2.7 Type Conversions

When an operator has operands of different types, they are converted to a common type according to a small number of rules. In general, the only automatic conversions are those that convert a “narrower'' operand into a “wider'' one without losing information, such as converting an integer into floating point in an expression like f + i. Expressions that don't make sense, like using a float as a subscript, are disallowed. Expressions that might lose information, like assigning a longer integer type to a shorter, or a floating-point type to an integer, may draw a warning, but they are not illegal.

A char is just a small integer, so chars may be freely used in arithmetic expressions. This permits considerable flexibility in certain kinds of character transformations. One is exemplified by this naive implementation of the function atoi, which converts a string of digits into its numeric equivalent.

/* atoi:  convert s to integer */int atoi(char s[]){    int i, n;    n = 0;    for (i = 0; s[i] >= '0' && s[i] <= '9'; ++i)        n = 10 * n + (s[i] - '0');    return n;}

As we discussed in Chapter 1, the expression

 s[i] - '0'

gives the numeric value of the character stored in s[i], because the values of '0', '1', etc., form a contiguous increasing sequence.

Another example of char to int conversion is the function lower, which maps a single character to lower case for the ASCII character set. If the character is not an upper case letter, lower returns it unchanged.

/* lower:  convert c to lower case; ASCII only */int lower(int c){    if (c >= 'A' && c <= 'Z')        return c + 'a' - 'A';    else        return c;}

This works for ASCII because corresponding upper case and lower case letters are a fixed distance apart as numeric values and each alphabet is contiguous – there is nothing but letters between A and Z. This latter observation is not true of the EBCDIC character set, however, so this code would convert more than just letters in EBCDIC.

The standard header <ctype.h>, described in Appendix B, defines a family of functions that provide tests and conversions that are independent of character set. For example, the function tolower is a portable replacement for the function lower shown above. Similarly, the test

 c >= '0' && c <= '9'

can be replaced by

isdigit(c)

We will use the <ctype.h> functions from now on.

There is one subtle point about the conversion of characters to integers. The language does not specify whether variables of type char are signed or unsigned quantities. When a char is converted to an int, can it ever produce a negative integer? The answer varies from machine to machine, reflecting differences in architecture. On some machines a char whose leftmost bit is 1 will be converted to a negative integer (“sign extension''). On others, a char is promoted to an int by adding zeros at the left end, and thus is always positive.

The definition of C guarantees that any character in the machine's standard printing character set will never be negative, so these characters will always be positive quantities in expressions. But arbitrary bit patterns stored in character variables may appear to be negative on some machines, yet positive on others. For portability, specify signed or unsigned if non-character data is to be stored in char variables.

Relational expressions like i > j and logical expressions connected by && and || are defined to have value 1 if true, and 0 if false. Thus the assignment

d = c >= '0' && c <= '9'

sets d to 1 if c is a digit, and 0 if not. However, functions like isdigit may return any non-zero value for true. In the test part of if, while, for, etc., “true'' just means “non-zero'', so this makes no difference.

Implicit arithmetic conversions work much as expected. In general, if an operator like + or * that takes two operands (a binary operator) has operands of different types, the “lower'' type is promoted to the “higher'' type before the operation proceeds. The result is of the integer type. Section 6 of Appendix A states the conversion rules precisely. If there are no unsigned operands, however, the following informal set of rules will suffice:

If either operand is long double, convert the other to long double.
Otherwise, if either operand is double, convert the other to double.
Otherwise, if either operand is float, convert the other to float.
Otherwise, convert char and short to int.
Then, if either operand is long, convert the other to long.

Notice that float=s in an expression are not automatically converted to =double; this is a change from the original definition. In general, mathematical functions like those in <math.h> will use double precision. The main reason for using float is to save storage in large arrays, or, less often, to save time on machines where double-precision arithmetic is particularly expensive.

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.

Conversions take place across assignments; the value of the right side is converted to the type of the left, which is the type of the result.

A character is converted to an integer, either by sign extension or not, as described above.

Longer integers are converted to shorter ones or to =char=s by dropping the excess high-order bits. Thus in

int  i;char c;?i = c;c = i;

the value of c is unchanged. This is true whether or not sign extension is involved. Reversing the order of assignments might lose information, however.

If x is float and i is int, then x = i and i = x both cause conversions; float to int causes truncation of any fractional part. When a double is converted to float, whether the value is rounded or truncated is implementation dependent.

