Was this in reference to literal audio archives? If so, I don't see any cons beside possible copyright infringement.
If you're talking about the codecs themselves, then I can do that.
<span>Pros:
</span>- Widespread acceptance. Supported in nearly all hardware devices, and continually adopted by newer ones.
- Faster decoding. Much more so than FLAC, Vorbis, etc.
- Relaxed licensing schedule.
<span>Cons:
</span><span>
</span>- Lower quality and efficiency than most modern codecs. (To be fair, never really noticed this one).
- Sometimes the maximum bitrate isn't enough.
- Pretty much void/unusable for high definition audio (higher than <span>48kHz).</span>
Answer:
//Here is the for loop in C.
for(n=10;n>0;n--)
{
printf("count =%d \n",n);
}
Explanation:
Since C is a procedural programming language.Here if a loop that starts with n=10; It will run till n becomes 0. When n reaches to 0 then loop terminates otherwise it print the count of n.
// here is code in C++.
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{ // variables
int n;
// for loop that runs 10 times
// when n==0 then loop terminates
for(n=10;n>0;n--)
{
cout<<"count ="<<n<<endl;
}
return 0;
}
Output:
count =10
count =9
count =8
count =7
count =6
count =5
count =4
count =3
count =2
count =1
A trademark can be anything! it could be be a hair tie, a flag and even a bag.
Answer:
a) 4 processes
b) 2 resources
c) R1: 2 instances
R2: 2 instances
d) R2
e) R1
f) R1
g) No resource
h) R1
i) R2
j) R2
k) No
l)No deadlock
Explanation:
You need to know that the resources that are required by the processes for completion are shown by the request edge and the resources allocated are shown by the allocation edge. And thus, we can find what resources are allocated to the process, and required for the completion accordingly. And here again, the cycle is created, and hence deadlock may or may not occur. However, we see that resources have multiple instances and get freed on time. And hence, deadlock does not occur.
Like,
P4 uses R2 and free one instance of R2.
P3 then uses one instance of R2 and free R2.
P3 then uses one instance of R3 and free R3.
P1 uses one instance of R1 and free R1 one instance
P1 then uses R2 and free R2.
P2 uses R1 and free R1.
Hence, all the processes are complete and deadlock does not occur.
