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# Conversion rules between unsigned operands and signed operand

Hi All,
I am trying to undestand "Type Conversions" from K&R book.I am not
able to understand the
bellow mentioned text
"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"
To understand it i wrote a small code

#include<stdio. h>
int main(void)
{
signed long sl = -5;
unsigned long ul= 5;
if (sl>ul)
{
printf("\n -5 is greater than 5\n");
}
else
{
printf("-5 is less than 5 \n");
}
return 0;
}

OP is
-5 is greater than 5
Could anybody help me to understand the conversion rules when
unsigned operands are involved ?

Regards,
Somenath

Jul 23 '07 #1
7 5055
somenath wrote:
>
Hi All,
I am trying to undestand "Type Conversions" from K&R book.I am not
able to understand the
bellow mentioned text
"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"
To understand it i wrote a small code

#include<stdio. h>
int main(void)
{
signed long sl = -5;
unsigned long ul= 5;
if (sl>ul)
{
printf("\n -5 is greater than 5\n");
}
else
{
printf("-5 is less than 5 \n");
}
return 0;
}

OP is
-5 is greater than 5
Could anybody help me to understand the conversion rules when
unsigned operands are involved ?
"rank" is the missing word here.
To compare two expressions, they must be converted to the same type.
The rule for comparing two exressions is simple,
and signed or unsigned types do not form any exceptions to this rule:
The lower ranking type is converted to the higher ranking type.

being machine dependent because of sizes, is wrong.
For example, suppose that int and long are both 32 bits.
Then -1L < 1U, because 1U, which is an unsigned int,
is *converted* (not promoted) to a signed long,
and it has absolutely nothing to do with whether or not
int and long are the same size.

"Integer promotions" on the other hand, which are distinct
from the kind of conversions performed on an operand
of a relational operator, are dependent upon whether
or not type int can represent all the positive values
of the lower ranking type.

N869
6.3.1 Arithmetic operands
6.3.1.1 Boolean, characters, and integers
[#1] Every integer type has an integer conversion rank
defined as follows:
-- No two signed integer types shall have the same rank,
even if they have the same representation.
-- The rank of a signed integer type shall be greater than
the rank of any signed integer type with less
precision.
-- 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.
-- The rank of any unsigned integer type shall equal the
rank of the corresponding signed integer type, if any.
-- The rank of any standard integer type shall be greater
than the rank of any extended integer type with the
same width.
-- The rank of char shall equal the rank of signed char
and unsigned char.
-- The rank of _Bool shall be less than the rank of all
other standard integer types.
-- The rank of any enumerated type shall equal the rank of
the compatible integer type (see 6.7.2.2).
-- The rank of any extended signed integer type relative
to another extended signed integer type with the same
precision is implementation-defined, but still subject
to the other rules for determining the integer
conversion rank.
-- For all integer types T1, T2, and T3, if T1 has greater
rank than T2 and T2 has greater rank than T3, then T1
has greater rank than T3.

/* BEGIN new.c */

#include<stdio. h>
#include<limits .h>

int main(void)
{
signed long sl = -5;
unsigned u = 5;

if (sl u) {
puts("-5 is greater than 5");
} else {
puts("-5 is less than 5");
}
printf("LONG_MA X is %ld\n", LONG_MAX);
printf("UINT_MA X is %u \n", UINT_MAX);
return 0;
}

/* END new.c */

On my machine the output of new.c is:
-5 is greater than 5
LONG_MAX is 2147483647
UINT_MAX is 4294967295

The unsigned type is converted to the long signed type,
even though on my machine,
they have the same size and number of bits.

--
pete
Jul 23 '07 #2
On Jul 23, 12:37 pm, pete <pfil...@mindsp ring.comwrote:
somenath wrote:
Hi All,
I am trying to undestand "Type Conversions" from K&R book.I am not
able to understand the
bellow mentioned text
"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"
To understand it i wrote a small code
#include<stdio. h>
int main(void)
{
signed long sl = -5;
unsigned long ul= 5;
if (sl>ul)
{
printf("\n -5 is greater than 5\n");
}
else
{
printf("-5 is less than 5 \n");
}
return 0;
}
OP is
-5 is greater than 5
Could anybody help me to understand the conversion rules when
unsigned operands are involved ?

