Answer:
One advantage of <u>byte-oriented APIs</u> is that all byte of the data sent by the API is retrieved but for <u>APIs with application-defined message boundaries</u>, some bytes of data could be lost in the streaming process.
One disadvantage of <u>byte-oriented APIs</u> is the time it takes to resend a byte increases its propagation delay, while the is an advantage of the <u>APIs with application-defined message boundaries</u> as it does not have to resend any lost bytes of data, making it faster.
Explanation:
Note that byte-oriented APIs uses the TCP socket which resends lost bytes of data and application-defined message boundaries APIs uses the UDP protocol to send data but does not resend if lost.
Explanation:
it can also be sent through flashdrive
//=indicating you to do the programming part on your own relating to the description provided against. This done because different programming languages require different coding for that.
n=integer value
n1=dummy storage for n
r=variable used to do the function
{
int n,r,n1,rev=0;
//do the coding here for storing the integer in the variable n
n1=n;
while(n>0){
r=n%10;
rev=(rev*10)+r;
n=n/10;
}
//now add a command for displaying the value of rev
}
this is just a logic i used for java
done.
Answer: provided in the explanation section
Explanation:
Given that:
Assume D(k) =║ true it is [1 : : : k] is valid sequence words or false otherwise
now the sub problem s[1 : : : k] is a valid sequence of words IFF s[1 : : : 1] is a valid sequence of words and s[ 1 + 1 : : : k] is valid word.
So, from here we have that D(k) is given by the following recorance relation:
D(k) = ║ false maximum (d[l]∧DICT(s[1 + 1 : : : k]) otherwise
Algorithm:
Valid sentence (s,k)
D [1 : : : k] ∦ array of boolean variable.
for a ← 1 to m
do ;
d(0) ← false
for b ← 0 to a - j
for b ← 0 to a - j
do;
if D[b] ∧ DICT s([b + 1 : : : a])
d (a) ← True
(b). Algorithm Output
if D[k] = = True
stack = temp stack ∦stack is used to print the strings in order
c = k
while C > 0
stack push (s [w(c)] : : : C] // w(p) is the position in s[1 : : : k] of the valid world at // position c
P = W (p) - 1
output stack
= 0 =
cheers i hope this helps !!!