2. Tell the exact content of the variables memory location in binary upon execution of the code, use sizeof(int)=2 and the ASCII value of ’A’= decimal 65= hexa 41, ’a’= decimal 97 = hexa 61 : int x, y; char z[4] = ”CaB” ; x = 63; y = -63;
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2. Tell the exact content of the variables memory location in binary upon execution of the code, use sizeof(int)=2 and the ASCII value of ’A’= decimal 65= hexa 41, ’a’= decimal 97 = hexa 61 :
int x, y;
char z[4] = ”CaB” ;
x = 63;
y = -63;
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- You are required to make changes in the below programs and introduce the use of compaction where required. #include<stdio.h> #include<conio.h> main() { int ms, bs, nob, ef,n, mp[10],tif=0; int i,p=0; clrscr(); printf("Enter the total memory available (in Bytes) -- "); scanf("%d",&ms); printf("Enter the block size (in Bytes) -- "); scanf("%d", &bs); nob=ms/bs; ef=ms - nob*bs; printf("\nEnter the number of processes -- "); scanf("%d",&n); for(i=0;i<n;i++) { printf("Enter memory required for process %d (in Bytes)-- ",i+1); scanf("%d",&mp[i]); } printf("\nNo. of Blocks available in memory -- %d",nob); printf("\n\nPROCESS\tMEMORY REQUIRED\t ALLOCATED\tINTERNAL FRAGMENTATION"); for(i=0;i<n && p<nob;i++) { printf("\n %d\t\t%d",i+1,mp[i]); if(mp[i] > bs) printf("\t\tNO\t\t---"); else { printf("\t\tYES\t%d",bs-mp[i]);tif = tif + bs-mp[i]; p++; } } if(i<n) printf("\nMemory is Full, Remaining Processes cannot be accomodated"); printf("\n\nTotal…6. Find the output of the given code if it is compiled on a 64-bit system. #include <stdio.h>int main(){ char *ptrToChar; int *ptrToInt; float *ptrToFloat; printf("%ld ", sizeof(ptrToChar)); printf("%ld ", sizeof(ptrToInt)); printf("%ld ", sizeof(ptrToFloat)); return 0;}A string of readings is stored in memory locations starting at XX70H, and the end of the string is indicated by the byte 0DH. Write a program to check each byte in the string, and count the number of bytes in the range of 30H to 39H (both inclusive). Store the result in memory locations starting from XX90H Data(H) 35, 2F, 30, 39, 3A, 37, 7F, 31, 0D, 32
- 7) Under computer integer arithmetic, the quotient J/K of two integers J and K is less than or equal to the usual quotient. True or false?5. Declare a 32-bit signed integer variable and initialize it with the value 234 (base 10). In memory, what are the hexadecimal values shown?Write a C code that finds the numbers of the Fibonacci sequence in a sequence entered by the user. Note that dynamic memory functions must be used if arrays are to be used.
- C++ Programming How many bits in a byte? multiples of 8 bit How many bytes in a double? __________ When assigning a float variable to an int variable, what is lost? __________ In the statement x = y; which variable is updated, x or y? _____ What include file is needed to make use of vector variables? ________ Give an example of a unary operator and a binary operator. _________ Write a line of code that declares an int type of variable and initializes it to the value 23. ____________________________ A variable declared outside of any function is called a __________ variable. You end a multi-line comment with ______________. What do you call the variable RATE in the following statement: const double RATE = 0.75; _________________________ What is the function performed by the cout object? ____________________ You can put several statements on the same line, true or false? _______ Write a single statement that outputs two variables x and y to the screen. _____________________ After the…Use the C programming language to write code that proclaims and sets a double, int, and char. Afterward, proclaim and set a pointer to the double, int, and char. The code must print the inscription of, and value stored in, and the memory size (in bytes) of all six variables. The "0x%x" formatting specifier outputs inscriptions in hexadecimal. The "%f" outputs floating values. The sizeof operator must determine the memory size allocated for all variables. SAMPLE OUTPUT: Char ___ 's inscription is 0x___Int ___'s inscription is 0x___Double ___'s inscription is 0x___Char* ___'s inscription is 0x___Int* ___'s inscription is 0x___Double* ___'s inscription is 0x___Char ___ 's value is ___Int ___'s value is ___Double ___'s value is ___Char* ___'s value is 0x___Int* ___'s value is 0x___Double* ___'s value is 0x___Char ___ 's size is ___ bytesInt ___'s size is ___ bytesDouble ___'s size is ___ bytesChar* ___'s size is ___ bytesInt* ___'s size is ___ bytesDouble* ___'s size is ___ bytesNeed help with this C program please! For this task, you will complete bit_ops.c by implementing the following three bit manipulation batch functions. You will want to use bitwise operations such as and (&), or (|), xor (^), not (~), left shifts (<<), and right shifts (>>). #include <stdio.h> #include <stdlib.h> // Note, the bits are counted from right to left. // Return the bit states of x within range of [start, end], in which both are inclusive. // Assume 0 <= start & end <= 31 unsigned * get_bits(unsigned x, unsigned start, unsigned end) { return NULL; // YOUR CODE HERE // Returning NULL is a placeholder // get_bits dynamically allocates an array a and set a[i] = 1 when (i+start)-th bit // of x is 1, otherwise siet a[i] = 0; // At last, get_bits returns the address of the array. } // Set the bits of x within range of [start, end], in which both are inclusive // Assume 0 <= start & end <= 31 void set_bits(unsigned * x,…
- Use the C programming language to write code that proclaims and sets a double, int, and char. Afterward, proclaim and set a pointer to the double, int, and char. The code must print the inscription of, and value stored in, and the memory size (in bytes) of all six variables. The "0x%x" formatting specifier outputs inscriptions in hexadecimal. The "%f" outputs floating values. The sizeof operator must determine the memory size allocated for all variables.What is a round-off error? Can integer operations cause round-off errors? Can floating-point operations cause round-off errors?Which of the following statements are true? A. A char typically uses twice as many bits as a short. B. A signed integer can represent more negative numbers than positive numbers. C. A 64-bit double uses twice as many bits for its exponents than a 32-bit float. D. A byte consists of 32 bit on 32-bit architectures and 64 bit on 64-bit architectures.