Chapter 5: Expressions
What's in Chapter 5?
Precedence and associativity
Unary operators
Binary operators
Assignment operators
Expression type and explicit
casting
Selection operator
Arithmetic overflow and
underflow
Most programming languages support the traditional concept of an expression as a combination of constants, variables, array elements, and function calls joined by various operators (+, , etc.) to produce a single numeric value. Each operator is applied to one or two operands (the values operated on) to produce a single value which may itself be an operand for another operator. This idea is generalized in C by including nontraditional data types and a rich set of operators. Pointers, unsubscripted array names, and function names are allowed as operands. And, as Tables 51 through 56 illustrate, many operators are available. All of these operators can be combined in any useful manner in an expression. As a result, C allows the writing very compact and efficient expressions which at first glance may seem a bit strange. Another unusual feature of C is that anywhere the syntax calls for an expression, a list of expressions, with comma separators, may appear.
The basic problem in evaluating expressions is deciding which parts of an expression are to be associated with which operators. To eliminate ambiguity, operators are given three properties: operand count, precedence, and associativity.
Operand count refers to the classification of operators as unary, binary, or ternary according to whether they operate on one, two, or three operands. The unary minus sign, for instance, reverses the sign of the following operand, whereas the binary minus sign subtracts one operand from another.
The following example converts the distance x in inches to a distance y in cm. Without parentheses the following statement seems ambiguous
y = 254*x/100;
If we divide first, then y can only take on values that are multiples of 254 (e.g., 0 254 508 etc.) So the following statement is incorrect.
y = 254*(x/100);
The proper approach is to multiply first then divide. To multiply first we must guarantee that the product 254*x will not overflow the precision of the computer. How do we know what precision the compiler used for the intermediate result 254*x? To answer this question, we must observe the assembly code generated by the compiler. Since multiplication and division associate left to right, the first statement without parentheses although ambiguous will actually calculate the correct answer. It is good programming style to use parentheses to clarify the expression. So this last statement has both good style and proper calculation.
y = (254*x)/100;
The issues of precedence and associativity were explained in Chapter 1. Precedence defines the evaluation order. For example the expression 3+4*2 will be 11 because multiplication as precedence over addition. Associativity determines the order of execution for operators that have the same precedence. For example, the expression 1032 will be 5, because subtraction associates left to right. On the other hand, if x and y are initially 10, then the expression x+=y+=1 will first make y=y+1 (11), then make x=x+y (21) because the operator += associates right to left. The table from chapter 1 is repeated for your convenience.
Precedence  Operators  Associativity 
highest  () []
. > ++(postfix)
(postfix) 
left to right 
++(prefix) (prefix)
!~ sizeof(type)
+(unary) (unary)
&(address) *(dereference) 
right to left  
*
/ % 
left to right  
+  
left to right  
<<
>> 
left to right  
<
<= > >= 
left to right  
== != 
left to right  
&  left to right  
^  left to right  
  left to right  
&&  left to right  
  left to right  
? :  right to left  
= +=
=
*= /=
%= <<=
>>= =
&= ^=

right to left  
lowest  ,  left to right 
Table 14: Precedence and associativity determine the order of operation
We begin with
the unary operators, which take a single input and give a single
output. In the following examples, assume all numbers are 16 bit signed
(short). The following variables are listed
short data; /* 32767 to +32767 */
short *pt; /* pointer to memory */
short flag; /* 0 is false, not zero is true */
operator

meaning

example  result 
~ 
binary complement  ~0x1234  0xEDCB 
! 
logical complement  !flag  flip 0 to 1 and notzero to 0 
& 
address of  &data  address in memory where data is stored 
 
negate  100  negative 100 
+ 
positive  +100  100 
++ 
preincrement  ++data  data=data+1, then result is data 
 
predecrement  data  data=data1, then result is data 
* 
reference  *pt  16 bit information pointed to by pt 
operator

meaning

example  result 
++ 
postincrement  data++  result is data, then data=data+1 
 
postdecrement  data  result is data, then data=data+1 
Next we list the binary arithmetic operators, which operate on two number inputs giving a single number result. The operations of addition, subtraction and shift left are the same independent of whether the numbers are signed or unsigned. As we will see later, overflow and underflow after an addition, subtraction and shift left are different for signed and unsigned numbers, but the operation itself is the same. On the other hand multiplication, division, and shift right have different functions depending on whether the numbers are signed or unsigned. It will be important, therefore, to avoid multiplying or dividing an unsigned number with a signed number.
operator

