Answer:
The code is not dereferencing the pointers. You have to place an asterisk in front of the pointer to read the value the pointer points to.
Explanation:
So "if (str1 != str2)" must be "if (*str1 != *str2)".
likewise:
while (*str1 != 0 && *str2 != 0)
and
result = (*str1 == *str2);
Using the computational language in JAVA it is possible to write a code that at least passes an unspecified number of integers
<h3>Writing code in JAVA</h3>
<em>class Exercise07_21</em>
<em>{</em>
<em>public static void main(String args[])</em>
<em>{</em>
<em>int sum=0;</em>
<em>for(int i=0;i<args.length;i++)</em>
<em>sum=sum+Integer.parseInt(args[i]);//converting string to integer and then adding it with sum variable and storing back in sum</em>
<em>System.out.println("The total is "+sum);//printing the sum</em>
<em>}</em>
<em>}</em>
See more about JAVA code at brainly.com/question/12975450
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Answer:
It can be 100001, 100000,000001,000000
Hence, answer is 4.
its also can be found through 2C2 *2 = 2!/0! *2 =2*2 =4
Explanation:
The answer is straight forward, we have 2 empty places and we can select from 2 items, and hence its 2C2, and now we can arrange them as well, and since we can arrange in 2 ways, hence a total number of possibility = 2 * 2C2 = 4, which is the required answer.
D I would think……… good luck
Answer:
Explanation:
The minimum depth occurs for the path that always takes the smaller portion of the
split, i.e., the nodes that takes α proportion of work from the parent node. The first
node in the path(after the root) gets α proportion of the work(the size of data
processed by this node is αn), the second one get (2)
so on. The recursion bottoms
out when the size of data becomes 1. Assume the recursion ends at level h, we have
(ℎ) = 1
h = log 1/ = lg(1/)/ lg = − lg / lg
Maximum depth m is similar with minimum depth
(1 − )() = 1
m = log1− 1/ = lg(1/)/ lg(1 − ) = − lg / lg(1 − )
