C语言编程经典指南：排版约定与操作详解

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Programming in C l9.2/2

OUCS 8 October 1996

Examples

AND XOR OR

0 & 0 = 0 0 ^ 0 = 0 0 | 0 = 0

1 & 0 = 0 1 ^ 0 = 1 1 | 0 = 1

0 & 1 = 0 0 ^ 1 = 1 0 | 1 = 1

1 & 1 = 1 1 ^ 1 = 0 1 | 1 = 1

For example, to obtain the two separate bytes of a two byte int (int may be more

than two bytes on your system, and not all computers have eight bit bytes, so this

example is not universal):

hiByte = (i >> 8) & 0xFF;

loByte = i & 0xFF;

If i = 1011010100110111 then

i >>8 = 0000000010110101 (shift I right by 8 bits).

(i>>8) & OxFF ANDs 0000000010110101 with 0000000011111111 giving

10110101 ( bits 9-16 of i )

i & 000000011111111 = 00110111 ( bits 1-8 of i).

OxFF is the hexadecimal representation of 11111111 .

See [6] for more on bitwise operators.

3.9 Promotions and Conversions ‡

3.9.1 Integral Promotions

Any value of type char or short int (signed or unsigned ) is converted to a

value of type int (if it can represent all the values of the original type) when it

appears in an expression. If int cannot store all the values of the original type then

unsigned int is used.

For example, assuming the following declarations:

char a, b;

then

a + b has type int

printf("%c",a); the type of the second argument is int

Operator Operand Result

~ 0010111 1101000

<< 0010111 0101110

>> 0010111 0001011

Programming in C l9.2/2

OUCS 10 October 1996

3.9.3 Problems

Whilst integral promotions and arithmetic conversions usually act to preserve

information it is possible to get caught out. Consider the following on a system where

the type of char is actually unsigned char .

#include <stdio.h>

int main()

{

char c = EOF;

if (c == EOF)

...

return 0;

}

On most systems EOF is a preprocessor symbolic constant with the int value -1.

On systems where negative numbers are stored using the two's complement system

(this includes the IBM PCs used in class), -1 is stored by setting every bit in the

variable to 1. When this is copied into c above, we end up with a byte of

11111111 . When the comparison is made, one operand is unsigned char , the

other is int. According to the integral promotions rule above, the unsigned char

value is converted to int; but unfortunately as the original type was unsigned

char the byte of 1s gets converted to the int value 255. Thus the test becomes

if (255 == -1) which, not surprisingly, fails. Had the type of char been

signed char then the problem would not have occurred, as the value -1 would be

preserved in the conversion to int. In fact, most programmers use int to store

characters anyway (except in strings).

3.10 Parsing Rules ‡

The C compiler parses a C program by splitting up the program into its constituent

words and operators. Each separate entity is called a token. The C compiler uses a

simple rule when deciding where to split the C source. The compiler bites off as large a

chunk of your program as possible which will form a legal C token. This mechanism is

called greedy lexical analysis. Possible ambiguities when using the increment

operators can be eliminated if the greedy rule is followed.

a = b+++c; /* a = (b++) + c */

a = b+ ++c; /* a = b + (++c) */

What about the following?

a = b+++++c; /* a = (b++) + (++c) */

There can be no ambiguity about this statement, the meaning given in the comment is

the only sensible one, but the greedy lexical analysis rule means that a C compiler

cannot arrive at the correct interpretation, instead it thinks the programmer meant

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