I am learning the c language using the K&R book. In the second chapter book, the author talks about implicit conversion. There book says this:
Conversion rules are more complicated when unsigned operands are involved. The problem is that
comparisons between signed and unsigned values are machine-dependent, because they depend on the sizes of the various integer types. For example, suppose that int is 16 bits and long is 32 bits. Then -1L < 1U, because 1U, which is an unsigned int, is promoted to a signed long. But -1L >
1UL because -1L is promoted to unsigned long and thus appears to be a large positive number.
I tried the code below in two different scenarios:
compiled on an x86 64bits platform and executed. Where sizeof(-1L) -> 8byte and sizeof(1U) -> 4 bytes
compiled on an x86 32bits platform and executed. Where sizeof(-1L) -> 4byte and sizeof(1U) -> 4 bytes
The code:
int main() {
if(-1L > 1U)
printf("true");
else
printf("false");
return 0;
}
The results:
x86 64bits: false
x86 32bits: true
so I'm getting two different OP in each case.
As author says, for 2 different data sizes one being 16 and the other 32, it holds good in my x86-64 case.
But im not able to understand why in the second case for 32 bits, I'm getting true.
As author says unsigned int is promoted to signed long int, if this is true then both
should be 4 bytes wide, then why is it printing true instead of false? As now both should be signed long.
As the author says it is machine dependent, then both long and int should have same byte size, so how the implicit conversion is happening here?
My understanding is that -1 is stored as two's complement i.e 0xFFFFFFFF > 0x1 so in the second case it should be true.
But this explanation contradicts the 1st case.
Please correct me if what I think is wrong, as I am new to implicit conversion.
Can anyone please explain this behaviour?
lets explain the rank system first
6.3.1 Arithmetic operand(c99 standard)
A) The rank of a signed integer type shall be greater than the rank of any signed integer
type with less precision(more bytes higher precision higher rank)
B) The rank of long long int shall be greater than the rank of long int, which shall be
greater than the rank of int, which shall be greater than the rank of short int, which
shall be greater than the rank of signed char.
C) The rank of any unsigned integer type shall equal the rank of the corresponding signed
integer type, if any.
(in other words if your system unsigned int is 32bits and your int is 32bits then the
ranks of these are the same.)
the above explains the rank.
now coming to arithmetic conversions.
6.3.1.8 Usual arithmetic conversions (c99 standard)
1)If both operands have the same type, then no further conversion is needed.
2)Otherwise, if both operands have signed integer types or both have unsigned integer
types, the operand with the type of lesser integer conversion rank is converted to the
type of the operand with greater rank.(similar to 1)
3)Otherwise, if the operand that has unsigned integer type has rank greater or equal to
the rank of the type of the other operand, then the operand with signed integer type is
converted to the type of the operand with unsigned integer type.
4)Otherwise, if the type of the operand with signed integer type can represent all of the
values of the type of the operand with unsigned integer type, then the operand with
unsigned integer type is converted to the type of the operand with signed integer type
5)Otherwise, both operands are converted to the unsigned integer type corresponding to the
type of the operand with signed integer type.
2) compiled on an x86 32bits platform and executed. Where sizeof(-1L) -> 4byte and sizeof(1U) -> 4 bytes
in your case look at statement 3 & C. the unsigned value(4bytes) has rank equal to the signed value(4btyes) therefore the singed value is converted to an unsigned value, when this happens the, the sign bit makes this look like a extremely large value. -1L > 1U therefore is true
1) compiled on an x86 64bits platform and executed. Where sizeof(-1L) -> 8byte and sizeof(1U) -> 4 bytes
in this case, the unsigned value rank is less than the rank of the singed value. look at 4).
the signed integer(8bytes) can represent any 4byte unsigned value. therefore the unsigned 4byte value is converted to a signed value.(this will preserve the sign bit, sign bit is 0)
therefore -1L > 1U is false
But im not able to understand why in second case in 32 bit its OP-->true. As author says unsigned int is promoted to signed long int if so then both are 4 byte wide, why its printing true instead of false.? since now both are signed long.
The auther says, that if int and long have different size, then unsigned int is promoted to signed long.
If int and long have the same size, then long is too small to hold all values of unsigned int and therefore both are converted to unsigned long.