"rank" is the missing word here.
To compare two expressions, they must be converted to the same type.
The rule for comparing two exressions is simple,
and signed or unsigned types do not form any exceptions to this rule:
The lower ranking type is converted to the higher ranking type.

being machine dependent because of sizes, is wrong.
For example, suppose that int and long are both 32 bits.
Then -1L < 1U, because 1U, which is an unsigned int,
is *converted* (not promoted) to a signed long,
and it has absolutely nothing to do with whether or not
int and long are the same size.

"Integer promotions" on the other hand, which are distinct
from the kind of conversions performed on an operand
of a relational operator, are dependent upon whether
or not type int can represent all the positive values
of the lower ranking type.

N869
6.3.1 Arithmetic operands
6.3.1.1 Boolean, characters, and integers
[#1] Every integer type has an integer conversion rank
defined as follows:
-- No two signed integer types shall have the same rank,
even if they have the same representation.
-- The rank of a signed integer type shall be greater than
the rank of any signed integer type with less
precision.
-- 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.
-- The rank of any unsigned integer type shall equal the
rank of the corresponding signed integer type, if any.
-- The rank of any standard integer type shall be greater
than the rank of any extended integer type with the
same width.
-- The rank of char shall equal the rank of signed char
and unsigned char.
-- The rank of _Bool shall be less than the rank of all
other standard integer types.
-- The rank of any enumerated type shall equal the rank of
the compatible integer type (see 6.7.2.2).
-- The rank of any extended signed integer type relative
to another extended signed integer type with the same
precision is implementation-defined, but still subject
to the other rules for determining the integer
conversion rank.
-- For all integer types T1, T2, and T3, if T1 has greater
rank than T2 and T2 has greater rank than T3, then T1
has greater rank than T3.

/* BEGIN new.c */

#include<stdio. h>
#include<limits .h>

int main(void)
{
signed long sl = -5;
unsigned u = 5;

if (sl u) {
puts("-5 is greater than 5");
} else {
puts("-5 is less than 5");
}
printf("LONG_MA X is %ld\n", LONG_MAX);
printf("UINT_MA X is %u \n", UINT_MAX);
return 0;

}

/* END new.c */

On my machine the output of new.c is:
-5 is greater than 5
LONG_MAX is 2147483647
UINT_MAX is 4294967295

The unsigned type is converted to the long signed type,
even though on my machine,
they have the same size and number of bits.

Hi Pete,
Many thanks for this information. But i am still not able to
understand the comparision.
As par my signed long has higher rank than unsigned int . so "u" will
be converted to
Signed long .Does it mean the value of "u" which is 5 will be
converted to -5 ?
Even then how "if (sl u)" is true ?

Regards,
Somenath
Jul 24 '07 #3
somenath wrote:
>
On Jul 23, 12:37 pm, pete <pfil...@mindsp ring.comwrote:
somenath wrote:
Hi All,
I am trying to undestand "Type Conversions" from K&R book.I am not
able to understand the
bellow mentioned text
"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"
To understand it i wrote a small code
#include<stdio. h>
int main(void)
{
signed long sl = -5;
unsigned long ul= 5;
if (sl>ul)
{
printf("\n -5 is greater than 5\n");
}
else
{
printf("-5 is less than 5 \n");
}
return 0;
}
OP is
-5 is greater than 5
Could anybody help me to understand the conversion rules when
unsigned operands are involved ?
"rank" is the missing word here.
To compare two expressions, they must be converted to the same type.
The rule for comparing two exressions is simple,
and signed or unsigned types do not form any exceptions to this rule:
The lower ranking type is converted to the higher ranking type.

being machine dependent because of sizes, is wrong.
For example, suppose that int and long are both 32 bits.
Then -1L < 1U, because 1U, which is an unsigned int,
is *converted* (not promoted) to a signed long,
and it has absolutely nothing to do with whether or not
int and long are the same size.

"Integer promotions" on the other hand, which are distinct
from the kind of conversions performed on an operand
of a relational operator, are dependent upon whether
or not type int can represent all the positive values
of the lower ranking type.

N869
6.3.1 Arithmetic operands
6.3.1.1 Boolean, characters, and integers
[#1] Every integer type has an integer conversion rank
defined as follows:
-- No two signed integer types shall have the same rank,
even if they have the same representation.
-- The rank of a signed integer type shall be greater than
the rank of any signed integer type with less
precision.
-- 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.
-- The rank of any unsigned integer type shall equal the
rank of the corresponding signed integer type, if any.
-- The rank of any standard integer type shall be greater
than the rank of any extended integer type with the
same width.
-- The rank of char shall equal the rank of signed char
and unsigned char.
-- The rank of _Bool shall be less than the rank of all
other standard integer types.
-- The rank of any enumerated type shall equal the rank of
the compatible integer type (see 6.7.2.2).
-- The rank of any extended signed integer type relative
to another extended signed integer type with the same
precision is implementation-defined, but still subject
to the other rules for determining the integer
conversion rank.
-- For all integer types T1, T2, and T3, if T1 has greater
rank than T2 and T2 has greater rank than T3, then T1
has greater rank than T3.

