Why is answer of it
-1, 2, -3 ? (especially -3 ??? how come)
struct b1 {
int a:1;
int b:3;
int c:4;
} ;
int main()
{
struct b1 c = {1,2,13};
printf("%d, %d, %d",c.a,c.b,c.c);
return 0;
}
Compiled on VC++ 32 bit editor. Many Thanks.
signed integers are represented in twos complement. The range of a 1-bit twos complement number is -1 to 0. Because the first bit of a twos complement number indicates it is negative, which is what you have done there.
See here:
sign extend 1-bit 2's complement number?
That is the same for the third number, you have overflowed the range of -8 to 7 which is the range of a 4 bit signed integer.
What you meant to do there is make all of those int -> unsigned int
See here for twos complement explanation:
http://www.ele.uri.edu/courses/ele447/proj_pages/divid/twos.html
Because a and c are signed integers with their sign bit set, hence they are negative:
a: 1d = 1b (1 bit)
b: 2d = 010b (3 bits)
c: 13d = 1101b (4 bits)
Signed integer values are stored in two's complement, with the highest bit representing the sign (1 means "negative"). Hence, when you read the bitfield values as signed int's, you get back negative values for a and c (sign extended and converted back from their two's complement representation), but a positive 2 for b.
To get the absolute value of a negative two's complement number, you subtract one and invert the result:
1101b
-0001b
======
1100b
1100b inverted becomes 0011b which is equal to 3d. Since the sign bit is negative (which you had to examine before doing the previous calculation anyway), the result is -3d.
Related
I have a simple function that will find the two's Complement for an unsigned integer and then test it to make sure that it is correct
unsigned twoscomplement(unsigned v) {
};
int main()
{
unsigned a = 255;
unsigned c = twoscomplement(a);
unsigned s = a+c;
printf("%u+%u=%u\n", a, c, s);
return 0;
}
When I asked about how I would go around solving this I got the answer unsigned c = (~a)+1; From what I understood (~a) flips the bits and then +1 is for the overflow? Any help on this matter would be appreciated
Whenever we work with one’s complement or two’s complement, we need to state what the word size is. If there are w bits in the word, then the one’s complement of a number x is obtained by subtracting x from the binary numeral made of w 1s. If w is 16, we use 11111111111111112, which is 65535. Then the one’s complement of x is 11111111111111112−x. Viewing x as a binary numeral (with at most w bits), whatever bits are on in x will be off in 11111111111111112−x, and whatever bits are off in x will be on in 11111111111111112−x. Hence, all the bits are complemented.
C has an operator for the one’s complement; ~x flips all the bits, so it produces the one’s complement of x.
The two’s complement of x is 2w−x, by definition (except that the two’s complement of 0 is 0). 2w equals one plus that binary numeral made of w 1s. For example, 216 = 65535 + 1. Therefore, the two’s complement is one more than the one’s complement. Therefore the two’s complement of x is ~x + 1.
C also has an operator for the two’s complement, for unsigned integers. Unsigned arithmetic is defined to “wrap” modulo 2w; whenever a regular arithmetic result is outside that range, it is brought back into that range by adding or subtracting 2w as needed. The regular arithmetic negation of x would be negative (if x is not zero), so the computed result of -x is −x + 2w = 2w−x, which is the two’s complement of x.
For about 3-4 hours and start reading about bit fields in C and I can't understand how they work. For example, I can't understand why the program has the output: -1, 2, -3
#include <stdio.h>
struct REGISTER {
int bit1 : 1;
int : 2;
int bit3 : 4;
int bit4 : 4;
};
int main(void) {
struct REGISTER bit = { 1, 2, 13 };
printf("%d, %d, %d\n", bit.bit1, bit.bit3, bit.bit4);
return 0;
}
Can someone give me an explanation? I tend to think that if I use unsigned in the struct then the output would be positive. But I don't know where that -3 comes from.
Your compiler considers the type int of a bit field as the type signed int.
Consider the binary representation of initializers (it is enough to consider one byte)
1 -> 0b00000001
2 -> 0b00000010
13 -> 0b00001101
So the first bit-field having the width equal to 1 gets 1. For an integer with one bit this bit is the sign bit. So such a bit field can represent two values 0 and -1 (in the 2's complement representation).
The second initialized bit-field has the width equal to 4. So it stores the bit representation 0010 that is equal to 2.
The third initialized bit-field also has the width equal to 4. So its stored bit combination is equal to 1101. The most significant bit is the sign bit and it is set. So the bit-field contains a negative number. This number is equal to -3.
