The arithmetic data types in C , are the integer types , such as
int or unsigned long , and the floating point types , such as
float , or long double .
Table of Contents
Conversion of the integer types
What is widening ?
Widening only applies to the signed and unsigned integer types . It does
not apply to other types , such as float or double .
Widening is not about converting from signed to unsigned , or from unsigned to signed , it is about expanding the signedness of an integer type , from a smaller type to a larger type , so from a smaller number of bits , to a larger one . The signedness of the integer type does not change .
For the signed type , widening is done by what is called :
sign extension . If the value of the signed type is negative , it is extended by filling the
new bits by the value 1 , if the value of the signed type is nonnegative , than it is
extended by filling the newly allocated bits , with 0 .
For the unsigned type , the newly allocated bits are filled
with 0 .
What is truncation ?
Truncation happens only for the integer types , that have the
same signedness , when passing from a larger integer type , to a smaller integer type , such as , when
passing from int to char , or from unsigned int to
unsigned char.
Be it is signed or unsigned , the larger type is made to fit the smaller type , by discarding the bits , of the larger type , that lies outside of the width of the smaller type , keeping only the lower order bits .
Truncation as described , which is keeping only a limited number of bits ,
doesn’t happen when converting from a floating point type , to an integer type , or
between floating point types .
When converting from a floating type , to an integer type , the fractional part is discarded , the
floating type bits are not made to fit the integer type width , but the floating type is transformed from
a floating point representation , to an integer representation .
The integer conversion procedure
Conversion of the integer types , consists of either
performing widening or truncation first , and later reinterpreting the bits .
So first truncation or widening is performed . Truncation or widening does not change the
signedness of the type , the type is only made to fit , a larger or narrower width , of the
same signedness .
After having performed widening or truncation , the gotten bits in the target width , are only
reinterpreted , as belonging to a new integer type , the target type .
The only exception to this rule , is when converting to the
_Bool type , any nonzero value is converted to 1 , and any zero value , is
converted to 0 .
Having explained , how conversion happens for the integer types , let us
explain the concept of integer rank , before explaining when conversion to another
integer type , takes place .
What is a rank ?
Each integer type in C has a rank .
unsigned , and signed integer types of the
same type , disregarding the signedness , have the same rank . For example
int , and unsigned int , have the same rank .
The order of the ranks of the integer types is , as follow :
_Bool < signed char < short < int < long < long long
The rank of any standard integer type , outlined above , is larger than the rank of any implementation defined , extended integer type , having the same width .
The char , and signed char integer types , have the same
rank .
The rank of an enum type is equal to its assigned implementation , defined integer type .
Integer promotion
Integer types , which have a rank smaller than int
, each is promoted , to either the int type , if the int type can represent all
its possible values , or to the unsigned int type , in the other case .
Integer types having a rank smaller than int , are
promoted when an operator expects , that one of its operand , to be of an arithmetic type .
Arithmetic types in C , are the integer or floating point types . An example of such an operator , is the
subtraction operator - , or the unary negation operator - .
Integer promotion , is not how the integer is converted to another type ,
it is when a integer is converted to another type , in this case it is because it has a
lower rank than int .
unsigned char x = 1;
signed char y = -x;
/*
x , is an unsigned char type ,
it has a rank lower than int .
The unary negation operator is
used , as such x must be promoted .
An int on 32 bits systems , have a
typical width of 32 bits , as such
it can store all the possible values
of an unsigned char , which is limited
to only 8 bits . As such the target
promotion type is an int .
Now that the target type is decided ,
which is an int , it has a larger
width than an unsigned char , as
such widening must be performed .
Widening does not change signedness ,
so the unsigned char is widened to an
unsigned int , by using zero fill .
The resulting value is 1 , and has a bit
representation of :
00000000000000000000000000000001
The resulting bits , are interpreted as if ,
as being of the type signed int , and the
negation operator is applied .
The result of the negation operator is -1 ,
which , has a bit representation of :
11111111111111111111111111111111
The value is to be stored in a signed
char type , both int and signed char ,
have the same signedness , as such
truncation is applied .
The width of a signed char type is
8 bits , as such the leading 24 bits
are discarded , and the result is :
11111111 , which is equal to -1 . */
Function call or return value
When making a function call , and an argument is of an integer type , different from the target parameter integer type , type conversion occurs .
The argument is converted to the parameter type , by first truncation , or widening , to the same width , and later on reinterpreting the bits as of the target parameter type .
#include<stdio.h>
void trivialFunction( unsigned int val){
if( val == 4294967295){
printf( "%u is equal to 4294967295\n" , val);
}
else{
printf( "%u is different from 4294967295\n" , val);}}
int main( void){
signed char x = -1 ;
trivialFunction( x);
unsigned char y = x ;
trivialFunction( y);}
/* Output
4294967295 is equal to 4294967295
255 is different from 4294967295
*/
/*
In the first call to trivialFunction ,
the passed argument is of type signed
char . trivialFunction expects its
argument , to be of the type unsigned
int , as such the argument must be
converted .