For binary arithmetic and relational operators:
If either operand has type long double, the other operand is converted to long double. Otherwise, if either operand has type double, the other operand is converted to double. Otherwise, if either operand has type float, the other operand is converted to float. Otherwise the integral promotions are performed on both operands.
(Integral promotion: A char, a short int, or an int bit-field, or their signed or unsigned varieties, or an enumeration type, may be used in an expression wherever an int or unsigned int may be used. If an int can represent all the values of the original type, the value is converted to an int; otherwise it is converted to an unsigned int.)
Then if either operand has type unsigned long int, the other operand is converted to unsigned long int. Otherwise, if one operand has type long int and the other has type unsigned int, if a long int can represent all values of an unsigned int the operand of type unsigned int is converted to long int; if a long int cannot represent all the values of an unsigned int, both operands are converted to unsigned long int. Otherwise, if either operand has type long int, the other operand is converted to long int. Otherwise, if either operand has type unsigned int, the other operand is converted to unsigned int. Otherwise, both operands have type int.
The sentence in bold explains your second case, where long int has the same width as unsigned int thus cannot hold all values of unsigned int.
(The above description lacks the type unsigned long long int and long long it, but the rules are basically the same.)
As all who answered above are correct, Just to add more clarity and my understanding writing here to get more clarity.
-->if one operand has type long int and the other has type unsigned int,
-->if a long int can represent all values of an unsigned int the operand of type unsigned int is converted to long int;
-->if a long int cannot represent all the values of an unsigned int, both operands are converted to unsigned long int.
So from above one operand has type long int i.e -1L and the other has type unsigned int i.e 1U
suppose sizeof -1L is --->8byte and sizeof 1U is 4 byte
0X0000-0XFFFFF values can be represented using in long int whose sizeof is 8 byte
so in this case long int can represent all values of an unsigned int i.e using 8byte ---> it can represent all the values unsigned int 1U.
so----> here operand of type unsigned int is converted to long int ---> -1L > 1U --> is false
coming 2nd case
if a long int cannot represent all the values of an unsigned int
i.e sizeof -1L -->4byte and sizeof 1U -->4byte
here long int cannot represent all the values i.e using 4 bytes--> it cannot represent all the values of unsigned int 1U. so both operands are converted to unsigned long int
-1L appears to large value since its unsigned now when compared to 1U.
i.e---->0xFFFFFFFF > 0x1 ---> its true
Related
... the integer promotions are performed on both operands.
Then the following rules are applied to the promoted operands:
If both operands have the same type, then no further conversion is needed.
Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank.
Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then the operand with signed integer type is converted to the type of the operand with unsigned integer type.
Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, then the operand with unsigned integer type is converted to the type of the operand with signed integer type.
Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type
Can someone explain me the difference between the last and penultimate points from above? Why is the last one needed? Aren't all the cases covered by the first 4? If someone can give an example would be perfect. Thanks
Otherwise, both operands are converted to the unsigned integer type
corresponding to the type of the operand with signed integer type
Let's consider a system where sizeof( long ) is equal to sizeof( unsigned int ) (for example the both are equal to 4).
In this case though the rank of the type long is greater than the rank pf the type unsigned int nevertheless an operand of the signed type long is unable to represent all values of the type unsigned int.
In this case the both operands used in a binary operation are converted to the type unsigned long.
Or another example when in some systems sizeof( long long ) is equal to sizeof( unsigned long ) and equal to 8. In this case again the both operands of these types will be converted to the type unsigned long long because the rank of long long is greater than the rank of the type unsigned long but not all values of operand of the type unsigned long can be represented by an object of the signed type long long.
That is this quote describes the situation when the rank pf an operand of a signed integer type is greater than the rank of an unsigned integer type but the operand of the signed integer type is unable to represent all values of the operand of the unsigned integer type.
The preceding quote
Otherwise, if the type of the operand with signed integer type can
represent all of the values of the type of the operand with unsigned
integer type, then the operand with unsigned integer type is converted
to the type of the operand with signed integer type.
describes the situation when the rank of a signed integer type is greater than the rank of an unsigned integer type and operand of the signed integer type can represent all values of operand of the unsigned integer type. For example when one operand of the type long and other of the type unsigned int and sizeof( long ) is equal to 8 and sizeof( unsigned int ) is equal tp 4.
