Calculator with user defined variables in C language - c

I hope you are having a nice day. Thank you for taking the time to read my question.
I am still a beginner in the C language. My professor has asked me to program a scientific calculator. The calculator should be able to read and store user-defined variables and work with them. for example, I should be able to enter a=5 b=9 etc. after that the program should calculate, for instance, a+1= 6, or a+b=14 and show it to the user. Of course, the user decides if the operation is addition, subtraction, division or multiplication.
The user should also be able to enter such input: e.g. c=5+9.
I have started working on the calculator, unfortunately, I have just been able to only allow the user to define one variable at a time and work with it.
For example:
a=7
7+a=14
That's all I could do. I asked my professor for help and he keeps telling me that I have to to teach the program how to separate between what is before the "=" and what's after it.
Thank you in advance for every help or piece of advice you give
This is the code I came up with
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
int main() {
char situation;
float operand1, operand2;
char i = 0;
char ch;
char op;
int operand = 0;
int j, digit, number[10] = {};
int j2 = 0;
char str[100] = { 0 };
while (1) {
printf("\n\na) To calculate choose me :)");
printf("\n\nb) If you want to calculate with variables, choose me:");
printf("\n\nc) To Exit choose me :(\n\n");
scanf_s("%c", &situation, 1);
while ((getchar()) != '\n');
switch (situation) {
case 'a':
printf("type in some arithmetic terms : \n");
scanf_s("%f", &operand1);
scanf_s("%c", &op, 1);
scanf_s("%f", &operand2);
while ((getchar()) != '\n');
switch (op) {
case '+':
printf("\n\n%.2f + %.2f = %.2f\n\n", operand1, operand2, operand1 + operand2);
break;
case '-':
printf("\n\n%.2f - %.2f = %.2f\n\n", operand1, operand2, operand1 - operand2);
break;
case '*':
printf("\n\n%.2f * %.2f = %.2f\n\n", operand1, operand2, operand1 * operand2);
break;
case '/':
printf("\n\n%.2f / %.2f = %.2f\n\n", operand1, operand2, operand1 / operand2);
break;
default:
printf("\n\nERROR!\n\n");
}
break;
case 'b':
printf("\n\nTo return to the main menu please enter any capital letter of your choosing\n\n");
printf("\n\nWhen calculating with a variable, please always write the variable on the right side, Thank you. You may start:\n\n");
do {
scanf_s("%s", &str, 99);
for (i = 0; i < 100; i++) {
if (str[i] == '=') {
for (j = 0; j < strlen(str); j++) {
ch = str[j];
if (ch >= '0' && ch <= '9') {
digit = ch - '0';
number[j2] = j2 * 10 + digit;
//printf("%d", number);
}
}
scanf_s("%d", &operand);
scanf_s("%c", &op, 1);
scanf_s("%d", &number[j2]);
while ((getchar()) != '\n');
switch (op) {
case '+':
printf("\n\n%d + %c = %d\n\n", operand, str[0], operand + number[j2]);
break;
case '-':
printf("\n\n % d - % c = % d\n\n", operand, str[0], operand - number[j2]);
break;
case '*':
printf("\n\n % d * % c = % d\n\n", operand, str[0], operand * number[j2]);
break;
case '/':
printf("\n\n % d / % c = % d\n\n", operand, str[0], operand / number[j2]);
break;
default:
printf("\n\nERROR!\n\n");
}
break;
}
}
} while (islower(str[0]));
while ((getchar()) != '\n');
break;
case 'c':
printf("\n\goodbye\n\n");
exit(0);
break;
default:
printf("\n\nThis is not an acceptable input. Please Try again!");
}
}
}

There are multiple problems in your code:
int number[10] = {}; is an invalid initializer in C. You should write:
int j, digit, number[10] = { 0 };
you should use double instead of float
scanf_s("%c", &situation, 1); is not portable: the Microsoft version of this function expects the size argument 1 as an UNSIGNED whereas the Standard C function defined as optional in Annex K specifies that the size argument 1 must be passed as a size_t, hence as (size_t)1. Avoid using this function and read user input as a line with fgets() and use the standard function sscanf() instead and do test the return value to detect and report invalid and/or missing input.
