Task 2 Get Binary 1: Implement the get_bin_1 function as directed in the comment above the function. If the directions are unclear, run the code, and look at the expected answer for a few inputs. In order to be able to compare your solution to the answer, you cannot simply print out the binary directly to the console. Instead, you have been provided with a very rudimentary stringbuilder implementation that allows you to build a string one character at a time, without the ability to erase. There are comments at the top of the code explaining how to use the stringbuilder, however you could also see how it is being used in task_2_check. Note: if you create a stringbuilder and append 'a', 'b', then 'c', the resulting string will be "abc" when printed. You are free to read through the stringbuilder implementation, but it is quite complicated in order to make it easy to use.
IN C PROGRAMMING
ONLY TASK 2
#include <stdio.h>
#include <stdbool.h>
#include <stdlib.h>
// BEGIN STRINGBUILDER IMPLEMENTATION
// This code is a very rudimentary stringbuilder-like implementation
// To create a new stringbuilder, use the following line of code
//
// stringbuilder sb = new_sb();
//
// If you want to append a character to the stringbuilder, use the
// following line of code. Replace whatever character you want to
// append where the 'a' is.
//
// sb_append_char(sb, 'a');
//
// Though there are some other functions that might be useful to you,
// the driver code provided uses the functions, so there is no need
// to use them manually.
typedef struct {
char** cars;
size_t* len;
size_t* alloc_size;
} stringbuilder;
stringbuilder new_sb() {
stringbuilder sb;
sb.cars = malloc(sizeof(char*));
*sb.cars = malloc(8*sizeof(char));
(*sb.cars)[0] = 0;
sb.len = malloc(sizeof(size_t));
*sb.len = 0;
sb.alloc_size = malloc(sizeof(size_t));
*sb.alloc_size = 8;
return sb;
}
void sb_append(stringbuilder sb, char a) {
int len = *sb.len;
if (len >= (*sb.alloc_size)-1) {
*sb.alloc_size = (*sb.alloc_size)*2;
char* newcars = malloc((*sb.alloc_size)*sizeof(char));
for (int i = 0; i < *sb.len; i++) {
newcars[i] = (*sb.cars)[i];
}
free(*sb.cars);
(*sb.cars) = newcars;
}
(*sb.cars)[len] = a;
len++;
(*sb.cars)[len] = 0;
*sb.len = len;
}
void delete_sb(stringbuilder sb) {
free(*sb.cars);
free(sb.cars);
free(sb.len);
free(sb.alloc_size);
}
bool sb_is_equal(stringbuilder sb1, stringbuilder sb2) {
if (*sb1.len != *sb2.len)
return false;
for (int i = 0; i < *sb1.len; i++) {
if ((*sb1.cars)[i] != (*sb2.cars)[i])
return false;
}
return true;
}
void print_sb(const stringbuilder sb) {
printf("%s", *sb.cars);
}
// END STRINGBUILDER IMPLEMENTATION
// ============================================================
// Write your solutions to the tasks below
const unsigned UNS_MAX = -1; // 1111...
const unsigned UNS_MIN = 0; // 0000...
const int INT_MAX = UNS_MAX >> 1; // 0111...
const int INT_MIN = ~INT_MAX; // 1000...
// Task 1
// For this function, you must return an integer holding the value
// x+1, however you may not use any constants or the symbol '+'
// anywhere in your solution. This means that
//
// return x - (-1);
//
// is not a valid soltion, because it uses the constant -1.
//
// Hint: Consider what internally happens when you do -x.
int plus_one(int x) {
return x;
}
// Task 2
// For this function, you must build a string that when printed,
// will output the entire binary representation of the integer x,
// no matter how many bits an integer is. You may NOT use
// division (/) or mod (%) anywhere in your code, and should
// instead rely on bitwise operations to read the underlying binary
// representation of x.
stringbuilder get_bin_1(int x) {
stringbuilder sb = new_sb();
sb_append(sb, '$');
return sb;
}
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