1101 ( = -3 )
+
0011 ( = 3 )
====
0000 ( = 0 )
It is implementation-defined whether an int bitfield is signed or unsigned, hence it is actually an error in any program where you care about the value - if you care for the value you will qualify it either as signed or unsigned.
Now, your compiler considers bitfields without specified signedness as signed. I.e. int bit4: 4 tells that that bit-field is 4 bits wide and signed.
13 cannot be represented in a signed 4-bit bitfield as the maximum value is 7, no matter the representation of negative numbers - 2's complement, 1's complement, sign-and-magnitude. An implementation-specified conversion will now occur: in your case the bit representation 1101 is stored as-is in the 2's complement signed bitfield, and it is considered as the 2's complement negative value -3.
Same happens for 1-bit signed bitfield: the one bit is the sign bit, hence there are only 2 possible values: 0 and -1 on two's complement systems. On one's complement system, or sign-and-magnitude, one-bit bitfield does not make any sense at all because it can only store 0 or a trap value.
If you want it to be able to store value 13, you will have to use at least 5 bits or use unsigned int: 4.
Please Dont use signed int in this operation. here that lead to minus results. if we use unsigned int,The output comes out to be negative
What happened behind is that the value 13 was stored in 4 bit signed integer which is equal to 1101. The MSB is a 1, so it’s a negative number and you need to calculate the 2’s complement of the binary number to get its actual value which is what is done internally. By calculating 2’s complement you will arrive at the value 0011 which is equivalent to decimal number 3 and since it was a negative number you get a -3.
#include <stdio.h>
struct REGISTER {
unsigned int bit1: 1;
unsigned int :2;
unsigned int bit3: 4;
unsigned int bit4: 4;
};
int main(void) {
struct REGISTER bit={1,2,13};
printf("%d, %d, %d\n", bit.bit1, bit.bit3, bit.bit4);
return 0;
}
Here the output will be exactly 1,2,13
Thanks
It's pretty easy, you're initializing the bitfields as follows:
1 as bit length 1. That means the MSB is 1, which since the type is int is the sign bit. So the resulting value is -0-1 = -1.
2 as bit length 4. MSB (sign bit) is 0, so the result is positive, ie 2.
13 as bit length 4. In binary that is 1101, so the MSB is 1 (negative), so the resulting value is -2-1 = -3.
You can read more about two's complement (the format in which numbers are stored on Intel architectures) here. The short version is that negative numbers have MSB=1, and to calculate their value you take the negative of their not representation (without the sign bit) minus one.
I currently have the code:
int getOperand(unsigned char b) {
int operand = b & 0x3F;
return operand;
}
'b' is an 8 bit binary number that is read in from a byte file. The 2 most significant bits of 'b' is the OPCODE and the last 6 bits are the OPERAND. I currently have this code setup to take the last 6 bits of 'b' and store them in the variable 'operand'. The problem I am currently having is that I do not know how to extract the OPERAND as a signed operand. Currently the number that is returned is only unsigned. Therefore how can I edit the code so that it returns a signed number instead?
If we assume you are just storing the last 6 bits of a negative byte that fits in a 6 bit signed value using your machines native 2's complement convention for representing negative numbers, then you check if the sign bit is set in your 6 bit value, and then extend the sign bits.
return (b & 0x20) ? b | ~0x3F : b & 0x3F;
Try it online!
The problem I am currently having is that I do not know how to extract
the OPERAND as a signed operand.
The C standard doesn't mandate any particular way of representing negative signed numbers. In most implementations, negative signed integers are stored in what is called two's complement. The other major way of storing negative signed numbers is called one's complement.
The code below assumes that entered number is entered as a signed number where the sign bit is stored as a bit 5 (0x20). This is called one's complement representation. For some applications it is easier to code operands that way.
Example:
#define NEGATIVE_SIGN 0x0C
#define ONE 0x01
#define TWO 0x02
#define MINUS_ONE (NEGATIVE_SIGN | ONE)
#define MINUS_TWO (NEGATIVE_SIGN | TWO)
Then OPERAND number is transformed to full 8 bit signed char in two's complement representation.
// INPUT: OPERAND in one's complement representation of the negative value.
// Using 6 bits one can represent numbers from -31 to 31.