Add these lines before including <stdio.h> at the top of your source file to prevent compiler warnings:
#ifdef _MSC_VER
#define _CRT_SECURE_NO_WARNINGS
#endif
scanf_s("%c", &op, 1); does not skip white space, such as spaces and newlines. You should write
scanf(" %c", &op);
while ((getchar()) != '\n'); is risky: this loop will run forever if the end of file occurs before a newline can be read.
instead of j < strlen(str), which may rescan the string at every iteration, you should write:
for (j = 0; str[j] != '\0'; j++)
i is used as an index, it should have type int or size_t, not char.
the format string in printf("\n\n % d - % c = % d\n\n", ...); is incorrect: the space between the % and the c is not supported. You probably meant to write this anyway:
printf("\n\n%d - %c = %d\n\n", operand, str[0], operand - number[j2]);
printf("\n\goodbye\n\n"); is incorrect: \g is an invalid escape sequence. You should write:
printf("\n\nGoodbye\n\n");
Here is a modified version using functions to parse the line and handle variable assignment separately from evaluating expressions:
#ifdef _MSC_VER
#define _CRT_SECURE_NO_WARNINGS
#endif
#include <ctype.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
/* 26 global variables */
double variables['z' - 'a' + 1];
/* the function skip_spaces() updates `*pos`, skipping any white
* space in `str` at the corresponding index
*/
void skip_spaces(const char *str, int *pos) {
while (isspace((unsigned char)str[*pos]))
*pos += 1;
}
/* the function trim_spaces() updates `*pos`, skipping any white
* space in `str` at the corresponding index. In addition, it
* removes trailing white space from the string, ie: newlines,
* spaces, TABs and other white space characters
*/
void trim_spaces(char *str, int *pos) {
int len = strlen(str);
while (len > *pos && isspace((unsigned char)str[len - 1]))
str[--len] = '\0';
skip_spaces(str, pos);
}
/* the function parse_operand() reads the next operand from `str`
* at index `*pos`. It recognises floating point numbers and
* variable names, which are replaced with their value. The value is
* stored into `*value`. `*pos` is updated past the operand and any
* white space after it.
*/
int parse_operand(char *str, int *pos, double *value) {
char *endp;
skip_spaces(str, pos);
char ch = str[*pos];
if (ch >= 'a' && ch <= 'z') {
*value = variables[ch - 'a'];
*pos += 1;
skip_spaces(str, pos);
return 1; // variable
}
*value = strtod(str + *pos, &endp);
if (endp > str + *pos) {
*pos = endp - str;
skip_spaces(str, pos);
return 2; // number
}
return 0;
}
/* parse_expression: parse an expression with basic operators,
* no precedence: the function expects at least one operand and
* keeps parsing and evaluating as long as there is a supported
* operator that follows. The result is stored into `*result`.
*/
int parse_expression(char *str, int *pos, double *result) {
double operand2;
char op;
if (!parse_operand(str, pos, result)) {
printf("missing operand: %s\n", str + *pos);
return 0;
}
while ((op = str[*pos]) == '+' || op == '-' || op == '*' || op == '/' || op == '%') {
*pos += 1;
if (!parse_operand(str, pos, &operand2)) {
printf("missing operand: %s\n", str + *pos);
return 0;
}
switch (op) {
case '+':
*result += operand2;
break;
case '-':
*result -= operand2;
break;
case '*':
*result *= operand2;
break;
case '/':
*result /= operand2;
break;
case '%':
*result = fmod(*result, operand2);
break;
}
}
return 1;
}
int main() {
char str[100];
printf("type some expressions:\n");
while (fgets(str, sizeof str, stdin)) {
double result, result2;
int pos = 0;
/* strip trailing whitespace, skip initial whitespace */
trim_spaces(str, &pos);
if (!str[pos]) {
/* stop on empty line */
break;
}
/* test for a variable assignment */