meaning

example  result 
+ 
addition  100+300  400 
 
subtraction  100300  200 
* 
multiplication  10*300  3000 
/ 
division  123/10  12 
% 
remainder  123%10  3 
<< 
shift left  102<<2  408 
>> 
shift right  102>>2  25 
The binary bitwise logical operators take two inputs and give a single result.
operator

meaning

example  result 
& 
bitwise and  0x1234&0x00FF  0x0034 
 
bitwise or  0x12340x00FF  0x12FF 
^ 
bitwise exclusive or  0x1234^0x00FF  0x12CB 
The binary Boolean operators take two Boolean inputs and give a single Boolean result.
operator

meaning

example  result 
&& 
and  0 && 1  0 (false) 
 
or  0  1  1 (true) 
Many
programmers confuse the logical operators with the Boolean operators.
Logical operators take two numbers and perform a bitwise logical
operation. Boolean operators take two Boolean inputs (0 and notzero)
and return a Boolean (0 or 1). In the program below, the
operation c=a&b;
will perform a bitwise
logical and of 0x0F0F and 0xF0F0
resulting in 0x0000. In the d=a&&b;
expression, the value a is considered as a true (because it is not
zero) and the value b also is considered a true (not zero). The Boolean
operation of true and true gives a true result (1).
short a,b,c,d;
void main(void){ a=0x0F0F; b=F0F0;
c = a&b; /* logical
result c will be 0x0000 */
d = a&&b; /* Boolean result d will
be 1 (true) */
}
Listing 51: Illustration of the difference between logical and Boolean operators
The binary relational operators take two number inputs and give a single Boolean result.
operator