/* BEGIN new.c */

#include<stdio. h>
#include<limits .h>

int main(void)
{
signed long sl = -5;
unsigned u = 5;

if (sl u) {
puts("-5 is greater than 5");
} else {
puts("-5 is less than 5");
}
printf("LONG_MA X is %ld\n", LONG_MAX);
printf("UINT_MA X is %u \n", UINT_MAX);
return 0;

}

/* END new.c */

On my machine the output of new.c is:
-5 is greater than 5
LONG_MAX is 2147483647
UINT_MAX is 4294967295

The unsigned type is converted to the long signed type,
even though on my machine,
they have the same size and number of bits.

Hi Pete,
Many thanks for this information. But i am still not able to
understand the comparision.
As par my signed long has higher rank than unsigned int . so "u" will
be converted to
Signed long .Does it mean the value of "u" which is 5 will be
converted to -5 ?
Even then how "if (sl u)" is true ?
The answer is simple: I was wrong.
The rules are more complicated.
Both sl and u get converted to type long unsigned in the
above expression on my machine.
The reason is because on my machine,
signed long can't represent the entire range of unsigned values.

The conversion in question,
is called "the usual arithmetic conversions".

N869
6.3.1.8 Usual arithmetic conversions
[#1] Many operators that expect operands of arithmetic type
cause conversions and yield result types in a similar way.
The purpose is to determine a common real type for the
operands and result. For the specified operands, each
operand is converted, without change of type domain, to a
type whose corresponding real type is the common real type.
Unless explicitly stated otherwise, the common real type is
also the corresponding real type of the result, whose type
domain is the type domain of the operands if they are the
same, and complex otherwise. This pattern is called the
usual arithmetic conversions:
First, if the corresponding real type of either operand
is long double, the other operand is converted, without
change of type domain, to a type whose corresponding
real type is long double.
Otherwise, if the corresponding real type of either
operand is double, the other operand is converted,
without change of type domain, to a type whose
corresponding real type is double.
Otherwise, if the corresponding real type of either
operand is float, the other operand is converted,
without change of type domain, to a type whose
corresponding real type is float.
Otherwise, 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.

--
pete
Jul 25 '07 #4
On Jul 25, 8:35 am, pete <pfil...@mindsp ring.comwrote:
somenath wrote:
On Jul 23, 12:37 pm, pete <pfil...@mindsp ring.comwrote:
somenath wrote:
Hi All,
I am trying to undestand "Type Conversions" from K&R book.I am not
able to understand the
bellow mentioned text
"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"
To understand it i wrote a small code
#include<stdio. h>
int main(void)
{
signed long sl = -5;
unsigned long ul= 5;
if (sl>ul)
{
printf("\n -5 is greater than 5\n");
}
else
{
printf("-5 is less than 5 \n");
}
return 0;
}
OP is
-5 is greater than 5
Could anybody help me to understand the conversion rules when
unsigned operands are involved ?
"rank" is the missing word here.
To compare two expressions, they must be converted to the same type.
The rule for comparing two exressions is simple,
and signed or unsigned types do not form any exceptions to this rule:
The lower ranking type is converted to the higher ranking type.
being machine dependent because of sizes, is wrong.
For example, suppose that int and long are both 32 bits.
Then -1L < 1U, because 1U, which is an unsigned int,
is *converted* (not promoted) to a signed long,
and it has absolutely nothing to do with whether or not
int and long are the same size.
"Integer promotions" on the other hand, which are distinct
from the kind of conversions performed on an operand
of a relational operator, are dependent upon whether
or not type int can represent all the positive values
of the lower ranking type.
N869
6.3.1 Arithmetic operands
6.3.1.1 Boolean, characters, and integers
[#1] Every integer type has an integer conversion rank
defined as follows:
-- No two signed integer types shall have the same rank,
even if they have the same representation.
-- The rank of a signed integer type shall be greater than
the rank of any signed integer type with less
precision.
-- 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.
-- The rank of any unsigned integer type shall equal the
rank of the corresponding signed integer type, if any.
-- The rank of any standard integer type shall be greater
than the rank of any extended integer type with the
same width.
-- The rank of char shall equal the rank of signed char
and unsigned char.
-- The rank of _Bool shall be less than the rank of all
other standard integer types.
-- The rank of any enumerated type shall equal the rank of
the compatible integer type (see 6.7.2.2).
-- The rank of any extended signed integer type relative
to another extended signed integer type with the same
precision is implementation-defined, but still subject
to the other rules for determining the integer
conversion rank.
-- For all integer types T1, T2, and T3, if T1 has greater
rank than T2 and T2 has greater rank than T3, then T1
has greater rank than T3.
/* BEGIN new.c */
#include<stdio. h>
#include<limits .h>
int main(void)
{
signed long sl = -5;
unsigned u = 5;
if (sl u) {
puts("-5 is greater than 5");
} else {
puts("-5 is less than 5");
}
printf("LONG_MA X is %ld\n", LONG_MAX);
printf("UINT_MA X is %u \n", UINT_MAX);
return 0;
}
/* END new.c */
On my machine the output of new.c is:
-5 is greater than 5
LONG_MAX is 2147483647
UINT_MAX is 4294967295
The unsigned type is converted to the long signed type,
even though on my machine,
they have the same size and number of bits.
Hi Pete,
Many thanks for this information. But i am still not able to
understand the comparision.
As par my signed long has higher rank than unsigned int . so "u" will
be converted to
Signed long .Does it mean the value of "u" which is 5 will be
converted to -5 ?
Even then how "if (sl u)" is true ?