if (str[pos] >= 'a' && str[pos] <= 'z' && str[pos + 1] == '=') {
/* variable assignment */
int v = str[pos] - 'a';
pos += 2;
if (parse_expression(str, &pos, &result) && !str[pos]) {
variables[v] = result;
printf("%s -> %.2f\n", str, result);
} else {
printf("invalid expression: %s\n", str);
}
} else {
/* other expression */
if (parse_expression(str, &pos, &result)) {
skip_spaces(str, &pos);
if (str[pos] == '\0') {
printf("%s -> %.2f\n", str, result);
} else
if (str[pos] == '=') {
/* comparison of expressions */
pos += 1;
if (parse_expression(str, &pos, &result2) && !str[pos]) {
if (result == result2) {
printf("%s -> true (%.2f == %.2f)\n", str, result, result2);
} else {
printf("%s -> false (%f != %f, delta: %e)\n",
str, result, result2, result2 - result);
}
} else {
printf("invalid expression: %s\n", str);
}
} else {
printf("invalid syntax: %s\n", str);
}
}
}
}
return 0;
}
Output:
b=2/3
b=2/3 -> 0.67
2/3=b
2/3=b -> true (0.67 == 0.67)
20/3=10*b
20/3=10*b -> false (6.666667 != 6.666667, delta: -8.881784e-16)

Related

Expression Tree - Operations with Characters

The user enters a string with an operation such as 4*5+2/3 and the code is supposed to make an expression tree out of it and the calculate said expression tree. I am having a problem where the program is making the expression tree with the decimal values of the ascii table instead of the actual numbers.
For example instead of 4*5+2/3, the program is storing and using 52 42 53 43 50 47 51 for the calculations. My desired run screen would be:
1 //this is the number of strings
4*5+2/3 //this is the string itself
20 //this is the result
However what I am getting is:
1 //this is the number of strings
4*5+2/3 //this is the string itself
2756 //this is the result
That is because the code is doing 52*53+50/51(because it is using the ascii values) and not 4*5+2/3.
I believe the reason for this is because I am storing 4*5+2/3 in a string of char and not in an array of int. I do not know if this is the case and would like some help.
You will not be able to run the following code as it is not complete but the whole program is five files and I do not know if I should put all of it here. I am new to both trees and StackOverflow.
This is my Make Expression Tree function and my Calculate Expression Tree function:
BTreeNode* MakeExpTree(char* exp, int len)
{
Stack stack;
BTreeNode * node, *right_node, *left_node;
InitStack(&stack);
for(int i = 0; i < len; i++){
if('0' <= exp[i] && exp[i] <= '9'){
node = CreateNode(exp[i]);
}
else{
right_node = PeekNode(&stack), Pop(&stack);
left_node = PeekNode(&stack), Pop(&stack);
node = CreateNode(exp[i]);
CreateRightSubtree(node, right_node);
CreateLeftSubtree(node, left_node);
}
PushNode(&stack, node);
}
return PeekNode(&stack);
}
int CalculateExpTree(BTreeNode* root)
{
int ret, op1, op2;
if(root == NULL){
return 0;
}
if(root->left_child == NULL && root->right_child == NULL){
return root->item;
}
op1 = CalculateExpTree(root->left_child);
op2 = CalculateExpTree(root->right_child);
switch(root->item){
case '+':
ret = op1 + op2;
break;
case '-':
ret = op1 - op2;
break;
case '*':
ret = op1 * op2;
break;
case '/':
ret = op1 / op2;
break;
case '#':
ret = op1 * pow( 2, op2);
break;
case '#':
ret = op1 / pow( 2, op2);
break;
}
return ret;
}
This is how I store the string from stdin in main function:
int main()
{
int num_exp, result, len = 0;
char input[10];
char IDK[129];
fgets(input, 9, stdin); //user enters number of strings
int m = sscanf(input, "%d", &num_exp);
char string[100][129] = { 0 };
char postfix[100][129] = { 0 };
for(int i = 0; i < num_exp; i++){
fgets(IDK, 129, stdin); //user enters string
int mm = sscanf(IDK, "%s", string[i]); //is this where the problem lies?