We reach (4) when one has a signed type and one has an unsigned type, and the signed type has a greater rank.
For example, this could be a long and a unsigned int.
What differentiates whether (4) or (5) is used is whether the signed type can represent all of the values of the unsigned type.
On a system with a 32 bit long and a 16 bit unsigned int, we'd use case 4
We'd use long.
On a system with a 32 bit long and a 32 bit unsigned int, we'd use case 5 because 4,294,967,295 can be represented by this unsigned int, but can't be represented by this long.
We'd use unsigned long.
When adding signed and unsigned values we follow these two rules (from https://stackoverflow.com/a/2280810/1073672)
Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then the operand with signed integer type is converted to the type of the operand with unsigned integer type.
Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, then the operand with unsigned integer type is converted to the type of the operand with signed integer type.
Which of the above rules will be triggered for:
unsigned int ui = 4;
ui = ui + 532;
The type of the constant 532 is int, which is signed.
Looking at the first rule, since unsigned int and int have the same rank, then the unsigned type's rank is greater than or equal to the signed type's. The first rule matches. The signed 532 is converted to unsigned int before the addition.
Even if the first rule hadn't matched, the addition cannot match the second rule, because a signed int cannot represent all the values of an unsigned int. (There are the same number of possible signed ints as unsigned ints, but for example, -1 is a signed int but not an unsigned int, therefore there must be at least one unsigned int that cannot be represented as a signed int.)
I have been reading the book The C Programming Language by K&R and came across this.
Conversion rules are more complicated when unsigned operands are involved. The problem
is that comparisons between signed and unsigned values are machine-dependent, because
they depend on the sizes of the various integer types. For example, suppose that int is 16 bits and long is 32 bits. Then -1L < 1U, because 1U, which is an unsigned int, is promoted to a signed long. But -1L > 1UL because -1L is promoted to unsigned long and thus appears
to be a large positive number.
From my beginning of C, i have little difficult regarding this Unsigned and Signed values. Can any one Explain how an unsigned value converted to signed value and viceversa with an explample.
You have touched the surface of usual arithmetic conversions with integral promotions (newer standards use term of integer promotions there).
What are these usual arithmetic conversions ?
In general when you have expression, that involves arithmetic or logical operator and operands don't match, then they need to be unified (as we say promoted) into common form. In C90 rules between integer operands (I am skipping intentionally rules for types "lower" that int here for simplicity) are in following gradation (note that there is no long long int type in C90):
int → unsigned int → long int → unsigned long int
However, there is one exception between unsigned int and long int. If these have the same size (in bits), then both operands of such types are promoted to common type unsigned long int. In any other case common type is that, which is on right side (e.g. when you have int and unsigned int operands, then first one is promoted to unsigned int). If both operands have the same type, then no promotion is made for types listed above.
How does this relate to -1L < 1U and -1L > 1UL ?
In your case it's assumed that sizeof(int) == 16 and sizeof(long) = 32, so for:
-1L < 1U second operand is promoted to long int
-1L > 1UL first operand is promoted to unsigned long
In former example value of expression is 1 as -1L < 1L. In latter case -1L is promoted to unsigned type by repeatedly adding or substracting n+1, when n is the largest value of type unsigned long (in your case n+1 == 2^32), which yields into large number (i.e. 2^32-1), thus value of whole expression is 1 (of type int) as well.
Hum... there is a longer (detailed) answer, and a shorter (approximate) answer...
...starting with the longer answer:
The standard (C99, 6.3.1) specifies an integer conversion rank, and where two integer operands have different rank, one will be converted up. The rank is related to the size of the integer; larger sizes have greater rank than smaller sizes. Where the sizes are the same there may be differences in rank -- but that's not important just now. However, a signed integer and an unsigned integer of the same size have the same rank.
It's worth remembering that signed int and unsigned int have the same size, by definition. They are examples of what the standard calls corresponding integer types (C99, 6.2.5). Each signed integer type has a corresponding unsigned one, and vice versa -- except for _Bool which doesn't have a sign (and, FWIW, has the lowest rank). As above, corresponding signed and unsigned types have the same rank.
