4. In this question, we consider the relative performance obtained by running a particular with various caches installed between the memory and the CPU. program Consider the following C function: 1 char a [512], b[512] add arrays (O { register int i; for (i = 0; i < 511; i = i+1) { a[i+1] = a[i] + b[i]; // Work!! } (The register keyword tells the C compiler to reserve a register for the variable i; let's assume it keeps i in RO. What does this mean for us? The variable i is not stored in main memory- it's stored in the CPU; this mean that it never has to be pulled from main memory into the cache when the CPU accesses it. The char keyword allocates character arrays at fixed locations in memory, using 1 byte per character.) Recall that in C, an array is a contiguous chunk of memory. Therefore, in this code, array a starts at some address (say, 0×00A), and ends at (start address) + n-sizeof (char) address, where n is the number of elements and sizeof (char) is the number of bytes a char takes in memory (and in this example, we're using char because it's 1 byte and makes calculating simple!). For the following questions, assume that the memory can read or write 1 byte at a time. (a) Suppose, when the code is actually run, the array a starts at physical memory address 0x0100, and b is right after (that is, adjacent to) a in memory. What are the physical addresses of the first 5 elements of a? The first 5 elements of b?
4. In this question, we consider the relative performance obtained by running a particular with various caches installed between the memory and the CPU. program Consider the following C function: 1 char a [512], b[512] add arrays (O { register int i; for (i = 0; i < 511; i = i+1) { a[i+1] = a[i] + b[i]; // Work!! } (The register keyword tells the C compiler to reserve a register for the variable i; let's assume it keeps i in RO. What does this mean for us? The variable i is not stored in main memory- it's stored in the CPU; this mean that it never has to be pulled from main memory into the cache when the CPU accesses it. The char keyword allocates character arrays at fixed locations in memory, using 1 byte per character.) Recall that in C, an array is a contiguous chunk of memory. Therefore, in this code, array a starts at some address (say, 0×00A), and ends at (start address) + n-sizeof (char) address, where n is the number of elements and sizeof (char) is the number of bytes a char takes in memory (and in this example, we're using char because it's 1 byte and makes calculating simple!). For the following questions, assume that the memory can read or write 1 byte at a time. (a) Suppose, when the code is actually run, the array a starts at physical memory address 0x0100, and b is right after (that is, adjacent to) a in memory. What are the physical addresses of the first 5 elements of a? The first 5 elements of b?
Computer Networking: A Top-Down Approach (7th Edition)
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![4. In this question, we consider the relative performance obtained by running a particular program
with various caches installed between the memory and the CPU.
Consider the following C function:
char a [512], b[512]
3 add_arrays () {
register int i;
for (i = 0; i < 511; i = i+1) {
a[i+1] = a [i] + b[i]; // Work!!
9 }
(The register keyword tells the C compiler to reserve a register for the variable i; let's assume
it keeps i in RO. What does this mean for us? The variable i is not stored in main memory-
it's stored in the CPU; this mean that it never has to be pulled from main memory into the
cache when the CPU accesses it.
The char keyword allocates character arrays at fixed locations in memory, using 1 byte per
character.)
Recall that in C, an array is a contiguous chunk of memory. Therefore, in this code, array a
starts at some address (say, 0x00A), and ends at (start address) + n'sizeof (char) address,
where n is the number of elements and sizeof (char) is the number of bytes a char takes
in memory (and in this example, we're using char because it's 1 byte and makes calculating
simple!).
For the following questions, assume that the memory can read or write 1 byte at a time.
(a) Suppose, when the code is actually run, the array a starts at physical memory address
Ox0100, and b is right after (that is, adjacent to) a in memory. What are the physical
addresses of the first 5 elements of a? The first 5 elements of b?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8fc544e9-30e1-4751-af23-a9bb72145ea9%2F3cc5d754-647f-4661-89dc-a5bdecdc3caa%2F1fecba5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4. In this question, we consider the relative performance obtained by running a particular program
with various caches installed between the memory and the CPU.
Consider the following C function:
char a [512], b[512]
3 add_arrays () {
register int i;
for (i = 0; i < 511; i = i+1) {
a[i+1] = a [i] + b[i]; // Work!!
9 }
(The register keyword tells the C compiler to reserve a register for the variable i; let's assume
it keeps i in RO. What does this mean for us? The variable i is not stored in main memory-
it's stored in the CPU; this mean that it never has to be pulled from main memory into the
cache when the CPU accesses it.
The char keyword allocates character arrays at fixed locations in memory, using 1 byte per
character.)
Recall that in C, an array is a contiguous chunk of memory. Therefore, in this code, array a
starts at some address (say, 0x00A), and ends at (start address) + n'sizeof (char) address,
where n is the number of elements and sizeof (char) is the number of bytes a char takes
in memory (and in this example, we're using char because it's 1 byte and makes calculating
simple!).
For the following questions, assume that the memory can read or write 1 byte at a time.
(a) Suppose, when the code is actually run, the array a starts at physical memory address
Ox0100, and b is right after (that is, adjacent to) a in memory. What are the physical
addresses of the first 5 elements of a? The first 5 elements of b?
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