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hoa [83]
2 years ago
13

What is the maximum possible value of num after the code has been run?

Computers and Technology
1 answer:
Artist 52 [7]2 years ago
8 0

Answer:

the maximum value from the first line is 20 - 0 = 20

and from the second 20-5 and 20+5 so the maximum is 25

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A Java main method uses the parameter (String[ ] variable) so that a user can run the program and supply "command-line" paramete
Firdavs [7]

Answer:

The answer is "Option a"

Explanation:

  • In java, the main function is the point of entry of every java program. Its syntax always starts "public static void main" with (String args[]), in which it can also be modified by the name of the string array name.
  • It also known as an entry point is the key process. In any program, it is the first method, that executes whenever you run a program. There is one main feature in a regular app that uses instances of certain classes to operate.
4 0
3 years ago
Please explain what Level 5 Automation is and give 2 examples of the technology.
arsen [322]

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8 0
2 years ago
Write a program to test the difference between %d and %i conversion
Gelneren [198K]
Void test(char *s)
{
  int i, d;
  sscanf(s, "%i", &i);
  printf("%s converts to %i using %%i\n", s, i);
  sscanf(s, "%d", &d);
  printf("%s converts to %d using %%d\n", s, d);
}

int main()
{
  test("123");
  test("0x123");
  return 0;
}

outputs:
123 converts to 123 using %i
123 converts to 123 using %d
0x123 converts to 291 using %i
0x123 converts to 0 using %d

As you can see, %i is capable of parsing hexadecimal, whereas %d is not. For printf they're the same.
7 0
3 years ago
A 4-way set associative cache has 64 blocks of 16 words. How many bits are there in the ""tag"" field of the 15-bit address for
Tanya [424]

Answer:

3 bits

Explanation:

Given a 4- way set associative cache that has 64 blocks of 16 words.

Therefore, the number of sets cache has:

\frac{64}{4} = 16

Now,

Cache data size is 16kB

The number of cache blocks can be calculated as:

16kB/16 = 1024 bytes/16 byte\times 16 = 256 cache blocks

Now,

Total sets = \frac{cache blocks}{associative sets}

Total sets = \frac{256}{4} = 64

Now,

2^{n} = 64

n = 6

For 15 bit address for the architecture, the bits in tag field is given by:

15 - (6 + 6) = 3 bits

Thus the tag field will have 3 bits

6 0
3 years ago
Answer this and you get free 100 points
Agata [3.3K]

Answer:

thanks my dude I needed it

5 0
3 years ago
Read 2 more answers
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