} //should I not be storing it in a char string?
for(int x = 0; x < num_exp; x++){
InfixToPostfix(string[x], postfix[x]); //converts strings from infix to postfix
}
BTreeNode* tree;
for(int k = 0; k < 129; k++){ //calculates length of string
if(postfix[0][k] == '\0'){
break;
}
len++;
}
tree = MakeExpTree(postfix[0], len); //makes expression tree
result = CalculateExpTree(tree); //calculates expression tree
//or is the problem in this function?
printf("%d \n", result);
return 0;
}
I am not understanding clearly what you are doing and it would help my learning too if you shared from what source are you learning these (we can chat in chat room) but I also tried to do similar thing and in my style it works partially (that is, sometimes when I insert larger number the answer is wrong but for small numbers generally correct).So you might get some help form my style of this code that calculates given string (A little help for me also would be appreciated commenters!!). My code:
#include <stdio.h>
#include <string.h>
#include "stackforcalc.h"
int isOperand(char b){
if(b>='0' && b<='9'){
return 1;
}else{
return 0;
}
}
int isOperator(char b){
if(b=='+' || b=='-' || b=='*' || b=='/'){
return 1;
}
return 0;
}
int getwt(char b){
int g=-1;
switch (b)
{
case '+':
case '-':
g=1;
break;
case '/':
case '*':
g=28787868;
break;
}
return g;
}
int higherprecedence(char a,char b){
int c=getwt(a);
int d=getwt(b);
return (c>=d)?1:0;
}
int infToPost(char *b,char *str){
int j=0;
for(int i=0;i<strlen(b);i++){
if(b[i]== ' ' || b[i]== ',' ){
continue;
}
else if(isOperator(b[i])){
str[j]=' ';
j++;
while(!empty() && gettop() != '(' && higherprecedence(gettop(),b[i])){
str[j]=gettop();
j++;
pop();
}
push(b[i]);
}
else if(isOperand(b[i])){
str[j]=b[i];
j++;
}
else if(b[i]=='('){
push(b[i]);
}
else if(b[i] ==')'){
while(!empty() && gettop() != '('){
str[i]=gettop();
j++;
pop();
}
pop();
}
}
while(!empty()){
str[j]=gettop();
j++;
pop();
}
}
int Evaluate(int t,char y,int r){
int ty;
switch(y){
case '+':
ty=t+r;
break;
case '-':
ty=r-t; //I inverted these.
break;
case '*':
ty=r*t;
break;
case '/': //I inverted these because
ty=r/t; //even though I did t/r it performed r/t.
break; //may be somewhere before the numbers were swapped
default:
ty=-1;
break;
}
return ty;
}
int calculatepostfix(char *c){
for(int i=0;i<strlen(c);i++){
if(c[i]==' ' || c[i]==','){
continue;
}
else if(isOperator(c[i])){
int op1=gettop2();
pop2();
int op2=gettop2();
pop2();
int oper=Evaluate(op1,c[i],op2);
push2(oper);
}
else if(isOperand(c[i])){
int res=0;
while(i<strlen(c) && isOperand(c[i])){
res=(res*10)+(c[i]-'0');
i++;
}
i--;
push2(res);
}
}
return gettop2();
}
int main(){
char b[65];
printf("\n \n**-- Calculator --**\n");
printf("Enter expression: ");
fgets(b,sizeof(b),stdin);
char str[50];
infToPost(b,str);
int tt =calculatepostfix(str);
printf("Your answer is: %d",tt);
}
The code in "stackforcalc.h" is
#ifndef stacycalc
#define stacycalc
#define maxsize 50
char a[maxsize];
int top=-1;
int abc[maxsize];
int to=-1;
void push2(int re){ abc[++to]=re; }
void push(char b){ a[++top]=b; }
void pop2(){ to--; }
void pop(){ top--;}
int gettop2(){ return (to==-1)?-1:abc[to]; }
char gettop(){ return (top==-1)?0:a[top]; }
int empty(){ return (top==-1)?1:0; }
#endif
That is because the code is doing 52*53+50/51 (because it is using the ascii values) and not 4*5+2/3.
Yes.
I believe the reason for this is because I am storing 4*5+2/3 in a string of char and not in an array of int.
No.
In C, int is just a bigger char. There's nothing magical about them; they both hold numbers.
There are a variety of ways to derive the integer value from an ASCII character. If you have exactly one character and don't want to use a library, you can "mask off" the bits: since the ASCII range for digits is 0x30 - 0x39,
static char string[] = "4";
int value = string[0] & 0x0F;
does the trick. For more complex operations, my favorite is sscanf(3), but many use atoi(3) or various flavors of strtol(3).