Since an argument of a function call is an expression, type conversion also takes place when arguments are passed to functions. In the absence of a function prototype, char and short become int, and float becomes double. This is why we have declared function arguments to be int and double even when the function is called with char and float.

Finally, explicit type conversions can be forced (“coerced'') in any expression, with a unary operator called a cast. In the construction

(type name) expression

the expression is converted to the named type by the conversion rules above. The precise meaning of a cast is as if the expression were assigned to a variable of the specified type, which is then used in place of the whole construction. For example, the library routine sqrt expects a double argument, and will produce nonsense if inadvertently handled something else. (sqrt is declared in <math.h>.) So if n is an integer, we can use

sqrt((double) n)

to convert the value of n to double before passing it to sqrt. Note that the cast produces the value of n in the proper type; n itself is not altered. The cast operator has the same high precedence as other unary operators, as summarized in the table at the end of this chapter.

If arguments are declared by a function prototype, as the normally should be, the declaration causes automatic coercion of any arguments when the function is called. Thus, given a function prototype for sqrt:

 double sqrt(double)

the call

root2 = sqrt(2)

coerces the integer 2 into the double value 2.0 without any need for a cast.

The standard library includes a portable implementation of a pseudo-random number generator and a function for initializing the seed; the former illustrates a cast:

unsigned long int next = 1;?/* rand:  return pseudo-random integer on 0..32767 */int rand(void){    next = next * 1103515245 + 12345;    return (unsigned int)(next/65536) % 32768;}?/* srand:  set seed for rand() */void srand(unsigned int seed){    next = seed;}

Exercise 2-3. Write a function htoi(s), which converts a string of hexadecimal digits (including an optional 0x or 0X) into its equivalent integer value. The allowable digits are 0 through 9, a through f, and A through F.

2.8 Increment and Decrement Operators

C provides two unusual operators for incrementing and decrementing variables. The increment operator ++ adds 1 to its operand, while the decrement operator — subtracts 1. We have frequently used ++ to increment variables, as in

if (c == '\n')    ++nl;

The unusual aspect is that ++ and — may be used either as prefix operators (before the variable, as in ++n), or postfix operators (after the variable: n++). In both cases, the effect is to increment n. But the expression ++n increments n before its value is used, while n++ increments n after its value has been used. This means that in a context where the value is being used, not just the effect, ++n and n++ are different. If n is 5, then

 x = n++;

sets x to 5, but

x = ++n;

sets x to 6. In both cases, n becomes 6. The increment and decrement operators can only be applied to variables; an expression like (i+j)++ is illegal.

In a context where no value is wanted, just the incrementing effect, as in

if (c == '\n')    nl++;

prefix and postfix are the same. But there are situations where one or the other is specifically called for. For instance, consider the function squeeze(s,c), which removes all occurrences of the character c from the string s.

/* squeeze:  delete all c from s */void squeeze(char s[], int c){   int i, j;?   for (i = j = 0; s[i] != '\0'; i++)       if (s[i] != c)           s[j++] = s[i];   s[j] = '\0';}

Each time a non-c occurs, it is copied into the current j position, and only then is j incremented to be ready for the next character. This is exactly equivalent to

if (s[i] != c) {    s[j] = s[i];    j++;}

Another example of a similar construction comes from the getline function that we wrote in Chapter 1, where we can replace

if (c == '\n') {    s[i] = c;    ++i;}

by the more compact

if (c == '\n')   s[i++] = c;

As a third example, consider the standard function strcat(s,t), which concatenates the string t to the end of string s. strcat assumes that there is enough space in s to hold the combination. As we have written it, strcat returns no value; the standard library version returns a pointer to the resulting string.

/* strcat:  concatenate t to end of s; s must be big enough */void strcat(char s[], char t[]){    int i, j;?    i = j = 0;    while (s[i] != '\0') /* find end of s */        i++;    while ((s[i++] = t[j++]) != '\0') /* copy t */        ;}

As each member is copied from t to s, the postfix ++ is applied to both i and j to make sure that they are in position for the next pass through the loop.

Exercise 2-4. Write an alternative version of squeeze(s1,s2) that deletes each character in s1 that matches any character in the string s2.

Exercise 2-5. Write the function any(s1,s2), which returns the first location in a string s1 where any character from the string s2 occurs, or -1 if s1 contains no characters from s2. (The standard library function strpbrk does the same job but returns a pointer to the location.)

2.9 Bitwise Operators

C provides six operators for bit manipulation; these may only be applied to integral operands, that is, char, short, int, and long, whether signed or unsigned.