The answer is simple: I was wrong.
The rules are more complicated.
Both sl and u get converted to type long unsigned in the
But does it mean -5 will be converted to 5 ? Even then how "if
(sl u)" is true ?

Jul 25 '07 #5
On Jul 25, 8:35 am, pete <pfil...@mindsp ring.comwrote:
somenath wrote:
On Jul 23, 12:37 pm, pete <pfil...@mindsp ring.comwrote:
somenath wrote:
Hi All,
I am trying to undestand "Type Conversions" from K&R book.I am not
able to understand the
bellow mentioned text
"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"
To understand it i wrote a small code
#include<stdio. h>
int main(void)
{
signed long sl = -5;
unsigned long ul= 5;
if (sl>ul)
{
printf("\n -5 is greater than 5\n");
}
else
{
printf("-5 is less than 5 \n");
}
return 0;
}
OP is
-5 is greater than 5
Could anybody help me to understand the conversion rules when
unsigned operands are involved ?
"rank" is the missing word here.
To compare two expressions, they must be converted to the same type.
The rule for comparing two exressions is simple,
and signed or unsigned types do not form any exceptions to this rule:
The lower ranking type is converted to the higher ranking type.
being machine dependent because of sizes, is wrong.
For example, suppose that int and long are both 32 bits.
Then -1L < 1U, because 1U, which is an unsigned int,
is *converted* (not promoted) to a signed long,
and it has absolutely nothing to do with whether or not
int and long are the same size.
"Integer promotions" on the other hand, which are distinct
from the kind of conversions performed on an operand
of a relational operator, are dependent upon whether
or not type int can represent all the positive values
of the lower ranking type.
N869
6.3.1 Arithmetic operands
6.3.1.1 Boolean, characters, and integers
[#1] Every integer type has an integer conversion rank
defined as follows:
-- No two signed integer types shall have the same rank,
even if they have the same representation.
-- The rank of a signed integer type shall be greater than
the rank of any signed integer type with less
precision.
-- 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.
-- The rank of any unsigned integer type shall equal the
rank of the corresponding signed integer type, if any.
-- The rank of any standard integer type shall be greater
than the rank of any extended integer type with the
same width.
-- The rank of char shall equal the rank of signed char
and unsigned char.
-- The rank of _Bool shall be less than the rank of all
other standard integer types.
-- The rank of any enumerated type shall equal the rank of
the compatible integer type (see 6.7.2.2).
-- The rank of any extended signed integer type relative
to another extended signed integer type with the same
precision is implementation-defined, but still subject
to the other rules for determining the integer
conversion rank.
-- For all integer types T1, T2, and T3, if T1 has greater
rank than T2 and T2 has greater rank than T3, then T1
has greater rank than T3.
/* BEGIN new.c */
#include<stdio. h>
#include<limits .h>
int main(void)
{
signed long sl = -5;
unsigned u = 5;
if (sl u) {
puts("-5 is greater than 5");
} else {
puts("-5 is less than 5");
}
printf("LONG_MA X is %ld\n", LONG_MAX);
printf("UINT_MA X is %u \n", UINT_MAX);
return 0;
}
/* END new.c */
On my machine the output of new.c is:
-5 is greater than 5
LONG_MAX is 2147483647
UINT_MAX is 4294967295
The unsigned type is converted to the long signed type,
even though on my machine,
they have the same size and number of bits.
Hi Pete,
Many thanks for this information. But i am still not able to
understand the comparision.
As par my signed long has higher rank than unsigned int . so "u" will
be converted to
Signed long .Does it mean the value of "u" which is 5 will be
converted to -5 ?
Even then how "if (sl u)" is true ?