Calculator with Decimal, Octal and Hexadecimal Numbers

I want to make a calculator that is capable of calculation with decimal numbers and is able to return the decimal values in their respective binary, octal or hexadecimal representation.
So far in the main method the program reads the command line and I can invoke the program by two ways.
The first way would be with 3 values:
"number1" "operator" "number2".
And the second way would be with 4 values:
"wished numeral system for the output" "number1" "operator" "number2".
Where for the wished numeral system output b would stand for for binary, o for octal and h for hexadecimal. In both ways the user should be able to input decimal, octal and hexadecimal numbers for the inputs number1 and number2.
#include "zahlen.h"
#include <stdio.h>
#include "stringTOint.h"
int main(int argc, char *argv[]) {
char o,op,sx[DIGITS+1],sy[DIGITS+1],sz[DIGITS+1];
int x,y,z;
char flag_x,flag_y;
/* 1) Read Commandline */
if (argc != 4 && argc != 5) {
printf("Aufruf: %s -o <x> <op> <y> \n",argv[0]);
return 1;
} else if(argc == 4) {
x = stringTOint(argv[1]);
op = argv[2][0];
y = stringTOint(argv[3]);
} else if(argc == 5) {
o = argv[1][0];
x = stringTOint(argv[2]);
op = argv[3][0];
y = stringTOint(argv[4]);
if(o != 'b' && o != 'o' && o != 'h') {
printf("Wrong Operation\n");
return 1;
}
}
/* 2) Solve the equation */
if(argc==4) {
printf("solve: %s %c %s \n", argv[1], op, argv[3]);
z = solve(x, op, y);
} else if(argc==5) {
printf("solve: %s %c %s \n", argv[2], op, argv[4]);
z = solve(x, op, y);
}
/* 3) Calculate the Representation of the wished Numeral System */
switch(o) {
case 'b':
intTObinaer(x, sx);
intTObinaer(y, sy);
intTObinaer(z, sz);
break;
case 'o':
intTOoctal(x,sx);
intTOoctal(y,sy);
intTOoctal(z,sz);
break;
case 'h':
intTOhexal(x,sx);
intTOhexal(y,sy);
intTOhexal(z,sz);
break;
default:
intTObinaer(x, sx);
intTObinaer(y, sy);
intTObinaer(z, sz);
break;
}
/* 4) Return the results */
printf("\n %s %d\n%c %s %d\n= %s %d\n", sx,x,op,sy,y,sz,z);
return 0;
}
The methods intTObinaer, intTOoctal and intTOhexal only differ by the base with which the decimal number gets divided.
intTObinaer(int i, char str[]) {
unsigned int zahl = i;
int j;
/* Fill Array with zeros */
int x = 0;
for (x; x < DIGITS+1; x++) {
str[x] = '0';
}
/*Calculate the Binary representation of the given Decimal integer */
for (j = DIGITS-1; j > 0; j--) {
/* This base gets changed to 8 or 16 for octal and hexal representation */
str[j] = (char) (zahl % 2) + '0';
zahl = zahl / 2;
if (zahl == 0) {
break;
}
}
/* Set the end of the Array */
str[DIGITS] = '\0';
}
The actual equation gets solved in the solve method, where the right operation for number1 and number2 gets chosen by an switchcase where the different cases can be selected by the char op that the user had input between the two numbers.
#include <stdio.h>
int solve(int x, char op, int y) {
int ergebnis = 0;
switch(op) {
case '+':
ergebnis = x + y;
break;
case '-':
ergebnis = x - y;
break;
case '*':
ergebnis = x * y;
break;
case '/':
ergebnis = x / y;
break;
case '&':
ergebnis = x & y;
break;
case '|':
ergebnis = x | y;
break;
default:
printf("Wrong input\n");
}
return ergebnis;
}
My question now is due to the fact the the user should be able to input different numeral systems(e.g. decimal, octal or hexadecimal) how can I identify the different numeral systems and then transfer them into decimal so that I can calculate the result. After that these decimal Numbers have to be converted back into the desired numeral system that the user wanted.