C is very fond of int and unsigned int. It will promote integers of lesser rank to int (or to unsigned int, if the promoted integer is unsigned and the same size as int). This is called integer promotion (which is a subset of all the possible conversions). For many operators, C will perform integer promotion on both operands before doing anything else. It will also perform integer promotion on arguments for things like printf().
Now we get to what are known as the usual arithmetic conversions (C99, 6.3.1.8), which for two integer operands are:
both arguments are promoted as required, so will be at least int or unsigned int
if they are now not the same type (type, not rank):
if both types are signed, or both types are unsigned, the lower rank is converted to the higher.
This is straightforward, and there is clearly no loss of signed-ness.
...otherwise, for different types, of different signed-ness, size matters:
if the signed operand has the greater size (and hence rank), the unsigned operand is converted to the type of the signed one.
This is also straightforward, the unsigned value can be represented, without loss, in the signed type.
if the unsigned operand has the greater size (and hence rank), the signed operand is converted to the type of the unsigned one.
So, for (say) unsigned long long and int (which one assumes are different sizes), the int is converted up, by adding ULLONG_MAX + 1 to it.
This is not so straightforward, the signed-ness is lost.
...otherwise, for different types, of different signed-ness, of the same size, the result will be two unsigned operands of their current size. The rank kicks in to decide what type the operands will be:
if the operands have the same rank, then the signed one is converted to the unsigned operand's type.
This is what happens when you have int and unsigned int operands, the int is converted (by adding UINT_MAX + 1 to it).
if the unsigned operand has the greater rank, then the signed operand is converted to the unsigned operand's type (as above).
otherwise, the signed operand has the greater rank, and both operands are converted to the unsigned type corresponding to the signed operand's type (ie to the unsigned type with the greater rank).
This all looks terrifyingly complicated :-( But, keep in mind that the size of an integer is the major component of its rank -- what makes it look complicated dealing with integers of the same size, but different signedness and type.
(For the picky: I have glossed over the wrinkle in the standard which allows for integer representations to include "padding" bits -- which is deeply exotic -- so, for size read width, where required.)
...but the essence of it is, in short:
After promotion to int or unsigned int:
if the operands have different sizes, the smaller is converted up, and if larger is unsigned but the smaller is not, the smaller loses its signed-ness.
if the operands have the same size, but different signed-ness, the signed one is converted to unsigned, losing signed-ness.
The conversion to unsigned (the loss of signed-ness) requires the addition of MAX+1, where MAX is the maximum value of the unsigned type being converted to.
Hence: -1L > 0UL... indeed -1L == ULONG_MAX.
Why is it so that
unsigned char k=-1
if(k==-1)
is false
unsigned int k=-1
if(k==-1)
is true
For the purpose of demonstration let's assume 8-bit chars and 32-bit ints.
unsigned char k=-1;
k is assigned the value 255.
if(k==-1)
The left-hand side of the == operator is an unsigned char. The right-hand side is an int. Since all possible values of an unsigned char can fit inside an int, the left-hand side is converted to an int (this is performed due the the integer promotions, quoted below). This results in the comparison (255 == -1), which is false.
unsigned int k=-1
k is assigned the value 4294967295
if(k==-1)
This time, the left-hand side (an unsigned int) cannot fit within an int. The standard says that in this case, both values are converted to an unsigned int. So this results in the comparison (4294967295 == 4294967295), which is true.
The relevant quotes from the standard:
Integer promotions: (C99, 6.3.1.1p2)
If an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int.
Usual arithmetic conversions: (6.3.1.8).
[For integral operands, ] the integer promotions are performed on both operands. Then the following rules are applied to the promoted operands:
- If both operands have the same type, then no further conversion is needed.
...
- Otherwise, if the operand that has unsigned integer type has rank greater or
equal to the rank of the type of the other operand, then the operand with
signed integer type is converted to the type of the operand with unsigned
integer type.
...
§6.3.1.1p2 of the C11 standard draft (n1570.pdf):
If an int can represent all values of the original type (as restricted
by the width, for a bit-field), the value is converted to an int;
otherwise, it is converted to an unsigned int. These are called the
integer promotions.58) All other types are unchanged by the integer
promotions.
In your second case, an int can't represent unsigned int k because that's out of range. Both operands end up being converted to unsigned int and compare equal.