First widening to the same width of
unsigned int takes place .
x is a signed char , as such it is
widened to a signed int , by sign
extension . The value of x is -1 ,
and it has a bit representation of
11111111 . The resulting value is -1 ,
and it has a bit representation of
11111111111111111111111111111111 .
This bit pattern , is next reinterpreted
as an unsigned int , so
11111111111111111111111111111111 ,
as an unsigned int , has a value of
4294967295 . This is why trivialFunction
prints : 4294967295 is equal to 4294967295
The variable y , has a type of unsigned
char . It is assigned the value of x ,
which is a signed char .
Both x , and y have the same width , as such
no widening occurs , only the bits of x ,
are reinterpreted as being unsigned .
The bits of x , has a value of -1 , which is
11111111 , when reinterpreted as unsigned ,
this yield the value of 255 .
Next trivialFunction is called with y ,
as a parameter .
Widening occurs , because y is an unsigned
char , and the function parameter is an
unsigned int . It is done by using
zero fill .
Hence the value of 1 which has a bit
representation of 11111111 , is widened
to the value 255 which has a bit
representation of :
00000000000000000000000011111111
The function prints :
255 is different from 4294967295
*/
When the return value of a function is different from its return type , the return value is converted to the function’s return type .
Assignment and initialization
When performing assignment or initialization to an integer
variable , using the = operator , and the expression to be assigned , is of a different
integer type , integer conversion takes place .
unsigned char x = 1 ; /* 1 is an integral literal , it is of the type int . The type int has a typical width of 32 bits , while an unsigned char has a typical width of 8 bits . The bits of the int type 00000000000000000000000000000001 are reinterpreted as being the bits of an unsigned int type . The result is : 00000000000000000000000000000001 The unsigned int value is truncated to 8 bits , and the value gotten is 00000001 , which is assigned to the unsigned char x .*/
For further information , about the type of integer literals , you can check this article .
Arithmetic operators
When one of the following operators , is being used , an integer conversion will occur .
* / % + - /* Multiplication , division , modulos , addition , subtraction .*/ < <= > >= == != /* Relational less , less or equal , larger , larger or equal , equal , not equal .*/ & ^ | /* Bitwise and , xor , or .*/ ?: /* The Ternary operator expression , must return a value of a specific type , the second and third operand , are converted to a same type . */ += -= *= /= %= &= ^= |= /* operate and assign operators */
In the case of these operators , a common type for the operands and the result must be determined . This is done as specified by the following table .
| Unsigned | Signed | Rank | Operands and Result type |
|---|---|---|---|
| uT | sT | uT >= sT | uT |
| uT | sT | uT < sT | sT , if sT is capable of holding , all the possible values of : uT . The unsigned type of sT , if sT is not capable of holding all the possible values of : uT . |
uT , means an unsigned type , such as :
unsigned intsT , means a signed type , such as : intThe result of relational operators such as < , is always 0 for true , and 1 for false , and it is always of the type int . |
|||
int si = -1 ;
unsigned int ui = 0 ;
ui = ui + si ;
/*
int , and unsigned int , have the same
rank , one is signed , and the other
is unsigned , as such both operands ,
must be converted to the unsigned int
type , and the result of the operation ,
is of an the unsigned int type .
ui is unsigned , so no conversion is
necessary .
To convert si , to the unsigned int type ,
and since both si and ui have the same
width , the bits of si , which are
11111111111111111111111111111111 ,
are kept as is , they are only reinterpreted
as belonging to a signed type , so now
they have a value of 4294967295 .
The addition is performed , and the result
of the addition operation , is as such :
4294967295 + 0 = 4294967295 , and is
of the type unsigned int .
ui is of type unsigned int , as such
no conversion is necessary , and the
result of 4294967295 is stored in ui .
*/
long long lli = -1 ;
ui = lli + ui ;
/*
lli is of the type long long ,
it has a higher rank than
unsigned int . long long
can hold all the values of
unsigned int , as such ,
both operands must be of
the type long long .
lli is of the type long long ,
so no conversion is necessary .
ui is an unsigned int , it has
a bit representation of
11111111111111111111111111111111
it is first extended to unsigned
long by using zero fill .
0000000000000000000000000000000011111111111111111111111111111111
After that , the gotten bits are
reinterpreted as being of type
long long . The gotten value , is
the same as the value of ui , which
is 4294967295 .
The addition is performed between ,
the two long long integer types , and
the result is -1 + 4294967295 , which
is equal to 4294967294 , and is of
the type long long .
ui is of the type unsigned int , the result
which is of the type long long must
be converted to unsigned int .
It is first converted to unsigned long long ,
the bits pattern does not change , so it remains
0000000000000000000000000000000011111111111111111111111111111110 .
Now that is is of the type unsigned long long ,
it is truncated to the type of ui , which is
unsigned int , and it has the format :
11111111111111111111111111111110 ,
and a value of : 4294967294 .
*/
long int li = -1 ;
li = li + ui ;
/*
ui is an unsigned int , whereas li
is a long int . A long int has
a higher rank than an unsigned
int . Assuming that on this machine ,
both long int and unsigned int have
a width of 32 bits , this means
that long int is not capable of
representing all the possible values
of the unsigned int type , as such
the operands , and the result
must be of the unsigned long type .
li bits are kept as is , and only
reinterpreted as being of the
type unsigned long , as such li
will have the value of 4294967295 .
ui is converted to unsigned long , by
widening. unsigned int , and unsigned long
have the same width , as such the bits
of ui remains the same ,
11111111111111111111111111111110 .