Looks like you only need to add two lines to do that:
#include "stdlib.h"
#define stringTOint(arg) ((int)strtol(arg,NULL,0))
Or better yet, replace those invocations of stringTOint() with corresponding strtol() invocations (and add the #include, of course).
strtol() uses the same prefixes as for C literals: 0 for octal, 0x for hex, no prefix is decimal.
I would like to suggest another approach to this problem.
Many of the parsing you perform can be performed directly by the sscanf function, the only case is the binary case that needs to be implemented differently.
The implementation follows 3 main step:
Parse the input using the sscanf function (or the ConvCharToBinfor binary values) and store the values in the variables a and b;
Perform the operation and store the result in the res variable;
Print the output result by using the printf parsing (or loop for the binary case).
An implementation would be the following:
#include<stdio.h>
#include<string.h>
typedef struct stack {
unsigned char data[32];
int size;
} stack_t;
int ConvCharToBin(char* input);
int main(int argc, char *argv[]) {
char numSys = 'd', op;
char** param = argv;
int a, b, res;
param++;
//error case
if(argc != 4 && argc != 5) {
//not a valid input
printf("Not a valid input");
return -1;
}
if(argc == 5) {
numSys = param[0][0];
param++;
}
op = param[1][0];
switch(numSys) {
case 'b':
a = ConvCharToBin(param[0]);
b = ConvCharToBin(param[2]);
break;
case 'd':
sscanf(param[0], "%d", &a);
sscanf(param[2], "%d", &b);
break;
case 'h':
sscanf(param[0], "%x", &a);
sscanf(param[2], "%x", &b);
break;
case 'o':
sscanf(param[0], "%o", &a);
sscanf(param[2], "%o", &b);
break;
default:
//no viable number system
return -1;
}
switch(op) {
case '+':
res = a + b;
break;
case '-':
res = a - b;
break;
case '/':
res = a / b;
break;
case '*':
res = a * b;
break;
case '&':
res = a & b;
break;
case '|':
res = a | b;
break;
default:
//no valid operand
printf("invalid operation\n");
return -1;
}
stack_t tmp;
tmp.size = 0;
int i;
switch(numSys) {
case 'b':
while (res) {
if (res & 1) {
tmp.data[tmp.size] = '1';
tmp.size++;
} else {
tmp.data[tmp.size] = '0';
tmp.size++;
}
res >>= 1;
}
for(i = tmp.size - 1; i >= 0; i--) {
printf("%c", tmp.data[i]);
}
printf("\n");
break;
case 'd':
printf("%d\n", res);
break;
case 'h':
printf("%x\n", res);
break;
case 'o':
printf("%o\n", res);
break;
}
return 0;
}
int ConvCharToBin(char* input) {
char* idx;
int res = 0x00000000;
for(idx = input; idx < input + strlen(input); idx++) {
res <<= 1;
if(*idx == '1') {
res |= 0x00000001;
}
}
return res;
}
The sscanf reads formatted data from a string (in you case the argv strings)
This can be parsed using the following:
%d for decimal;
%x for hexadecimal;
%o for octal.
Unfortunately there is no C standard for parsing binary using sscanf, so this is done apart using the stdout.
I would also point out that this implementation has two limitation
Input/output limited to 32 bit unsigned (so from 0 to 4294967295), but with some slight modifications it can be extended;
No error checking for the input values, this can also be easily implemented.

How do I evaluate decimals & floating point numbers in scientific e notation - e.g. {1.23e4}. in postfix?

I have written my code to evaluate from postfix to result. However, I am stuck at how to do it when the postfix is going to be in decimals & floating point numbers in scientific e notation - e.g. {1.23e4}. Any specific suggestion would be highly appreciated. Thanks.