The answer is simple: I was wrong.
The rules are more complicated.
Both sl and u get converted to type long unsigned in the
But does it mean -5 will be converted to 5 ? Even then how "if
(sl u)" is true ?

Jul 25 '07 #6
somenath wrote:
>
On Jul 25, 8:35 am, pete <pfil...@mindsp ring.comwrote:
somenath wrote:
On Jul 23, 12:37 pm, pete <pfil...@mindsp ring.comwrote:
somenath wrote:
Hi All,
I am trying to undestand "Type Conversions" from K&R book.I am not
able to understand the
bellow mentioned text
"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"
To understand it i wrote a small code
#include<stdio. h>
int main(void)
{
signed long sl = -5;
unsigned long ul= 5;
if (sl>ul)
{
printf("\n -5 is greater than 5\n");
}
else
{
printf("-5 is less than 5 \n");
}
return 0;
}
OP is
-5 is greater than 5
Could anybody help me to understand the conversion rules when
unsigned operands are involved ?
"rank" is the missing word here.
To compare two expressions, they must be converted to the same type.
The rule for comparing two exressions is simple,
and signed or unsigned types do not form any exceptions to this rule:
The lower ranking type is converted to the higher ranking type.
being machine dependent because of sizes, is wrong.
For example, suppose that int and long are both 32 bits.
Then -1L < 1U, because 1U, which is an unsigned int,
is *converted* (not promoted) to a signed long,
and it has absolutely nothing to do with whether or not
int and long are the same size.
"Integer promotions" on the other hand, which are distinct
from the kind of conversions performed on an operand
of a relational operator, are dependent upon whether
or not type int can represent all the positive values
of the lower ranking type.
N869
6.3.1 Arithmetic operands
6.3.1.1 Boolean, characters, and integers
[#1] Every integer type has an integer conversion rank
defined as follows:
-- No two signed integer types shall have the same rank,
even if they have the same representation.
-- The rank of a signed integer type shall be greater than
the rank of any signed integer type with less
precision.
-- 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.
-- The rank of any unsigned integer type shall equal the
rank of the corresponding signed integer type, if any.
-- The rank of any standard integer type shall be greater
than the rank of any extended integer type with the
same width.
-- The rank of char shall equal the rank of signed char
and unsigned char.
-- The rank of _Bool shall be less than the rank of all
other standard integer types.
-- The rank of any enumerated type shall equal the rank of
the compatible integer type (see 6.7.2.2).
-- The rank of any extended signed integer type relative
to another extended signed integer type with the same
precision is implementation-defined, but still subject
to the other rules for determining the integer
conversion rank.
-- For all integer types T1, T2, and T3, if T1 has greater
rank than T2 and T2 has greater rank than T3, then T1
has greater rank than T3.
/* BEGIN new.c */
#include<stdio. h>
#include<limits .h>
int main(void)
{
signed long sl = -5;
unsigned u = 5;
if (sl u) {
puts("-5 is greater than 5");
} else {
puts("-5 is less than 5");
}
printf("LONG_MA X is %ld\n", LONG_MAX);
printf("UINT_MA X is %u \n", UINT_MAX);
return 0;
}
/* END new.c */
On my machine the output of new.c is:
-5 is greater than 5
LONG_MAX is 2147483647
UINT_MAX is 4294967295
The unsigned type is converted to the long signed type,
even though on my machine,
they have the same size and number of bits.
Hi Pete,
Many thanks for this information. But i am still not able to
understand the comparision.
As par my signed
long has higher rank than unsigned int . so "u" will
be converted to
Signed long .Does it mean the value of "u" which is 5 will be
converted to -5 ?
Even then how "if (sl u)" is true ?
The answer is simple: I was wrong.
The rules are more complicated.
Both sl and u get converted to type long unsigned in the

But does it mean -5 will be converted to 5 ?
No. The converted expression is equal to ((long unsigned)-5).
Even then how "if
(sl u)" is true ?
The rule for converting a negative value to an unsigned integer type
is to add (1 + UNSIGNED_INTEGE R_TYPE_MAX) repeatedly
to the negative value until it is not negative any more.