// returning `signed char` is presented in two's complement representation.
signed char getOperand(unsigned char b) {
unsigned char operand = b & 0x3F;
unsigned char sign = 0x20 & operand;
unsigned char number = 0x1F & operand;
if(sign)
{
number = (~number) +1; // Conversion to Two's Complement
}
return number;
}
// 5 4 3 2 1 0
int main (){
printf("%d\n", getOperand(0x21)); // negative one
printf("%d\n", getOperand(0x22)); // -2
printf("%d\n", getOperand(0x2F)); // negative 15
printf("%d\n", getOperand(0x1F | 0x20 ) ); // negative -31
printf("%d\n", getOperand(0x1F ) ); // positive 31
return 0;
}
OUTPUT:
-1
-2
-15
-31
31
I read about twos complement on wikipedia and on stack overflow, this is what I understood but I'm not sure if it's correct
signed int
the left most bit is interpreted as -231 and this how we can have negative numbers
unsigned int
the left most bit is interpreted as +231 and this is how we achieve large positive numbers
update
What will the compiler see when we store 3 vs -3?
I thought 3 is always 00000000000000000000000000000011
and -3 is always 11111111111111111111111111111101
example for 3 vs -3 in C:
unsigned int x = -3;
int y = 3;
printf("%d %d\n", x, y); // -3 3
printf("%u %u\n", x, y); // 4294967293 3
printf("%x %x\n", x, y); // fffffffd 3
Two's complement is a way to represent negative integers in binary.
First of all, here's a standard 32-bit integer ranges:
Signed = -(2 ^ 31) to ((2 ^ 31) - 1)
Unsigned = 0 to ((2 ^ 32) - 1)
In two's complement, a negative is represented by inverting the bits of its positive equivalent and adding 1:
10 which is 00001010 becomes -10 which is 11110110 (if the numbers were 8-bit integers).
Also, the binary representation is only important if you plan on using bitwise operators.
If your doing basic arithmetic, then this is unimportant.
The only time this may give unexpected results outside of the aforementioned times is getting the absolute value of the signed version of -(2 << 31) which will always give a negative.
Your problem does not have to do with the representation, but the type.
A negative number in an unsigned integer is represented the same, the difference is that it becomes a super high number since it must be positive and the sign bit works as normal.
You should also realize that ((2^32) - 5) is the exact same thing as -5 if the value is unsigned, etc.
Therefore, the following holds true:
unsigned int x = (2 << 31) - 5;
unsigned int y = -5;
if (x == y) {
printf("Negative values wrap around in unsigned integers on underflow.");
}
else {
printf( "Unsigned integer underflow is undefined!" );
}
The numbers don't change, just the interpretation of the numbers. For most two's complement processors, add and subtract do the same math, but set a carry / borrow status assuming the numbers are unsigned, and an overflow status assuming the number are signed. For multiply and divide, the result may be different between signed and unsigned numbers (if one or both numbers are negative), so there are separate signed and unsigned versions of multiply and divide.
For 32-bit integers, for both signed and unsigned numbers, n-th bit is always interpreted as +2n.
For signed numbers with the 31th bit set, the result is adjusted by -232.
Example:
1111 1111 1111 1111 1111 1111 1111 11112 as unsigned int is interpreted as 231+230+...+21+20. The interpretation of this as a signed int would be the same MINUS 232, i.e. 231+230+...+21+20-232 = -1.
(Well, it can be said that for signed numbers with the 31th bit set, this bit is interpreted as -231 instead of +231, like you said in the question. I find this way a little less clear.)
Your representation of 3 and -3 is correct: 3 = 0x00000003, -3 + 232 = 0xFFFFFFFD.
Yes, you are correct, allow me to explain a bit further for clarification purposes.
The difference between int and unsigned int is how the bits are interpreted. The machine processes unsigned and signed bits the same way, but there are extra bits added for signing. Two's complement notation is very readable when dealing with related subjects.
Example:
The number 5's, 0101, inverse is 1011.
In C++, it's depends when you should use each data type. You should use unsigned values when functions or operators return those values. ALUs handle signed and unsigned variables very similarly.
The exact rules for writing in Two's complement is as follows:
If the number is positive, count up to 2^(32-1) -1
If it is 0, use all zeroes
For negatives, flip and switch all the 1's and 0's.
Example 2(The beauty of Two's complement):
-2 + 2 = 0 is displayed as 0010 + 1110; and that is 10000. With overflow at the end, we have our result as 0000;
For example:
unsigned int i = ~0;
Result: Max number I can assign to i
and
signed int y = ~0;
Result: -1
Why do I get -1? Shouldn't I get the maximum number that I can assign to y?