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <stdlib.h>
#define SIZE 50 /* Size of Stack */
double s[SIZE];
int top=-1; /* Global declarations */
int flag=0;
double pop()
{ /* Function for POP operation */
return(s[top--]);
}
double push(double elem)
{ /* Function for PUSH operation */
if(flag==1){
int num;
num=pop();
s[++top]=elem+10*num;
}
else if(flag==0){
s[++top]=elem;
flag=1;
}
}
void main()
{ /* Main Program */
char pofx[50],ch;
int i=0;
double op1,op2;
printf("Enter the Postfix Expression:");
fgets(pofx,100,stdin);
while( (ch=pofx[i++]) != '\n')
{
if(isdigit(ch)) push(ch-'0'); /* Push the operand */
else if(ch==' ')
flag=0;
else
{ /* Operator,pop two operands */
flag=0;
op2=pop();
op1=pop();
switch(ch)
{
case '+':push(op1+op2);break;
case '-':push(op1-op2);break;
case '*':push(op1*op2);break;
case '/':push(op1/op2);break;
case '^':push(pow(op1,op2));break;
default:
printf("Input invalid ... give proper input\n");
return 0;
}
}
}
printf("Result: %lf\n",s[top]);
}
How do I evaluate decimals & floating point numbers in scientific e notation ... (?)
To convert a string into FP value use strtod() #M Oehm
Yet code has other problems in that the operator symbols '-' and '+' may also begin valid value tokens like -123.45.
// insufficient test to determine if the next part of the string is a number or operator.
if(isdigit(ch))
push(ch-'0');
Use strtod() to convert text to double and determine if the next part of the string is a double.
Alternative code:
const char *st = pofx;
while (*st) {
char *end; //location to store end of FP parsing
double value = strtod(st, &end);
if (end > st) {
push(value);
st = end;
} else if (isspace((unsigned char) *st)) {
st++;
} else {
switch (*st) {
case '+':push(pop() + pop());break; // pop order irrelevant
case '-':{ double t = pop(); push(pop() - t);break; } // pop order relevant
case '*':push(pop() * pop());break;
...
default: {
printf("Input invalid operator: character code %d\n", *st);
return 0;
}
} // end switch
st++;
}
}
Re-write push()
void push(double elem) {
if (top + 1 >= SIZE) {
printf("Stack overflow\n");
return;
}
s[++top] = elem;
}
Wrong argument to fgets()
char pofx[50];
// fgets(pofx,100,stdin); // 100??
fgets(pofx, sizeof pofx, stdin); // better
The function strtod from <stdlib.h> will parse a double for you. It takes the string to parse and a pointer to a string, that will begin with the first unparsed character. (There's a similar function for long integers, strtol.)
Here's an example of how this might work in your case. (The code just prints out the tokens and doesn't do any calculations.)
#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>
int is_number(const char *p)
{
if (*p == '-' || *p == '+') p++;
if (*p == '.') p++;
return isdigit((unsigned char) *p);
}
int main(void)
{
const char *line = " .1 20.1* 1.0e-3+2e+7 * -1.0*";
const char *p = line;
while (isspace((unsigned char) *p)) p++;
while (*p) {
if (is_number(p, after_op)) {
char *end;
double x = strtod(p, &end);
if (p == end) {
puts("Illegal number");
p++;
} else {
printf("Number %g\n", x);
p = end;
}
} else {
int c = *p++;
switch (c) {
case '+': puts("Operator add"); break;
case '-': puts("Operator sub"); break;
case '*': puts("Operator mult"); break;
case '/': puts("Operator div"); break;
case '^': puts("Operator pow"); break;
default: printf("Illegal char '%c'\n", c);
}
}
while (isspace((unsigned char) *p)) p++;
}
return 0;
}
This is still very crude, mind you. Strings like 20x21 will be parsed as number 20, unknown character x and number 21. While this code is bad at detecting and reporting errors (which really should be done, but it's left as an exercise, yadda, yadda), it works for valid input.
[Edit: I've incorporated chux's suggestions and also made the code compatible to read numbers with explicit signs, but that means that the binary operators - and + cannot immediately be followed by a digit. Pick your poison. I've also allowed the abridged version that leaves out a leading zero, e.g. .12, a format that I'm not very fond of.]

C Calculator With getchar/putchar

First time ever C programming class. I have to create a C calculator using putchar/getchar. The code below if what I have to far and loops to ask the user for input. The issue I am having is how to account for space/multiple spaces before input, space/multiple spaces between the digits and operand, no spaces between operand and digits, and spaces after input by the user is complete.
Right now the code will work with one space between the digits and operand.
