Compile the pseudo-x86 program below into x86, placing all variables on the stack. movq $1, x movq $4, tmp_1 addq $5, tmp_1 movq tmp_1, tmp_2 addq x, tmp_2 movq tmp_2, %rdi callq print_int Part (a): for each variable in the program, assign the variable a home on the stack. • x: • tmp_1 : • tmp_2 : Part (b): what is the size (in bytes) of the stack frame for this program?
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- IN C PROGRAMMING LANGUAGE: Please write a pointer version of strncmp() named pstr_ncmp(char *s, char *t, int n) which compares the first n characters.IN C PROGRAMMING LANGUAGE: Please write a pointer version of strncpy() named pstr_ncpy(char *dest, char *src, int n) which copies n characters from src and copies them into dest.Using pointers create a c program with a function that asks for a maximum number of inputs and counts how many even numbers are there in the array. Example: Enter max inputs: 3 Enter input 1: 561 Enter input 2: 780 Enter input 3: 943 Even count: 1 Even numbers: 780 Address: x
- Assignment for Computer Architecture: N Factual by Recusion *please have comments in the code* You are to write a program in MIPS that computes N! using recursion. Remember N! is the product of all the numbers from 1 to N inclusive, that is 1 x 2 x 3 x (N – 1) x N. It is defined as 1 for N = 0 and is undefined for values less than 0. The programs first requests the user to input the value of N (display a prompt first so the user knows what to do). If the input value is less than 0, the program is to display “N! undefined for values less than 0” and then requests the user to input the value of N again. If the value input is non-negative, it is to compute N! using a recursive function, that is one that calls itself. You are to have your name, the assignment number, and a brief description of the program in comments at the top of your program. Since this is an assembly language program, I expect to see comments on almost every line of code in the program. Also make the…Solution Floating point representation: It is defined as the representation of floating numbers. It includes sign bit, exponent, and mantissa bits. Based on precision it has 2 types. 1. For IEEE 754 single-precision floating-point numbers, what is the exponent of a denormalized floating-point number in decimal? Solution: In IEEE 754 single-precision, exponent bits are 8. Therefore exponent = 2 ^(n-1) -1 = 2^(8-1) -1 = 127 OPTION D 2. For IEEE 754 single-precision floating-point numbers, how many bits for mantissa? Solution: In IEEE 754 single-precision, there are 23 bits for mantissa. sign = 1 bit exponent = 8 bits mantissa = 23 bits 3. For IEEE 754 single-precision floating-point numbers, which of the following is an example of NAN? Solution: In IEEE 754 single-precision, NAN is a special value where all exponents bits are 1's and the mantissa is non zero. a. 1 111 1 111 0000 0000 1101 0000 0000 0000 Exponent is not all 1's. Not a NAN b. 0 111 1 111 1000 0000…in c++ 1. Consider an array of 20 elements is stored in the memory of 40 bytes from 200 to 238. Assume array index starts from 3. Find out the address of 10th index element.
- Question 1: Convert the following infix expression into postfix one: (A+B^D)/(D-F)+G PS: you have to illustrate the different steps of this conversion one by one. Token #Scan Postfix Expression Stack 1 2 3 4 15 6 7 18 9 10 11 12 13 14 15 The final solution is: Question 2: Evaluate the postfix expression ABC D+ * / Where A = 66, B = 1, C = 12 and D=10The return address of a function can be displayed via a series of commands. Keep in mind that any changes you make to the stack must not impede the return of the procedure to its callerCode in C Code in the file IO: /************************************************************* This program prints a degree-to-radian table using a for- loop structure. The results are printed to a file and the the screen. *************************************************************/ #include <stdio.h> #define PI 3.141593 #define FILENAME "tableD2R.dat" int main(void) { /* Declare variables. */ double radians; FILE *fileout; /* Open file. */ fileout = fopen(FILENAME,"w"); if (fileout == NULL) printf("Error opening input file. \n"); else { /* Print radians and degrees in a loop. */ printf("Degrees to Radians \n"); for (int degrees=0; degrees<=360; degrees+=10) { radians = degrees*PI/180; printf("%6i %9.6f \n",degrees,radians); fprintf(fileout,"%6i %9.6f \n",degrees,radians); } /* Exit program. */ }
- Please help me in c++ program declare two characker arrays and takes thier input and add /0 at the end of both then store value of first char array in third array which terminates terminates array when reads /0 at end of first char array and start printing 2nd char array in same 3rd string and print /0 at the end of third string. Output: Enter first string: this is /0 Enter seond string: cat /0 third string: this is cat /0FOR AES CIPHER FILL IN THE CODE FOR ENCRYPTION PROCESS IN C++ FOR ENCRYPTION FUNCTION, MIX COLUMNS AND SHIFT ROW #include "aes_128_common.h" #ifndef AES_128_ENCRYPT_H #define AES_128_ENCRYPT_H byte* encrypt_aes_128(byte* plaintext, byte* key); void shift_rows(byte* byte_array); void mix_columns(byte* byte_array); #endif //AES_128_ENCRYPT_H #include #ifndef AES_128_COMMON_H #define AES_128_COMMON_H typedef unsigned int byte; byte* string_to_byte_array(std::string str); void print_byte_array(byte* byte_array, size_t length); void print_state(byte* byte_array); void newline(); void add_round_key(byte* byte_array); byte substitute_byte(byte byte_to_substitute); void substitute_bytes(byte* byte_array); #endif //AES_128_COMMON_H #include "aes_128_encrypt.h" // Encryption function byte* encrypt_aes_128(byte* plaintext, byte* key) { byte* ciphertext = NULL; // Fill this function return ciphertext; } // Shift rows void shift_rows(byte* byte_array) { // Fill this function…a. Given the variables valuesArray and arrayPtr defined below, assign the variable arravPtr to the valuesArray and do pointer arithmetic to print out (using a cout statement) each element of valuesArray. int valuesArray[3] = {66, 23, 92}; int *arrayPtrị b. The address of valuesArrav[0] is Ox7ffc125d3e20. What is the address of these elements of the array? valuesArray[1] address is: valuesArray[2] address is: ww m me