meaning

example  result 
== 
equal  100 == 200  0 (false) 
!= 
not equal  100 != 200  1 (true) 
<  less than  100 < 200  1 (true) 
<=  less than or equal  100 <= 200  1 (true) 
>  greater than  100 > 200  0 (false) 
>=  greater than or equal  100 >= 200  0 (false) 
Some
programmers confuse assignment equals with the relational
equals. In the following example, the first if
will execute the subfunction()
if a is equal to zero (a is
not modified). In the second case, the variable b is set to zero, and
the subfunction()
will never be executed because
the result of the equals assignment is the value (in this case the 0
means false).
short a,b;
void program(void){
if(a==0) subfunction(); /* execute
subfunction if a is zero */
if(b=0) subfunction(); /* set b
to zero, never execute subfunction */
}
Listing 52: Illustration of the difference between relational and assignment equals
Before looking at the kinds of expressions we can write in C, we will first consider the process of evaluating expressions and some general properties of operators.
The assignment
operator is used to store data into variables. The syntax is variable=expression;
where variable
has been previously defined.
At run time, the result of the expression is saved into the variable.
If the type of the expression is different from the variable, then the
result is automatically converted. For more information about types and
conversion, see expression
type and explicit casting. The assignment operation
itself has a result, so the assignment operation can be nested.
short a,b;
void initialize(void){
a = b = 0; /* set both variables to zero
*/
}
Listing 53: Example of a nested assignment operation
The
read/modify write assignment operators are convenient. Examples are
shown below.
short a,b;
void initialize(void){
a += b; /*
same as a=a+b */
a = b; /*
same as a=ab */
a *= b; /*
same as a=a*b */
a /= b; /*
same as a=a/b */
a %= b; /*
same as a=a%b */
a <<=
b; /* same as a=a<<b */
a <<=
b; /* same as a=a<<b */
a >>=
b; /* same as a=a>>b */
a = b; /*
same as a=ab */
a &=
b; /* same as a=a&b */
a ^= b; /*
same as a=a^b */
}
Listing 54 List of all read/modify/write assignment operations
Most compilers
will produce the same code for the short and long version of the
operation. Therefore you should use the read/modify/write operations
only in situations that make the software easier to understand.
void function(void){
PORTA = 0x01; /* set PA0 high
*/
PORTB &=~ 0x80; /* clear PB7 low
*/
PORTC ^= 0x40; /* toggle PC6 */
}
Listing 55 Good examples of read/modify/write assignment operations
Expression Types and Explicit Casting
We saw earlier that numbers are represented in the computer using a wide range of formats. A list of these formats is given in Table 5.7. Notice that for the 6811 and 6812, the int and short types are the same. On the other hand with the Intel Pentium, the int and long types are the same. This difference may cause confusion, when porting code from one system to another. I suggest you use the int type when you are interested in efficiency and don't care about precision, and use the short type when you want a variable with a 16bit precision.
type  range  precision  example variable 
unsigned char  0 to 255  8 bits  unsigned char uc; 
char  127 to 127  8 bits  char sc; 
unsigned int  0 to 65535U  16 bits  unsigned int ui; 
int  32767 to 32767  16 bits  int si; 
unsigned short  0 to 65535U  16 bits  unsigned short us; 
short  32767 to 32767  16 bits  short ss; 
long  2147483647L to 2147483647L  32 bits  long sl; 
An obvious
question arises, what happens when two numbers of different types are
operated on? Before operation, the C compiler will first convert one or
both numbers so they have the same type. The conversion of one type
into another has many names:
automatic conversion,
implicit conversion,
coercion,
promotion, or
widening.
There are three ways to consider this issue. The first way to think about this is if the range of one type completely fits within the range of the other, then the number with the smaller range is converted (promoted) to the type of the number with the larger range. In the following examples, a number of type1 is added to a number of type2. In each case, the number range of type1 fits into the range of type2, so the parameter of type1 is first promoted to type2 before the addition.
type1  type2  example  
unsigned char  fits inside  unsigned short  uc+us is of type unsigned short 
unsigned char  fits inside  short  uc+ss is of type short 
unsigned char  fits inside  long  uc+sl is of type long 
char  fits inside  short  sc+ss is of type short 
char  fits inside  long  sc+sl is of type long 
unsigned short  fits inside  long  us+sl is of type long 
short  fits inside  long  ss+sl is of type long 
The second way to consider mixed precision operations is that in most cases the compiler will promote the number with the smaller precision into the other type before operation. If the two numbers are of the same precision, then the signed number is converted to unsigned. These automatic conversions may not yield correct results. The third and best way to deal with mixed type operations is to perform the conversions explicitly using the cast operation. We can force the type of an expression by explicitly defining its type. This approach allows the programmer to explicitly choose the type of the operation. Consider the following digital filter with mixed type operations. In this example, we explicitly convert x and y to signed 16 bit numbers and perform 16 bit signed arithmetic. Note that the assignment of the result into y, will require a demotion of the 16 bit signed number into 8 bit signed. Unfortunately, C does not provide any simple mechanisms for error detection/correction (see overflow and underflow.)
char y; // output of the filter
unsigned char x; // input of the filter
void filter(void){
y = (12*(short)x + 56*(short)y)/100;
}
Listing 56: Examples of the selection operator
We apply an explicit cast simply by preceding the number or
expression with parentheses surrounding the type. In this next digital
filter all numbers are of the same type. Even so, we are worried that
the intermediate result of the multiplications and additions might
overflow the 16bit arithmetic. We know from digital signal processing