-5 gets converted to 4294967291LU

--
pete
Jul 25 '07 #7
On Jul 25, 3:35 pm, pete <pfil...@mindsp ring.comwrote:
somenath wrote:
On Jul 25, 8:35 am, pete <pfil...@mindsp ring.comwrote:
somenath wrote:
On Jul 23, 12:37 pm, pete <pfil...@mindsp ring.comwrote:
somenath wrote:
Hi All,
I am trying to undestand "Type Conversions" from K&R book.I am not
able to understand the
bellow mentioned text
"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"
To understand it i wrote a small code
#include<stdio. h>
int main(void)
{
signed long sl = -5;
unsigned long ul= 5;
if (sl>ul)
{
printf("\n -5 is greater than 5\n");
}
else
{
printf("-5 is less than 5 \n");
}
return 0;
}
OP is
-5 is greater than 5
Could anybody help me to understand the conversion rules when
unsigned operands are involved ?
"rank" is the missing word here.
To compare two expressions, they must be converted to the same type.
The rule for comparing two exressions is simple,
and signed or unsigned types do not form any exceptions to this rule:
The lower ranking type is converted to the higher ranking type.
being machine dependent because of sizes, is wrong.
For example, suppose that int and long are both 32 bits.
Then -1L < 1U, because 1U, which is an unsigned int,
is *converted* (not promoted) to a signed long,
and it has absolutely nothing to do with whether or not
int and long are the same size.
"Integer promotions" on the other hand, which are distinct
from the kind of conversions performed on an operand
of a relational operator, are dependent upon whether
or not type int can represent all the positive values
of the lower ranking type.
N869
6.3.1 Arithmetic operands
6.3.1.1 Boolean, characters, and integers
[#1] Every integer type has an integer conversion rank
defined as follows:
-- No two signed integer types shall have the same rank,
even if they have the same representation.
-- The rank of a signed integer type shall be greater than
the rank of any signed integer type with less
precision.
-- 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.
-- The rank of any unsigned integer type shall equal the
rank of the corresponding signed integer type, if any.
-- The rank of any standard integer type shall be greater
than the rank of any extended integer type with the
same width.
-- The rank of char shall equal the rank of signed char
and unsigned char.
-- The rank of _Bool shall be less than the rank of all
other standard integer types.
-- The rank of any enumerated type shall equal the rank of
the compatible integer type (see 6.7.2.2).
-- The rank of any extended signed integer type relative
to another extended signed integer type with the same
precision is implementation-defined, but still subject
to the other rules for determining the integer
conversion rank.
-- For all integer types T1, T2, and T3, if T1 has greater
rank than T2 and T2 has greater rank than T3, then T1
has greater rank than T3.
/* BEGIN new.c */
#include<stdio. h>
#include<limits .h>
int main(void)
{
signed long sl = -5;
unsigned u = 5;
if (sl u) {
puts("-5 is greater than 5");
} else {
puts("-5 is less than 5");
}
printf("LONG_MA X is %ld\n", LONG_MAX);
printf("UINT_MA X is %u \n", UINT_MAX);
return 0;
}
/* END new.c */
On my machine the output of new.c is:
-5 is greater than 5
LONG_MAX is 2147483647
UINT_MAX is 4294967295
The unsigned type is converted to the long signed type,
even though on my machine,
they have the same size and number of bits.
Hi Pete,
Many thanks for this information. But i am still not able to
understand the comparision.
As par my signed
long has higher rank than unsigned int . so "u" will
be converted to
Signed long .Does it mean the value of "u" which is 5 will be
converted to -5 ?
Even then how "if (sl u)" is true ?
The answer is simple: I was wrong.
The rules are more complicated.
Both sl and u get converted to type long unsigned in the
But does it mean -5 will be converted to 5 ?

No. The converted expression is equal to ((long unsigned)-5).
Even then how "if
(sl u)" is true ?

The rule for converting a negative value to an unsigned integer type
is to add (1 + UNSIGNED_INTEGE R_TYPE_MAX) repeatedly
to the negative value until it is not negative any more.

-5 gets converted to 4294967291LU
But is this rules applies to largest negative number such as
-2147483648 ? then what will be the result when it is converted to
unsigned integer type?
Jul 30 '07 #8

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