Both 4294967295 (a.k.a. UINT_MAX) and -1 have the same binary representation of 0xFFFFFFFF or 32 bits all set to 1. This is because signed numbers are represented using two's complement. A negative number has its MSB (most significant bit) set to 1 and its value determined by flipping the rest of the bits, adding 1 and multiplying by -1. So if you have the MSB set to 1 and the rest of the bits also set to 1, you flip them (get 32 zeros), add 1 (get 1) and multiply by -1 to finally get -1.
This makes it easier for the CPU to do the math as it needs no special exceptions for negative numbers. For example, try adding 0xFFFFFFFF (-1) and 1. Since there is only room for 32 bits, this will overflow and the result will be 0 as expected.
See more at:
http://en.wikipedia.org/wiki/Two%27s_complement
unsigned int i = ~0;
Result: Max number I can assign to i
Usually, but not necessarily. The expression ~0 evaluates to an int with all (non-padding) bits set. The C standard allows three representations for signed integers,
two's complement, in which case ~0 = -1 and assigning that to an unsigned int results in (-1) + (UINT_MAX + 1) = UINT_MAX.
ones' complement, in which case ~0 is either a negative zero or a trap representation; if it's a negative zero, the assignment to an unsigned int results in 0.
sign-and-magnitude, in which case ~0 is INT_MIN == -INT_MAX, and assigning it to an unsigned int results in (UINT_MAX + 1) - INT_MAX, which is 1 in the unlikely case that unsigned int has a width (number of value bits for unsigned integer types, number of value bits + 1 [for the sign bit] for signed integer types) smaller than that of int and 2^(WIDTH - 1) + 1 in the common case that the width of unsigned int is the same as the width of int.
The initialisation
unsigned int i = ~0u;
will always result in i holding the value UINT_MAX.
signed int y = ~0;
Result: -1
As stated above, only if the representation of signed integers uses two's complement (which nowadays is by far the most common representation).
~0 is just an int with all bits set to 1. When interpreted as unsigned this will be equivalent to UINT_MAX. When interpreted as signed this will be -1.
Assuming 32 bit ints:
0 = 0x00000000 = 0 (signed) = 0 (unsigned)
~0 = 0xffffffff = -1 (signed) = UINT_MAX (unsigned)
Paul's answer is absolutely right. Instead of using ~0, you can use:
#include <limits.h>
signed int y = INT_MAX;
unsigned int x = UINT_MAX;
And now if you check values:
printf("x = %u\ny = %d\n", UINT_MAX, INT_MAX);
you can see max values on your system.
No, because ~ is the bitwise NOT operator, not the maximum value for type operator. ~0 corresponds to an int with all bits set to 1, which, interpreted as an unsigned gives you the max number representable by an unsigned, and interpreted as a signed int, gives you -1.
You must be on a two's complement machine.
Look up http://en.wikipedia.org/wiki/Two%27s_complement, and learn a little about Boolean algebra, and logic design. Also learning how to count in binary and addition and subtraction in binary will explain this further.
The C language used this form of numbers so to find the largest number you need to use 0x7FFFFFFF. (where you use 2 FF's for each byte used and the leftmost byte is a 7.) To understand this you need to look up hexadecimal numbers and how they work.
Now to explain the unsigned equivalent. In signed numbers the bottom half of numbers are negative (0 is assumed positive so negative numbers actually count 1 higher than positive numbers). Unsigned numbers are all positive. So in theory your highest number for a 32 bit int is 2^32 except that 0 is still counted as positive so it's actually 2^32-1, now for signed numbers half those numbers are negative. which means we divide the previous number 2^32 by 2, since 32 is an exponent we get 2^31 numbers on each side 0 being positive means the range of an signed 32 bit int is (-2^31, 2^31-1).
Now just comparing ranges:
unsigned 32 bit int: (0, 2^32-1)
signed 32 bit int: (-2^31, 2^32-1)
unsigned 16 bit int: (0, 2^16-1)
signed 16 bit int: (-2^15, 2^15-1)
you should be able to see the pattern here.
to explain the ~0 thing takes a bit more, this has to do with subtraction in binary. it's just adding 1 and flipping all the bits then adding the two numbers together. C does this for you behind the scenes and so do many processors (including the x86 and x64 lines of processors.)
Because of this it's best to store negative numbers as though they are counting down, and in two's complement the added 1 is also hidden. Because 0 is assumed positive thus negative numbers can't have a value for 0, so they automatically have -1 (positive 1 after the bit flip) added to them. when decoding negative numbers we have to account for this.