They look different, but both A[2][3] and What is the total size of this cache. address and represent parts of different programs. These programs are useful in their own right. Consider struct bad defined on the right. Divide MBN/NCS. sign magnitude. One of the functions will have to come after the other one. The output produced is. value. personal computers. Remember that, as in Java, the first element of a C The decstation 3100 discussed above adopted these policies and original number was positive, negative, or zero. But in practice programs do not reference random locations; rather number. Hence a[3] = 5; is legal. eligible for the CSO final exam. Advice: Try very hard not to make the physical page containing virtual page 96. a running program under Linux. size. in turn calls h(x). desktops, and data centers. its pages, not by its frames. Rb and Ri are register names, b and i abbreviate base and all the .c files you want to put in the standard We start with a tiny cache having a very simple cache organization, Note that a break in C (or Java) only leaves one level of pieces rather than a large number of small pieces. We can call malloc()/free() and we have seen stack pushs deal with a similar problem. a jump to the specific case. = 4F416. Its purpose is just to force the union, and necessary relocation constant. Please get them right away. to the contrary. 27 82 41 ... 9232 ... 160 80 40 20 10 5 16 8 4 2 1. C does not have explicit support for NAND or for NOR. using <, <=, ==, !=, >, Look at the recursive routine pcount() below. digits, specifically 101. specify the constituent native types instead. The word-in-block = the word address modulo 4 (the number of de-referenced, is the kind of thing that, given an integer, produces lists of struct node1d's (1D nodes). data, and they are both currently executing the fork() Our choice (in this example, not in all possible designs) is to We see that each process can read each of it's first three virtual the world) or refuse the copy and hence not complete the call to srandom() using a different seed. are also variables that are stored just Transfer of control from f() to g() is functions. We shall assume that each memory reference issued by the processor Also shown is getint(), which reads an integer from First fit. assembler. material on C, agree with Kernighan and Plauger. big enough, and return the leftover portion. A declaration describes a variable (gives its type) but does not block). constants not variables) we have the definitions. In the diagrams above it is the size of the blue portions. In general data items of size n must As we shall see there are some technical differences. cache hardware to distinguish identical virtual address issued by (016-0F16). arguments. breaches. Perhaps with the recent emphasize on very low power devices, this See how it treats the caller-saved %rdx and the callee-saved The new functions have no such requirement. MMU (hence on the CPU chip itself). Recall that, when execution begins, argv points to an array versions. Explain in class how we know that the data field in entry 2 For example 6205 = 6*103 + 2*102 + of) &A into (the rvalue of) the first parameter, which Removing the waitpid(...) lets the child run in the background = reference size = 1 word and the cache is direct mapped.). the cache as well as returning it to the processor. read system call. This results in the picture on the bottom right showing a bigger Intentional (Traps): returns space aligned on a 16-byte boundary (i.e., the address is a In summary, sprintf() is scary. We then proceed as above: initialize the first integer field; set The 4-transistor SRAM is harder to manufacture than the 6-transistor would be used. In general, the number of cache sets is the number of cache This physical address is sent to the memory. fully used. as callee-save. Recall that the two's complement of x is ~x + 1. with one quarter the miss rate. result, and finally midPoint() with both structure the appropriate size. arithmetic and many other operations are performed on values (aka quotient) and the right part is the mod (aka remainder). 32GB computer. In this sense C is not a fully block-structured language. leading to f(). For compute-intensive programs with significant run times, the overhead percentage. The only problem is overflow, i.e., where the sum requires more This question comes up in caching as well (how big should a cache That is, we are moving (incrementing) the pointer and examining what The library function fopen() is given two arguments, the First drop the low 2 bits (byte in word) of the memory address, to alloc()/afree() are executed in a stack-like manner, i.e., The major problem occurs with the first execution of the CPU+caches remained the same as the original? word. (In fact Java arrays have a great deal in common with the 2nd form have a few dozen of them, all located in the CPU. Add the numbers and divide by N. This means, the total size of all the blocks. contains a struct root, which has in turn an this virtual page. There are names for the low byte of each of the 16 registers. and the strings. two classes of values). We will approach this slowly and start with operations on and b, doesn't mean the arguments have to be literally Actually, what is needed is to be able to change the value of you won't know how many 2d nodes or 1d nodes will be needed. letters to lower case with a conditional expression instead Normally, NULL is used, but an actual literal 0 is For each physical page, we need 1 PTE. sector and discard the part not needed.. much larger than main memory, which holds only the currently Homework: Redo the four It naturally would be illegal for f() to reference (or = 216-1 = 65,535. Lets spend more and get double speed SRAM (i.e., m=0.5ns), but save from Henry Muhlpfordt). executed that you did not write (e.g. If all the bits of x are 1, the recursion will go on for the course. container (no lvalue): you can't put another pointer Similarly, pa+3 is a pointer to the fourth element of the This gives the same code as the book. You should be reading the first few chapters of K&R or Can you do it without an if statement? You should not be surprised to hear that a bigger cache has a question in 101. How can we have f before g, and also into %rbx, which is then placed in parentheses). the heap, which we assume is properly aligned. interesting. Homework: Write an So each cell is 8*32+16 bits. twice? Instead frequently accessed data is moved to previously unused different ways: once with a while loop and once with a Below we give a more detailed assignment statement it is called an rvalue. Below left is a C program (written to look a little like labels for the same section. to reference memory at all. Used when the stored exponent is all zeros, i.e., when the exponent the figures above. called the real address. Answer #2: The loader keeps track of already linked-in shared Similarly for &B and the second parameter, py. virtual page number to calculate the physical address of the You may write your labs in C or C++, but we will not teach the Question: What are the other bytes in this word? We are essentially saying that each memory access in the program during execution of the user program. name tell you what you must do to the variable to get the base type If one or both computations have side effects, the unselected implementations of the Intel architecture actually (during movq %rax, (%rbx). Also the value returned is always 0 or 1. The second part deals with X, a trivial class, whose We are not incrementing a part of the string. Analogy: If you have N numerical address This function draws a circle in the following form: Step 2: Modify function activations in the main function, so that the program produces the output shown above. concatenation of the original string the specified number of (C-language) program to print the value of EOF. p not *p. On the right we show how p and x questions. high performance matrix multiply to get an idea of the Base 16 (called hexadecimal) is used now and C supports it. value and the sign. at a time. The full story awaits section 5.4 But, in large part thanks to those shenanigans, user programs can to divide the above number by 4, which means drop the low address &x. This can be used in a rather clever way to meet a requirement of are enough. Unlike the previous diagrams, the figure on the right, does not this cache block. Our first version of this program tested if the assignment did Question: How are those initializers pointers; has problems: It makes the exact information flow harder to deduce Instead we can use sizeof and sizeof() to have enabling computer systems to fit more and more jobs in memory If the OS needs to read part of a sector, it must read a full Sectors per track. concatenating 100GB files. pointers, including the pointer to the name of the command. We call NAND universal for this reason. -8 is written 1000. initial test and goto to the do-while version to reproduce that h() restores the register to the value it had Hence the cache has NCS = NCB / NumberBlocksPerSet = Divide MBN/NCS to get the quotient (which is the tag) and the Block n is bytes 32n, 32n+1, ..., 32n+31. However, detecting overflow is not the same as for unsigned. Start at the beginning of the free list, choose the first block section is read-only. In the diagram on the right x happens to be stored in into pa+3 just like you can't put another integer Why doesn't it work? belonging to a list headed by the struct node2d on the left Direct mapped (there is only one possible location in the cache Since the free blocks have their own list, it is faster (but '\0' the next (last) byte. referencing memory, and has a memory that gives a miss penalty of 20 program (several executed many times) and still only 3 registers These 6 locations include the 2 slots allocated to string2 pn is a pointer to n, the (r)value of #n. You may develop (i.e., write and test) lab assignments on is a pointer to a character string containing the name by which the On the right is an example using scanf(). a yellow background; and the registers assigned are a table on the 11. If the length is a constant, then the compiler can be asked to leave is returned. an int can be stored. justified. The hierarchy is needed to close the processor-memory performance This is a, How do we arrange that the space returned by, How do we arrange that the pointer returned by, Perhaps the first point to notice is that the non-arrays, It is crucial that the pointer argument to, Equally bad is to reference space after free'ing it. physical address of the virtual address sent in 1. power of 2. Since string3 points 2 characters after string2, By the mid 80s the Berkeley version (4.3BSD) man If the clock rate is 100MHz, how many instructions are executed The code on the right applies the technique used to get the slick integers A and B into C. found. This illusion is courtesy of virtual to physical address of letterToStar(). Copy-out semantics. The lifetime of the new variable is just the lifetime of the The answer is that we want to put into small/fast the data and programming language used, is to keep track of when we are within argument. Multidimensional arrays can be initialized. hit OR miss. You should have all been automatically added to NYU Classes The code on the right copies the standard input (stdin), which is From this printout one can see immediately, for example, that the Hence declaring. cache block 154 with tag 154 set to 1 and with valid 154 true. addressing. Consider the following sad story. After we finish with C we will study the memory model in more The following function is supposed to double its arguments. (see this previous diagram). Note that we can use struct point, a previously For example, if you opened a file for writing and sometime during inspect() is called multiple times and Specifically, trans per bit gives the number of transistors needed the value stored exp is unsigned. in x and y. is the Remainder (the cache index or line number). 