Create F2_15.c #15
Create F2_15.c #15
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The application you will be writing for is intended to run on a small, inexpensive embedded device that does not have a floating point unit. This means that the processor can only do integer arithmetic. However, the device does have a display and needs to display the result of floating point arithmetic to the user. Because this is a small inexpensive processor there is no support for strings or any mathematical library functions. You must write all the code yourselves performing all the floating point math with integers only and no strings! Character arrays, or C strings, can't tell you their size but they do end with a '\0' character.
The requirements for the code are to write one of the functions presented below. All the functions return a boolean value indicating the success or failure of the operation. Improper data is the most likely cause for a function to fail. You may add your own additional helper functions (I had at least a half dozen) but you must include them for the review. Do not change the interface to these functions.
These two functions take as parameters two sets of mantissa and characteristic and a char array to hold the result of the arithmetic operations. The result of the add or subtract should be converted into char's and placed in the result array. The array must end with a '\0'. The 'len' parameter tells how many characters can be placed on the result array. For these functions to return true you must at a minimum store the characteristic of the result. If the result is a non-integer place as many of the digits as will fit in the result after a decimal point.
```c
bool add(int c1, int n1, int d1, int c2, int n2, int d2, char result[], int len);
bool subtract(int c1, int n1, int d1, int c2, int n2, int d2, char result[], int len);
...
...
char answer[100];
int c1, n1, d1;
int c2, n2, d2;
c1 = 1;
n1 = 1;
d1 = 2;
c2 = 2;
n2 = 2;
d2 = 3;
//if the C string could hold at least the characteristic if(add(c1, n1, d1, c2, n2, d2, answer, 100))
{
//display string with answer 4.166666666...
}
else
{
//display error message
}
'''
|
I am working on this |
Things to improveLooking at the code, there are a few areas of improvement. Some of these include: I have a few recommendations for these problems: (2) Lack of helper functions int pos = 0;
if (numeratorSum < 0) {
result[pos++] = '-';
numeratorSum = -numeratorSum;
}
int characteristicSum = numeratorSum / commonDenominator;
int temp = characteristicSum;
if (temp == 0) {
result[pos++] = '0';
} else {
int divisor = 1;
while (temp / divisor >= 10) {
divisor *= 10;
}
while (divisor > 0) {
result[pos++] = '0' + temp / divisor;
temp %= divisor;
divisor /= 10;
}
}
result[pos++] = '.';
numeratorSum %= commonDenominator;
if (numeratorSum == 0) {
result[pos++] = '0';
} else {
int divisor = commonDenominator / 10;
while (divisor > 0 && numeratorSum / divisor == 0) {
result[pos++] = '0';
divisor /= 10;
}
while (divisor > 0) {
result[pos++] = '0' + numeratorSum / divisor;
numeratorSum %= divisor;
divisor /= 10;
}
}
result[pos] = '\0';(3) Lack of buffer overflow preventive measures Looking at the subtract function, these same changes need to be made, with variables such as numeratorSum being renamed to numeratorDiff, so I recommend making these changes in that function too. I also recommend implementing measures to handle buffer underflow, so I recommend including checks that ensure positional variables are greater than or equal to 0. (4) Improper division handling int add(int c1, int n1, int d1, int c2, int n2, int d2, char result[], int len) {
int commonDenominator = d1 * d2;
if(commonDenominator != 0)
{
int numerator1 = c1 * commonDenominator + n1 * d2;
int numerator2 = c2 * commonDenominator + n2 * d1;
//Include rest of add function code here...int subtract(int c1, int n1, int d1, int c2, int n2, int d2, char result[], int len) {
int commonDenominator = d1 * d2;
if(commonDenominator != 0)
{
int numerator1 = c1 * commonDenominator + n1 * d2;
int numerator2 = c2 * commonDenominator + n2 * d1;
//Include rest of subtract function here...Pros and ConsProsThere are some pros in this code. Some of these include proper variable naming, which makes it easier to understand the code. ConsThe only cons that come to mind are those listed in Things to improve. Thank you for reading this, feel free to contact me if you have any questions, comments, or concerns. |
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Few suggestions that can improve the code:
- Create a helper function to format the result as a string to reduce duplication.
- Ensure proper handling of edge cases such as zero denominators and buffer overflow.
- Use more descriptive variable names to improve clarity.
Refactored version of the code -
#include <stdio.h>
#include <stdlib.h>
void formatResult(int numerator, int denominator, char result[], int len) {
int pos = 0;
int characteristic = numerator / denominator;
// Handle negative results
if (numerator < 0) {
result[pos++] = '-';
numerator = -numerator;
}
// Handle characteristic part
int temp = characteristic;
if (temp == 0) {
result[pos++] = '0';
} else {
int divisor = 1;
while (temp / divisor >= 10) {
divisor *= 10;
}
while (divisor > 0) {
result[pos++] = '0' + temp / divisor;
temp %= divisor;
divisor /= 10;
}
}
// Handle fractional part
result[pos++] = '.';
numerator %= denominator;
if (numerator == 0) {
result[pos++] = '0';
} else {
int divisor = denominator / 10;
while (divisor > 0 && numerator / divisor == 0) {
result[pos++] = '0';
divisor /= 10;
}
while (divisor > 0) {
result[pos++] = '0' + numerator / divisor;
numerator %= divisor;
divisor /= 10;
}
}
result[pos] = '\0';
}
int add(int c1, int n1, int d1, int c2, int n2, int d2, char result[], int len) {
// Check for zero denominators
if (d1 == 0 || d2 == 0) {
snprintf(result, len, "Error: Zero denominator");
return -1;
}
// Calculate common denominator and numerators
int commonDenominator = d1 * d2;
int numerator1 = c1 * commonDenominator + n1 * d2;
int numerator2 = c2 * commonDenominator + n2 * d1;
int numeratorSum = numerator1 + numerator2;
// Format the result
formatResult(numeratorSum, commonDenominator, result, len);
return 1;
}
int subtract(int c1, int n1, int d1, int c2, int n2, int d2, char result[], int len) {
// Check for zero denominators
if (d1 == 0 || d2 == 0) {
snprintf(result, len, "Error: Zero denominator");
return -1;
}
// Calculate common denominator and numerators
int commonDenominator = d1 * d2;
int numerator1 = c1 * commonDenominator + n1 * d2;
int numerator2 = c2 * commonDenominator + n2 * d1;
int numeratorDiff = numerator1 - numerator2;
// Format the result
formatResult(numeratorDiff, commonDenominator, result, len);
return 1;
}
|
I am working on this |
I already worked on this. Were there none left? |
The application you will be writing for is intended to run on a small, inexpensive embedded device that does not have a floating point unit. This means that the processor can only do integer arithmetic. However, the device does have a display and needs to display the result of floating point arithmetic to the user. Because this is a small inexpensive processor there is no support for strings or any mathematical library functions. You must write all the code yourselves performing all the floating point math with integers only and no strings! Character arrays, or C strings, can't tell you their size but they do end with a '\0' character.
The requirements for the code are to write one of the functions presented below. All the functions return a boolean value indicating the success or failure of the operation. Improper data is the most likely cause for a function to fail. You may add your own additional helper functions (I had at least a half dozen) but you must include them for the review. Do not change the interface to these functions.
These two functions take as parameters two sets of mantissa and characteristic and a char array to hold the result of the arithmetic operations. The result of the add or subtract should be converted into char's and placed in the result array. The array must end with a '\0'. The 'len' parameter tells how many characters can be placed on the result array. For these functions to return true you must at a minimum store the characteristic of the result. If the result is a non-integer place as many of the digits as will fit in the result after a decimal point.