As there is no sign extension in unsigned promotion the result is defferent.
unsigned char k is promoted from unsigned char (value=255) to int (value=255).
In the case of unsigned int k, -1 is promoted from int (value=-1) to unsigned int (value=2^32-1).
unsigned char k = -1;
-1 is an integer literal of type int, which is equivalent to signed int. When you assign a large signed integer to a smaller unsigned type, the result will get truncated in undefined ways, the C standard does not guarantee what will happen.
In the real world outside the C standard, this is what's most likely (assuming 32-bit CPU, two's complement):
-1 is 0xFFFFFFFF. The least significant byte of 0xFFFFFFFF will get assigned to k. k == 255. Try to print it using printf("%u") and see for yourself.
In the case of unsigned int k=-1, -1 is still a signed int. But it gets implicitly (silently) promoted to an unsigned int when it is stored in k. When you later compare k == -1, the right side -1 will once more get promoted to an unsigned type, and will compare equal with the data stored in k.
I was reading in the C99 standard about the usual arithmetic conversions.
If both operands have the same type, then no further conversion is
needed.
Otherwise, if both operands have signed integer types or both have
unsigned integer types, the operand with the type of lesser integer
conversion rank is converted to the type of the operand with greater
rank.
Otherwise, if the operand that has unsigned integer type has rank
greater or equal to the rank of the type of the other operand, then
the operand with signed integer type is converted to the type of the
operand with unsigned integer type.
Otherwise, if the type of the operand with signed integer type can
represent all of the values of the type of the operand with unsigned
integer type, then the operand with unsigned integer type is converted
to the type of the operand with signed integer type.
Otherwise, both operands are converted to the unsigned integer type
corresponding to the type of the operand with signed integer type.
So let's say I have the following code:
#include <stdio.h>
int main()
{
unsigned int a = 10;
signed int b = -5;
printf("%d\n", a + b); /* 5 */
printf("%u\n", a + b); /* 5 */
return 0;
}
I thought the bolded paragraph applies (since unsigned int and signed int have the same rank. Why isn't b converted to unsigned ? Or perhaps it is converted to unsigned but there is something I don't understand ?
Thank you for your time :-)
Indeed b is converted to unsigned. However what you observed is that b converted to unsigned and then added to 10 gives as value 5.
On x86 32bit this is what happens
b, coverted to unsigned, becomes 4294967291 (i.e. 2**32 - 5)
adding 10 becomes 5 because of wrap-around at 2**32 (2**32 - 5 + 10 = 2**32 + 5 = 5)
0x0000000a plus 0xfffffffb will always be 0x00000005 regardless of whether you are dealing with signed or unsigned types, as long as only 32 bits are used.
Repeating the relevant portion of the code from the question:
unsigned int a = 10;
signed int b = -5;
printf("%d\n", a + b); /* 5 */
printf("%u\n", a + b); /* 5 */
In a + b, b is converted to unsigned int, (yielding UINT_MAX + 1 - 5 by the rule for unsigned-to-signed conversion). The result of adding 10 to this value is 5, by the rules of unsigned arithmetic, and its type is unsigned int. In most cases, the type of a C expression is independent of the context in which it appears. (Note that none of this depends on the representation; conversion and arithmetic are defined purely in terms of numeric values.)
For the second printf call, the result is straightforward: "%u" expects an argument of type unsigned int, and you've given it one. It prints "5\n".
The first printf is a little more complicated. "%d" expects an argument of type int, but you're giving it an argument of type unsigned int. In most cases, a type mismatch like this results in undefined behavior, but there's a special-case rule that corresponding signed and unsigned types are interchangeable as function arguments -- as long as the value is representable in both types (as it is here). So the first printf also prints "5\n".
Again, all this behavior is defined in terms of values, not representations (except for the requirement that a given value has the same representation in corresponding signed and unsigned types). You'd get the same result on a system where signed int and unsigned int are both 37 bits, signed int has 7 padding bits, unsigned int has 11 padding bits, and signed int uses a 1s'-complement or sign-and-magnitude representation. (No such system exists in real life, as far as I know.)
It is converted to unsigned, the unsigned arithmetic just happens to give the result you see.
The result of unsigned arithmetic is equivalent to doing signed arithmetic with two's complement and no out of range exception.