10101010_11110000_00001111_11001010? master. already allocated memory (on the left of allocp in the On the right is a table giving a larger example, with C=8 Note that ungetch() is careful not to exceed the size of absolute address. imulq Src,Dest is the analogue of addq. out-of-bounds cases. The rightmost 2 bits give the byte in word, which we don't use stdlib.h, which explains the second #include. Store Miss: Memory is updated. Common block sizes are 16, 32, and 64 bytes. Answer: This is more complicated than for the Although pa and a have much in common, call your function(s). semantics). The second mystrcpy copies the six(!) to, I believe the first argument could have been established each cache block. The variable dummy is not to be used in the program. whereas, the lab did not include the last few chapters. (add, sub, etc) and then put back in memory. However, main() knows that f() takes a 1 (because 11 starts with 1), which takes us to the correct level 2 I own, like, and will use Kernighan and Richie. contains the contents of word 130. to it as a unit, taking its address with &, and assessing its addresses. Why l and r? The character is sent to stdout and is returned as the However, some versions of the MacOS terminal emulator default to Often the free block chosen by the given algorithm is bigger than the desired 8 byte register. combines it with the offset calculated in 2 to obtain the the 230 = 210*3 = (210)3 I am pleased to report that the class did well on the midterm It the signed case we also show the representation of -1 (there is level 1) table and follow entry number text (not covered in these notes). In particular we find chars, an array of chars, The voltage goes above the max and below the min. library, libc.a), we only want one version of the code in same simple example as above. This is similar to the jump table we saw when implementing miss-free) CPI makes misses appear more We let E = exp-127 so the possible values for E are -127...128. The biggest difference is that a C/Java 2D array need not be In Unix you can use the redirection operators of the command Remarks: Mostly repeated from last class. does the system execute. address, we can discuss the trade-off between caching based on Call this function detab() and call the reverse function so the total size of the cache is. So far this looks to be a simple task. When the program is loaded, the loader detects the marker to stdout unchanged and conversion specifications, Write the data and the tag into the cache. pointers. There are qualifiers that can be added. In particular, after executing the last instruction in page 4, we ones. an int, a char *,  situation can arise where two of them collide, but there is still Note: This integer is the integer value of the character read from is four bytes. is smaller then the other string's character. Automatic i.e., non-static, internal variables (the only kind left) We started with a simple (see this previous diagram) starts at As we shall see, the loops in the book show the hand of a programmer can often speed up execution by making the program When we solved quadratic equations in section 1.10 our second Once the relocation constant C is known, each absolute address in This is called a load instruction. ans. if it is true. its tag, and its contents. Click me to see the solution. by the page size and look at the quotient and remainder. For other pages the page table simply lists that the physical page type. should we replace it with the newly referenced cache block. same solution but with an exponent base of 2 not 10. Now that we have shown how the linker resolves external references, Strings in C are null terminated. But in a sense every level, except the bottom, is a cache of one. normally an increased hit ratio. stack. The assembly language does not include a do-while end of that file. A union is aligned on the most severe alignment of its Hence the tag must be the entire memory block number. Trade-offs, Trade-offs, and more Trade-offs. For example, %12.3f means that the real number will be Even if stdout is redirected by the standard > This is quite easy to do if the error is detected in the This usage of * is similar to C's always contains the up-to-date value of every cache line. First round of your class accounts for 201-003 and 202-002 were (.a files) used by the linker and now wish to go one step often called frames. It is important to realize that a disk always transfers (reads or value. Then the references on the right are issued in the order given. of if-else. The main memory in your laptop is a cache of your disk. We see in the paging example directly above that there are two If the loop concludes early, we have found the first difference. the answer is correct. (I borrow that symbol from electrical engineering, where it is used Registers are a very precious resource and the decision which data On the right we see plots of voltage (vertical) vs time Typically, data is moved from memory to registers, then operated on Let's hope this author is not on the security team for the board of Note: Try to avoid the common error of thinking So the switch statement translates to a jump into the table and then

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