Answer:
27 ounces
Explanation:
You can convert 2 pounds into 32 ounces then subtract the 5 ounces that she ate from that 32.
Answer:
// here is code in the C++.
#include <bits/stdc++.h>
using namespace std;
// main function
int main() {
// variable to store the input
int birth_month,birth_year;
cout<<"enter birth month:";
// read the birth month
cin>>birth_month;
cout<<"enter birth year:";
// read the birth year
cin>>birth_year;
// print the output
cout<<birth_month<<"/"<<birth_year<<endl;
return 0;
}
Explanation:
Declare two variables to store the birth month and birth year.Read the inputs from the user and assign it the variables.Print the birth month and year separated by a slash "/".
Output:
enter birth month:1
enter birth year:2000
1/2000
Answer:
Automatic and Explicit Buffering.
In the case of explicit buffering, the length of the queue is provided while in automatic buffering the queue size needs to be indefinite. In automatic buffering there is no need to block the sender while coping the message. While in explicit buffering the sender is blocked for the space in queue.
No memory is being wasted in explicit buffering.
Send by Copy and Send by Reference.
By using the send by copy method, the receiver is not able to change the state of parameter while send by reference allow. The advantage of using the send by reference method is that it allows to change a centralized application to its distributed version.
Fixed-sized and Variable-sized Messages.
In fixed size messaging refers, the buffer size is fixed. This means only a particular number of messages can only be supported by the fixed size buffer. The drawback of the fixed size messages is that they must be a part of fixed size buffer. These are not a part of variable size buffer. The advantage of variable size message is that the length of the message is variable means not fixed. The buffer length is unknown. The shared memory is being used by the variable size messages.
Explanation:
Solution :
#include 
#include 
#include 
//Converts 
to binary string.
* hexadecimal
Binary(char* hexdec)
{
long 
= 0;
char *string = 
(sizeof(char) * 9);
while (hexdec[i]) {
//Simply assign binary string for each hex char.
switch (hexdec[i]) {
strcat(string, "0000");
break;
strcat(string, "0001");
break;
strcat(string, "0010");
break;
strcat(string, "0011");
break;
strcat(string, "0100");
break;
strcat(string, "0101");
break;
strcat(string, "0110");
break;
strcat(string, "0111");
break;
strcat(string, "1000");
break;
strcat(string, "1001");
break;
case 'A':
case 'a':
strcat(string, "1010");
break;
case 'B':
case 'b':
strcat(string, "1011");
break;
case 'C':
case 'c':
strcat(string, "1100");
break;
case 'D':
case 'd':
strcat(string, "1101");
break;
case 'E':
case 'e':
strcat(string, "1110");
break;
case 'F':
case 'f':
strcat(string, "1111");
break;
default:
printf("\nInvalid hexadecimal digit %c",
hexdec[i]);
string="-1" ;
}
i++;
}
return string;
}
int main()
{ //Take 2 strings
char *str1 =hexadecimalToBinary("FA") ;
char *str2 =hexadecimalToBinary("12") ;
//Input 2 numbers p and n.
int p,n;
scanf("%d",&p);
scanf("%d",&n);
//keep j as length of str2
int j=strlen(str2),i;
//Now replace n digits after p of str1
for(i=0;i<n;i++){
str1[p+i]=str2[j-1-i];
}
//Now, i have used c library strtol
long ans = strtol(str1, NULL, 2);
//print result.
printf("%lx",ans);
return 0;
}