Adding 4294967295 + 4294967294 =
8589934589 . The result is of
the type unsigned long , it cannot
fit the width of the unsigned
long type , which has a max
value of 4294967295 , as such
overflow has occurred .
The modulo of the result , with
regards to 2 to the power of
the number of bits ,
of the unsigned long type is taken .
This is equal to
8589934589 % 4294967296 =
4294967293 , which is
11111111111111111111111111111101
The result must be stored in li ,
li is of the long type , the gotten
value is of the type unsigned long ,
the bits are kept as is , and the
result is only reinterpreted as being
a signed long , as such the value
of li is -3 .
*/
For information on overflow of the signed , and unsigned integer types , you can check this , and this article .
Cast operator
Explicit conversion happens , when using the
casting operator : (Type) expression . As an example , explicitly casting the
int literal 1 , to the long type .
(long) 1
When using the cast operator , this is called
explicit casting , all other cases of conversion , are called implicit casting .
Conversion of the floating point types
Converting from one floating type , to another
Each floating point type , has a different range , and precision . As such , when passing from floating point types , with higher precision and range , to floating point types with smaller precision and range , a loss of precision , an underflow , or an overflow , can occur .
double dbl = 16777217 ; float fl = dbl ; /* Loss of precision , fl value is 16777216 .*/ dbl = 1.2e-50 ; fl = dbl ; /* Underflow occurs , the behavior is implementation defined , in this case , fl has a value of 0 .*/ dbl = 3.4e100; fl = dbl ; /* Overflow occurs , the behavior is implementation defined , fl has a value of positive infinity */
When passing from a floating point type , with a smaller range and precision , to a floating point type , with a higher range and precision , no precision loss , underflow , or overflow , occurs . The precision and range are preserved .
float fl = 1.099511627776e12f; double dbl = fl ; /* precision and range are preserved , dbl has a value of : 1.099511627776e12f , which is 2 to the power 40 .*/
Converting from a floating point type , to an integer type
The process of converting a floating point type , to an integer type , can be thought of , as if , the number is first converted to the decimal notation , then the fractional part is discarded , then it is represented in its signed or unsigned representation .
double dbl = 1.3f; unsigned char uc = (unsigned char) x ; // uc is equal to 1 dbl = -1.3f; uc = dbl ; // uc is equal to 255
If the floating point number , is too large to be represented in an integer type , the behavior is not defined by the C standard , it is defined by the implementation .
double dbl = 3.4E30 ; unsigned long ul = dbl ; //ul is equal to 0 unsigned int ui = dbl ; //ui is equal to 0 unsigned char uc = dbl ; //uc is equal to 0 long li = dbl ; //li is equal to -9223372036854775808 int si = dbl ; //si is equal to -2147483648 signed char sc = dbl ; //sc is equal to 0
Converting from an integer type , to a floating point type
When converting an integer type , to a floating point type , the integer value can always be represented , but there might be a loss of precision .
int si = 16777216 ; float fl = si ; // fl is equal to 16777216 si = 16777217 ; float fl = si ; // fl is equal to 16777216
When does the conversion occurs ?
When a floating point type is involved , conversion occurs , when a function call is made , and the passed argument is of a different type than the function parameter , or when the return value of a function , is of different type , than its return type .
It also happens , when performing floating point variables , assignment and initialization . Finally , it
occurs when using arithmetic operators , or when done explicitly , using explicit casting , such as
(float) 1.0 .
Rank of floating point types
The rank of a floating point type , is always higher than the rank of an integer type , so an integer type , is always converted to a floating point type .
As for the floating point types , they have the following ranks :
float < double < long double
If an arithmetic operation , involves two floating point types of different ranks , the floating point type with the lower rank , is converted to the floating point type with the higher rank .
The type of the result of an operation involving a floating point type , is the same type , as the one determined by the ranking algorithm .
Floating point types promotion
As with the integer type promotion , when performing arithmetic operations , a floating point type can be promoted , to a type with a higher precision and range .
This is not to be confused with how conversion happens , or in which rank , an arithmetic operation involving a floating point type , is to be performed , It is always performed using the higher operand rank .
The promotion rule , is defined in the macro
FLT_EVAL_METHOD , defined in the header float.h .
If FLT_EVAL_METHOD is set to 0 , then arithmetic
operations are done in the type of the widest operand , so if both operands are float , the
arithmetic operation is done using the float type , so no promotion occurs .
If FLT_EVAL_METHOD is set to 1 , then arithmetic
operations are performed , by promoting the operands to long double , if any operand is of
the long double type , otherwise operands are promoted to the double type , even
if both operands are of the float type .
If FLT_EVAL_METHOD is set to 2 , then arithmetic
operations are performed , by promoting the operands , to the long double type .
If FLT_EVAL_METHOD is set to -1 , then the behavior
is not defined .