that the final result will always fit into the 16bit variable. For
more information on the design and analysis of digital filters, see
Chapter 15 of Embedded Microcomputer Systems: Real Time
Interfacing by Jonathan W. Valvano. In this example, the cast
(long)
will specify the calculations be
performed in 32bit precision.
// y(n) = [113*x(n) + 113*x(n2)  98*y(n2)]/128,
channel specifies the A/D channel
short x[3],y[3]; // MACQs containing current and previous
#define OC5 0x20
void interrupt 13 TOC5handler(void){
TFLG1 = OC5; // ack
OC5F
TC5 = TC5+8333; // fs=240Hz
y[2]=y[1]; y[1]=y[0]; // shift MACQ
x[2]=x[1]; x[1]=x[0];
x[0] = A2D(channel); // new data
y[0] =
(113*((long)x[0]+(long)x[2])98*(long)y[2])>>7;}
Listing 57: We can use a cast to force higher precision arithmetic
We saw in Chapter 1, casting was used to assign a symbolic name to an I/O port. In particular the following define casts the number 0x0000 as a pointer type, which points to an unsigned 8 bit data. More about pointers can be found in Chapter 7.
#define PORTA *(unsigned char volatile *)(0x0000)
The selection operator takes three input parameters and yields one output result. The format is
Expr1 ? Expr2 : Expr3
The first
input parameter is an expression, Expr1, which yields a boolean (0 for
false, not zero for true). Expr2 and Expr3 return values that are
regular numbers. The selection operator will return the result of Expr2
if the value of Expr1 is true, and will return the result of Expr3 if
the value of Expr1 is false. The type of the expression is determined
by the types of Expr2 and Expr3. If Expr2 and Expr3 have different
types, then the usual promotion is applied. The resulting time is
determined at compile time, in a similar manner as the Expr2+Expr3
operation, and not at run time depending on the value of Expr1. The
following two subroutines have identical functions.
short
a,b;
void sub1(void){
a = (b==1) ? 10 : 1;
}
void sub2(void){
if(b==1)
a=10;
else
a=1;
}
Listing 58: Examples of the selection operator
Arithmetic Overflow and Underflow
An important issue when performing arithmetic calculations on integer values is the problem of underflow and overflow. Arithmetic operations include addition, subtraction, multiplication, division and shifting. Overflow and underflow errors can occur during all of these operations. In assembly language the programmer is warned that an error has occurred because the processor will set condition code bits after each of these operations. Unfortunately, the C compiler provides no direct access to these error codes, so we must develop careful strategies for dealing with overflow and underflow. It is important to remember that arithmetic operations (addition, subtraction, multiplication, division, and shifting) have constraints when performed with finite precision on a microcomputer. An overflow error occurs when the result of an arithmetic operation can not fit into the finite precision of the result. We will study addition and subtraction operations in detail, but the techniques for dealing with overflow and underflow will apply to the other arithmetic operations as well. We will consider two approaches
avoiding the error
detecting the error then correcting the result
For example when two 8 bit numbers are added, the sum may not fit back into the 8 bit result. We saw earlier that the same digital hardware (instructions) could be used to add and subtract unsigned and signed numbers. Unfortunately, we will have to design separate overflow detection for signed and unsigned addition and subtraction.
All microcomputers have a condition code register which contain bits which specify the status of the most recent operation. In this section, we will introduce 4 condition code bits common to most microcomputers. If the two inputs to an addition or subtraction operation are considered as unsigned, then the C bit (carry) will be set if the result does not fit. In other words, after an unsigned addition, the C bit is set if the answer is wrong. If the two inputs to an addition or subtraction operation are considered as signed, then the V bit (overflow) will be set if the result does not fit. In other words, after a signed addition, the V bit is set if the answer is wrong. The Freescale 6805 does not have a V bit, therefore it will be difficult to check for errors after an operation on signed numbers.
bit
name
meaning after addition or
subtraction
N
negative
result is negative
Z
zero
result is zero
V
overflow
signed overflow
C
carry
unsigned overflow
For an 8 bit unsigned number, there are only 256 possible values, 0 to 255. We can think of the numbers as positions along a circle. There is a discontinuity at the 0255 interface, everywhere else adjacent numbers differ by ±1. If we add two unsigned numbers, we start at the position of the first number a move in a clockwise direction the number of steps equal to the second number. For example, if 96+64 is performed in 8 bit unsigned precision, the correct result of 160 is obtained. In this case, the carry bit will be 0 signifying the answer is correct. On the other hand, if 224+64 is performed in 8 bit unsigned precision, the incorrect result of 32 is obtained. In this case, the carry bit will be 1, signifying the answer is wrong.
Figure 51: 8 bit unsigned addition.
For subtraction, we start at the position of the first number a move in a counterclockwise direction the number of steps equal to the second number. For example, if 16064 is performed in 8 bit unsigned precision, the correct result of 96 is obtained (carry bit will be 0.) On the other hand, if 3264 is performed in 8 bit unsigned precision, the incorrect result of 224 is obtained (carry bit will be 1.)
Figure 52: 8 bit unsigned subtraction.
In general, we see that the carry bit is set when we cross over from 255 to 0 while adding or cross over from 0 to 255 while subtracting.
Observation: The carry bit, C, is set after an unsigned add or subtract when the result is incorrect.