&bitwise AND""^bitwise exclusive OR#ERRORleft shift>>right shift~one's complement (unary)

The bitwise AND operator & is often used to mask off some set of bits, for example

n = n & 0177;

sets to zero all but the low-order 7 bits of n.

The bitwise OR operator | is used to turn bits on:

x = x | SET_ON;

sets to one in x the bits that are set to one in SET_ON.

The bitwise exclusive OR operator ^ sets a one in each bit position where its operands have different bits, and zero where they are the same.

One must distinguish the bitwise operators & and | from the logical operators && and ||, which imply left-to-right evaluation of a truth value. For example, if x is 1 and y is 2, then x & y is zero while x && y is one.

The shift operators << and >> perform left and right shifts of their left operand by the number of bit positions given by the right operand, which must be non-negative. Thus x << 2 shifts the value of x by two positions, filling vacated bits with zero; this is equivalent to multiplication by 4. Right shifting an unsigned quantity always fits the vacated bits with zero. Right shifting a signed quantity will fill with bit signs (“arithmetic shift'') on some machines and with 0-bits (“logical shift'') on others.

The unary operator ~ yields the one's complement of an integer; that is, it converts each 1-bit into a 0-bit and vice versa. For example

x = x & ~077

sets the last six bits of x to zero. Note that x & ~077 is independent of word length, and is thus preferable to, for example, x & 0177700, which assumes that x is a 16-bit quantity. The portable form involves no extra cost, since ~077 is a constant expression that can be evaluated at compile time.

As an illustration of some of the bit operators, consider the function getbits(x,p,n) that returns the (right adjusted) n-bit field of x that begins at position p. We assume that bit position 0 is at the right end and that n and p are sensible positive values. For example, getbits(x,4,3) returns the three bits in positions 4, 3 and 2, right-adjusted.

/* getbits:  get n bits from position p */unsigned getbits(unsigned x, int p, int n){    return (x >> (p+1-n)) & ~(~0 << n);}

The expression x >> (p+1-n) moves the desired field to the right end of the word. ~0 is all 1-bits; shifting it left n positions with ~0<<n places zeros in the rightmost n bits; complementing that with ~ makes a mask with ones in the rightmost n bits.

Exercise 2-6. Write a function setbits(x,p,n,y) that returns x with the n bits that begin at position p set to the rightmost n bits of y, leaving the other bits unchanged.

Exercise 2-7. Write a function invert(x,p,n) that returns x with the n bits that begin at position p inverted (i.e., 1 changed into 0 and vice versa), leaving the others unchanged.

Exercise 2-8. Write a function rightrot(x,n) that returns the value of the integer x rotated to the right by n positions.

2.10 Assignment Operators and Expressions

An expression such as

 i = i + 2

in which the variable on the left side is repeated immediately on the right, can be written in the compressed form

i += 2

The operator += is called an assignment operator.

Most binary operators (operators like + that have a left and right operand) have a corresponding assignment operator op===, where op is one of

 +   -   *   /   %      &   ^   |

If expr1 and expr2 are expressions, then

expr1 op= expr

is equivalent to

expr1  = (expr1) op (expr2)

except that expr1 is computed only once. Notice the parentheses around expr2:

x *= y + 1

means

x = x * (y + 1)

rather than

x = x * y + 1

As an example, the function bitcount counts the number of 1-bits in its integer argument.

/* bitcount:  count 1 bits in x */int bitcount(unsigned x){    int b;?    for (b = 0; x != 0; x >>= 1)        if (x & 01)            b++;    return b;}

Declaring the argument x to be an unsigned ensures that when it is right-shifted, vacated bits will be filled with zeros, not sign bits, regardless of the machine the program is run on.

Quite apart from conciseness, assignment operators have the advantage that they correspond better to the way people think. We say “add 2 to i'' or “increment i by 2'', not “take i, add 2, then put the result back in i''. Thus the expression i + 2= is preferable to i = i+2. In addition, for a complicated expression like

yyval[yypv[p3+p4] + yypv[p1]] += 2

the assignment operator makes the code easier to understand, since the reader doesn't have to check painstakingly that two long expressions are indeed the same, or to wonder why they're not. And an assignment operator may even help a compiler to produce efficient code.

We have already seen that the assignment statement has a value and can occur in expressions; the most common example is

while ((c = getchar()) != EOF)          ...

