:
int msb = 0;
while(x) { // while there are still bits
x >>= 1; // right-shift the argument
msb++; // each time we right shift the argument, increment msb
}
I suppose sizeof(char) is one byte. Then when I write following code,
#include<stdio.h>
int main(void)
{
char x = 10;
printf("%d", x<<5);
}
The output is 320
My question is, if char is one byte long and value is 10, it should be:
0000 1010
When I shift by 5, shouldn't it become:
0100 0001
so why is output 320 and not 65?
I am using gcc on Linux and checked that sizeof(char) = 1
In C, all intermediates that are smaller than int are automatically promoted to int.
Therefore, your char is being promoted to larger than 8 bits.
So your 0000 1010 is being shifted up by 5 bits to get 320. (nothing is shifted off the top)
If you want to rotate, you need to do two shifts and a mask:
unsigned char x = 10;
x = (x << 5) | (x >> 3);
x &= 0xff;
printf("%d", x);
It's possible to do it faster using inline assembly or if the compiler supports it, intrinsics.
Mysticial is right. If you do
char x = 10;
printf("%c", x);
It prints "#", which, if you check your ASCII table, is 64.
0000 1010 << 5 = 0001 0100 0000
You had overflow, but since it was promoted to an int, it just printed the number.
Because what you describe is a rotate, not a shift. 0 is always shifted in on left shifts.