For an 8 bit signed number, the possible values range from 128 to 127. Again there is a discontinuity, but this time it exists at the 128127 interface, everywhere else adjacent numbers differ by ±1. The meanings of the numbers with bit 7=1 are different from unsigned, but we add and subtract signed numbers on the number wheel in a similar way (e.g., addition of a positive number moves clockwise.) Adding a negative number is the same as subtracting a positive number hence this operation would cause a counterclockwise motion. For example, if 32+64 is performed, the correct result of 32 is obtained. In this case, the overflow bit will be 0 signifying the answer is correct. On the other hand, if 96+64 is performed, the incorrect result of 96 is obtained. In this case, the overflow bit will be 1 signifying the answer is wrong.
Figure 53: 8 bit signed addition.
For subtracting signed numbers, we again move in a counterclockwise
direction. Subtracting a negative number is the same as adding a
positive number hence this operation would cause a clockwise motion.
For example, if 3264 is performed, the correct result of 32 is
obtained (overflow bit will be 0.) On the other hand, if 9664 is
performed, the incorrect result of 96 is obtained (overflow bit will be
1.)
Figure 54: 8 bit signed subtraction.
In general, we see that the overflow bit is set when we cross over from 127 to 128 while adding or cross over from 128 to 127 while subtracting.
Observation: The overflow bit, V, is set after a signed add or subtract when the result is incorrect.
Another way to determine the overflow bit after an addition is to consider the carry out of bit 6. The V bit will be set of there is a carry out of bit 6 (into bit 7) but no carry out of bit 7 (into the C bit). It is also set if there is no carry out of bit 6 but there is a carry out of bit 7. Let X7,X6,X5,X4,X3,X2,X1,X0 and M7,M6,M5,M4,M3,M2,M1,M0 be the individual binary bits of the two 8 bit numbers which are to be added, and let R7,R6,R5,R4,R3,R2,R1,R0 be individual binary bits of the 8 bit sum. Then, the 4 condition code bits after an addition are shown in Table 5.10.
Table 5.10. Condition code bits after an 8 bit addition operation.
Let the result R be the result of the subtraction XM. Then, the 4 condition code bits are shown in Table 5.11.
Table 511. Condition code bits after an 8 bit subtraction operation.
Common Error: Ignoring overflow (signed or unsigned)
can result in significant errors.
Observation: Microcomputers have two sets of conditional branch instructions (if statements) which make program decisions based on either the C or V bit.
Common Error: An error will occur if you unsigned conditional branch instructions (if statements) after operating on signed numbers, and viceversa.
There are some applications where arithmetic errors are not possible. For example if we had two 8 bit unsigned numbers that we knew were in the range of 0 to 100, then no overflow is possible when they are added together.
Typically the numbers we are processing are either signed or unsigned (but not both), so we need only consider the corresponding C or V bit (but not both the C and V bits at the same time.) In other words, if the two numbers are unsigned, then we look at the C bit and ignore the V bit. Conversely, if the two numbers are signed, then we look at the V bit and ignore the C bit. There are two appropriate mechanisms to deal with the potential for arithmetic errors when adding and subtracting. The first mechanism, used by most compilers, is called promotion. Promotion involves increasing the precision of the input numbers, and performing the operation at that higher precision. An error can still occur if the result is stored back into the smaller precision. Fortunately, the program has the ability to test the intermediate result to see if it will fit into the smaller precision. To promote an unsigned number we add zero’s to the left side. In a previous example, we added the unsigned 8 bit 224 to 64, and got the wrong result of 32. With promotion we first convert the two 8 bit numbers to 16 bits, then add.
We can check the 16 bit intermediate result (e.g., 228) to see if the answer will fit back into the 8 bit result. In the following flowchart, X and M are 8 bit unsigned inputs, X16, M16, and R16 are 16 bit intermediate values, and R is an 8 bit unsigned output. The oval symbol represents the entry and exit points, the rectangle is used for calculations, and the diamond shows a decision. Later in the book we will use parallelograms and trapezoids to perform input/output functions.
Figure 55: Promotion can be used to avoid overflow and underflow.To promote a signed number, we duplicate the sign bit as we add binary digits to the left side. Earlier, we performed the 8 bit signed operation 9664 and got a signed overflow. With promotion we first convert the two numbers to 16 bits, then subtract.
We can check the 16 bit intermediate result (e.g., 160) to see if the answer will fit back into the 8 bit result. In the following flowchart, X and M are 8 bit signed inputs, X16, M16, and R16 are 16 bit signed intermediate values, and R is an 8 bit signed output.
Figure 56: Promotion can be used to avoid overflow and underflow.The other mechanism for handling addition and subtraction errors is called ceiling and floor. It is analogous to movements inside a room. If we try to move up (add a positive number or subtract a negative number) the ceiling will prevent us from exceeding the bounds of the room. Similarly, if we try to move down (subtract a positive number or add a negative number) the floor will prevent us from going too low. For our 8 bit addition and subtraction, we will prevent the 0 to 255 and 255 to 0 crossovers for unsigned operations and 128 to +127 and +127 to 128 crossovers for signed operations. These operations are described by the following flowcharts. If the carry bit is set after an unsigned addition the result is adjusted to the largest possible unsigned number (ceiling). If the carry bit is set after an unsigned subtraction, the result is adjusted to the smallest possible unsigned number (floor.)
Figure 57: In assembly language we can detect overflow and underflow.If the overflow bit is set after a signed operation the result is adjusted to the largest (ceiling) or smallest (floor) possible signed number depending on whether it was a 128 to 127 cross over (N=0) or 127 to 128 cross over (N=1). Notice that after a signed overflow, bit 7 of the result is always wrong because there was a cross over.
Figure 58: In assembly language we can detect overflow and underflow.
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