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reduction submodule has the longest delay, 150 ns; this delay then determines the delay of a stage in the pipeline The five (sub)modules in the critical path can be partitioned into three fairly even stages with delays of 125 ns (partial product generation), 150 ns (partial product reduction), and 125 ns (final reduction, normalization, and rounding) The resultant three-stage pipelined design is shown in Figure 25 In determining the actual clocking rate of the pipelined design, we must consider clocking requirements Assuming that edge-triggered registers are used for buffering between pipeline stages, we must add the clock-edge-to-register-output delay of 17 ns and the setup time of 5 ns to the stage delay of 150 ns This results in the minimum clocking period of 172 ns Therefore, instead of clocking at the rate of 25 MHz, the new pipelined design can be clocked at the rate of 58 MHz This represents a factor of 23 increase in throughput Note, however, that f\ the latency for performing each multiplication has increased slightly, from 400 to | 516 ns ^ The only additional hardware required for the pipelined design is the edgetriggered register chips for buffering between pipeline stages On top of the original 175 IC chips, an additional 82 IC chips are required Using chip count as a measure of hardware complexity, the total of 257 IC chips represents an increase of 4 5 % in terms of hardware complexity This 4 5 % increase in hardware cost resulted in a 130% increase in performance Clearly, this three-stage pipelined design of the floating-point multiplier is a win over the original nonpipelined design This example assumes board-level implementations nsing off-the-shelf parts Given today's chip technology, this entire design can be easily implemented as a small module on a chip While a board-level implementation of the floating-point multiplier may be viewed as outdated, the purpose of this example is to succinctly illustrate the effectiveness of pipelining using a published specific design with actual latency and hardware cost parameters In fact, the upper curve in Figure 23 reflects the parameters from this example.

Since the program just started, the input buffer is empty. scanf() will wait until characters appear in the input buffer, which will happen as soon as you type some characters and press Return. Type your first name, and press Return.

Sort the list by Name, and locate these les: EXPLOREREXE exercises DESKTOPINI WININI This is the Windows Explorer application you re using for these This contains the con guration data for your desktop

scanf() will ignore white space characters in the input buffer. For example, if you type a few spaces

and tabs and then press Return, scanf() will still sit there, waiting for some real input. Try it!

Step 3 Although MS-DOS is no longer used as an operating system, some of the original MS-DOS applications (commands) are still very much alive These are now launched from the command-line interface in Windows as well as when you need to invoke the Recovery Console As a PC technician, you re likely to need one or more of these command-line tools; for now, you should just learn where to nd them a b While still in Windows Explorer, nd the C:\WINDOWS\System32 folder and open it This folder stores all the command-line applications Sort the details list by Type, scroll down to the MS-DOS applications, and locate these les (see Figure 4-13): FORMATCOM This is used to prepare hard drives and oppy diskettes for storing data

Once you type your name, scanf() will copy the characters, a byte at a time, into the char array pointed to by name. Remember, because name was declared as an array, name points to the first of the 50 bytes allocated for the array. If you type in the name Dave, scanf() will place the four characters 'D', 'a', 'v', and 'e' in the first four of the 50 bytes allocated for the array. Next, scanf() will set the fifth byte to a value of 0 to terminate the string properly (see Figure 8-11). Since the string is properly terminated by the 0 in name[4], we don t really care about the value of the bytes name[5] through name[49]. Next, we pass name on to printf(), asking it to print the name as part of a welcoming message. The %s tells printf() that name points to the first byte of a zero-terminated string. printf() will step through memory, one byte at a time, starting with the byte that name points to. printf() will print each byte in turn until it hits a byte with a value of 0. The zero byte marks the end of the string.

printf( "Welcome, %s.\n", name );

Source Waser and Flynn, 1982

D a v e / 0

EDITCOM This text editor program can create and modify con guration les in the command-line mode The Windows equivalent is Notepad

Figure 8-11. Here s the array name after the string "Dave" is copied to it. Notice that name[4] has a value of 0.

If name[4] didn t contain a zero, the string wouldn t be properly terminated. Passing a nonterminated string to printf() is a sure way to confuse printf(). printf() will step through memory one byte at a time, printing a byte and looking for a zero. It will keep printing bytes until it happens to encounter a byte set to 0. Remember, C strings must be terminated!

If you re using Vista, you may have to add the Type column Right-click the column bar as shown in Figure 4-14 and check Type Step 4 When working with modern-day Windows, it s important to know the key system les involved in the boot process and fundamental core les Use the Search for les and folders tool to locate and record the absolute path of the following system les For a list of all command-line commands, open the Command Prompt and type help