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# CS 2211a ASSIGNMENT 3 solved

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## Description

Part one: 70%
The goal of the exercise is to implement a simple calculator, called ”evaluate.c”, to evaluate
simple arithmetic expression.
(1) An arithmetic expression is an expression that results in a numeric value. We consider
numeric value to be real or floating point numbers. A simple arithmetic expression
involves numeric values connected by arithmetic operators. In this exercise, numeric
value will be real or floating point numbers and operators will be +, −, ∗, and /.
For example, 4 + 35 is a simple arithmetic expression and 2.3 ∗ 4 − 5 − -7.8 + 9 / 3
is another simple arithmetic expression.
In evaluating arithmetic expression, ∗ and / have higher precedence than + and −.
With operators of the same precedence, the order of evaluation is from left to right. We
will use s exp to represent simple arithmetic expression, m exp to represent simple
arithmetic expression in which all operators are ∗ or /, l op to represent operators +
and −, h op to represent operators ∗ and /, and num to represent numeric value.
In the following, simple arithmetic expression is represented recursively, where | represent OR relationship. This recursive definition would be useful in designing your
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solution for this exercise.
s exp → m exp
s exp → s exp l op m exp
m exp → num
m exp → m exp h op num
l op → + | −
h op → ∗ | /
(2) In your program, first the user is asked to input a simple arithmetic expression. In
the inputted simple arithmetic expression, there could be space characters before or
after a number or an operator, The input numbers could be either integers or floating
numbers and we assume that the user will always enter valid numbers. Your program
should handle non-valid single character input for operators.
(3) After user input, the program will calculate and print the numeric value of the inputted
simple arithmetic expression. The program does not read the whole expression before
its calculation. The calculation proceeds while reading numbers and operators of the
inputted simple arithmetic expression.
In evaluating simple arithmetic expression, + and − have the same precedence and
the evaluation order is from left to right, and ∗ and / have the same precedence and
the evaluation order is from left to right.
We will use one non-recursive function and one recursive function to perform the evaluation. The implementation of these functions should follow the recursive definition for
simple expression in (1). To guide you toward this goal, we provide template functions.
We ask you to use these templates and fill in the missing code.
// Input: none, the s_expression will be read in from stdin
// Effect: the s_expression is evaluated using
// a while loop or a do while loop:
// repeatedly get num with m_exp() and then get op with
// get_op() while op is ’+’ or ’-’; when op is ’\n’, exit loop.
// Output: this function returns the value of the s_expression
float s_exp(void) {
}
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// Input: ’sub_exp’: the value of the current sub m_expression
// to the left of ’op’ location in stdin.
// ’op’ : an operator, ’*’ or ’/’. ’op’ could also be
// ’+’, ’-’, or ’\n’ indicating the end of the m_expression.
// “+’, ’-’, or ’\n’ should be pushed back to stdin.
// the rest of the m_expression will be read in from stdin
// Effect: the m_expression is evaluated using recursion:
// get next_num with get_num() and then get next_op with get_op()
// use ’sub_exp op next_num’ and ’next_op’ to do recursive call
// Output: this function returns the value of the current m_expression
float m_exp(float sub_exp, char op) {
}
The following two functions should also be used to simplify the programming task.
// Input: none, read from stdin
// Effect: get the next operator of the expression
// this operator can be +, -, *, /, or ’\n’
// ’\n’ indicates the end of the expression input
// Output: return the next operator of the expression.
char get_op() {
}
// Input: none, read from stdin
// Effect: get the next numeric value of the expression
// Output: return the next numeric value of the expression.
float get_num() {
}
To push back one character in ch to stdin, use ungetc(ch, stdin). ungetc() is a library
function in stdin.h that can push back a character to a specified stream.
When an error is detected, you can exit your program by a function call exit(EXIT F AILURE).
To use this function, the required header is:
#include < stdlib.h >
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(4) Then the user is asked if she/he wants to continue. Two characters can be entered, each
corresponding to a possible action.
• Y for continuing inputting a simple arithmetic expression
• N for quit
(5) Evaluate and print values, or report input format errors, for the following simple arithmetic expressions with your program.
• 5
• 3 ∗ 4
• 4 + 3
• 2.6 / 2 − 1.5 ∗ 10
• -2.0 ∗ 2 + 1.5 ∗ 10 − 100
• 3.5 − -1.5 ∗ 10 / 3.0 + 2 ∗ 3
• 3.5 − 1.5 % 10 − 2.0
Part two: 30%
The goal of this exercise is to implement a C program, called ”ss integer.c”, to display integer
in seven-segment display format.
(1) Calculators, watches, and other electronic devices often rely on seven-segment displays
for numerical output. To form a digit, such devices “turn on” some of the seven
segments while leaving others “off”.
_ _ _ _ _ _ _ _
| | | _| _| |_| |_ |_ | |_| |_|
|_| | |_ _| | _| |_| | |_| _|
(2) In your program, the user is asked to input an integer. After reading the input, the
program will output the inputted integer using seven-segment displays.
You should use a three dimensional array of characters to store the 10 digits.
Here is what the array may look like:
const char segements[10][3][3] =
{ { {’ ’, ’_’, ’ ’}, {’|’, ’ ’, ’|’}, {’|’, ’_’, ’|’} }, … };
Only the initialization for segments[0] is given and you should fill in the rest.
Do not forget the sign of the integer.
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(3) Then the user is asked if she/he wants to continue. Two characters can be entered, each
corresponding to a possible action.
• Y for continuing inputting an integer
• N for quit
(4) Testing your program by inputing several representative integers.