The other assignment operators (+=, -=, etc.) can also occur in expressions, although this is less frequent.

In all such expressions, the type of an assignment expression is the type of its left operand, and the value is the value after the assignment.

Exercise 2-9. In a two's complement number system, x & (x-1)= deletes the rightmost 1-bit in x. Explain why. Use this observation to write a faster version of bitcount.

2.11 Conditional Expressions

The statements

if (a > b)    z = a;else    z = b;

compute in z the maximum of a and b. The conditional expression, written with the ternary operator “=?:='', provides an alternate way to write this and similar constructions. In the expression

expr1 ? expr2 : expr3

the expression expr1 is evaluated first. If it is non-zero (true), then the expression expr2 is evaluated, and that is the value of the conditional expression. Otherwise expr3 is evaluated, and that is the value. Only one of expr2 and expr3 is evaluated. Thus to set z to the maximum of a and b,

   z = (a > b) ? a : b;    /* z = max(a, b) */

It should be noted that the conditional expression is indeed an expression, and it can be used wherever any other expression can be. If expr2 and expr3 are of different types, the type of the result is determined by the conversion rules discussed earlier in this chapter. For example, if f is a float and n an int, then the expression

(n > 0) ? f : n

is of type float regardless of whether n is positive.

Parentheses are not necessary around the first expression of a conditional expression, since the precedence of ?: is very low, just above assignment. They are advisable anyway, however, since they make the condition part of the expression easier to see.

The conditional expression often leads to succinct code. For example, this loop prints n elements of an array, 10 per line, with each column separated by one blank, and with each line (including the last) terminated by a newline.

for (i = 0; i < n; i++)    printf("%6d%c", a[i], (i%10==9 || i==n-1) ? '\n' : ' ');

A newline is printed after every tenth element, and after the n-th. All other elements are followed by one blank. This might look tricky, but it's more compact than the equivalent if-else. Another good example is

printf("You have %d items%s.\n", n, n==1 ? "" : "s");

Exercise 2-10. Rewrite the function lower, which converts upper case letters to lower case, with a conditional expression instead of if-else.

2.12 Precedence and Order of Evaluation

Table 2.1 summarizes the rules for precedence and associativity of all operators, including those that we have not yet discussed. Operators on the same line have the same precedence; rows are in order of decreasing precedence, so, for example, *, /, and % all have the same precedence, which is higher than that of binary + and -. The “operator'' () refers to function call. The operators -> and . are used to access members of structures; they will be covered in Chapter 6, along with sizeof (size of an object). Chapter 5 discusses * (indirection through a pointer) and & (address of an object), and Chapter 3 discusses the comma operator.

Table 2.1: Precedence and Associativity of Operators

Unary & +, -, and * have higher precedence than the binary forms.

Note that the precedence of the bitwise operators &, ^, and | falls below == and !=. This implies that bit-testing expressions like

if ((x & MASK) == 0) ...

must be fully parenthesized to give proper results.

C, like most languages, does not specify the order in which the operands of an operator are evaluated. (The exceptions are &&, ||, ?:, and ,.) For example, in a statement like

x = f() + g();

f may be evaluated before g or vice versa; thus if either f or g alters a variable on which the other depends, x can depend on the order of evaluation. Intermediate results can be stored in temporary variables to ensure a particular sequence.

Similarly, the order in which function arguments are evaluated is not specified, so the statement

printf("%d %d\n", ++n, power(2, n));   /* WRONG */

can produce different results with different compilers, depending on whether n is incremented before power is called. The solution, of course, is to write

 ++n;      printf("%d %d\n", n, power(2, n));

Function calls, nested assignment statements, and increment and decrement operators cause “side effects'' – some variable is changed as a by-product of the evaluation of an expression. In any expression involving side effects, there can be subtle dependencies on the order in which variables taking part in the expression are updated. One unhappy situation is typified by the statement

a[i] = i++;

The question is whether the subscript is the old value of i or the new. Compilers can interpret this in different ways, and generate different answers depending on their interpretation. The standard intentionally leaves most such matters unspecified. When side effects (assignment to variables) take place within an expression is left to the discretion of the compiler, since the best order depends strongly on machine architecture. (The standard does specify that all side effects on arguments take effect before a function is called, but that would not help in the call to printf above.)

The moral is that writing code that depends on order of evaluation is a bad programming practice in any language. Naturally, it is necessary to know what things to avoid, but if you don't know how they are done on various machines, you won't be tempted to take advantage of a particular implementation.