Meaning, 10 + 5 works. However, for example, 10+5 does not and ____5 + 10 (where __ = space), or 10+5______, or 10_________+ 10 do not work.
Any advice or help on how to account for multiple spaces in between the digits and operand and before after any user input is so very greatly appreciated.
Thank you so very much for any and all help with the current code. Really do appreciate your help and time!!
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <math.h>
int add(int input1, char operand, int input2);
int subtract(int input1, char operand, int input2);
int mod(int input1, char operand, int input2);
int multiply(int input1, char operand, int input2);
int divide(int input1, char operand, int input2);
char cont(void);
int main()
{
int answer = 0;
int ch = 0;
int input1 = 0;
char operand = 0;
int input2 = 0;
int function = 0;
char flag;
do {
input1 = 0, input2 = 0, operand = 0;
printf("\nPlease enter a calculation to be made.\n");
while (((ch = getchar()) != ' ') && (ch != EOF) && (ch != '\n')){
if (ch == '-') {
printf("\nError: no negatives allowed.\n");
}
else if (!isdigit(ch)){
printf("\nError: number not inputted (first number).\n");
}
else {
input1 = (input1 * 10) + (ch - '0');
}
}
while (((ch = getchar()) != ' ') && (ch != EOF) && (ch != '\n')){
switch (ch){
case '+':
operand = '+';
break;
case '-':
operand = '-';
break;
case '%':
operand = '%';
break;
case '*':
operand = '*';
break;
case '/':
operand = '/';
break;
default:
printf("Error: input is not one of the allowed operands.");
break;
}
}
while (((ch = getchar()) != ' ') && (ch != '\n')){
if (ch == '-') {
printf("\nError: no negatives allowed.\n");
}
else if (!isdigit(ch)){
printf("\nError: number not inputted (second number).\n");
}
else {
input2 = (input2 * 10) + (ch - '0');
}
}
printf("%d", input1);
putchar(' ');
printf("%c", operand);
putchar(' ');
printf("%d", input2);
putchar(' ');
putchar('=');
putchar(' ');
if (operand == '+'){
answer = add(input1, operand, input2);
printf("%d", answer);
}
else if (operand == '-'){
answer = subtract(input1, operand, input2);
printf("%d", answer);
}
else if (operand == '%'){
answer = mod(input1, operand, input2);
printf("%d", answer);
}
else if (operand == '*'){
answer = multiply(input1, operand, input2);
printf("%d", answer);
}
else if (operand == '/'){
answer = divide(input1, operand, input2);
printf("%d", answer);
}
flag = cont();
}
while (flag == 'y' || flag == 'Y');
return 0;
}
int add(int input1, char operand, int input2){
return input1 + input2;
}
int subtract(int input1, char operand, int input2){
return input1 - input2;
}
int mod(int input1, char operand, int input2){
return input1 % input2;
}
int multiply(int input1, char operand, int input2){
return input1 * input2;
}
int divide(int input1, char operand, int input2){
return input1 / input2;
}
char cont()
{
char flag;
printf("\nDo you want to process another calculation (y/n)? ");
scanf("%c%*c", &flag);
return flag;
}
Ok! the solution to this problem involves understanding how getchar() works. It also includes a concept of how input from standard input [stdin] works.
So, whenever input to variables is taken using scanf()/getchar() etc. either operations are required in order to alert the compiler that the input for the variable is confirmed. One is new line \n(Pressing Enter) and the other is white-space. They will alert that the input is finalized.
This is explained by input 10 + 5 [Press Enter]. The first space ensures ch has values 10 and thus input1 has a value 10. Then + come through and after the space the operand gets the value of +. Later, Enter is pressed after writing 5. This will save value 5 in input2. The rest of the operations happen perfectly.
In other cases where we have so many spaces ____5 + 10 / 10+5______ / 10_________+ 10 we must look at how the code will follow on such input.
_____5+10 The first space will miss the first while loop because space is an exit condition. The control then goes to the second while loop, where second space will also be an exit condition for the loop. The same happens to the third while loop and it also exits. Thus, the answer will be input1 operand input2 and all these have default values i.e. 0, 0 (which is blank when scaled down to character) and 0 respectively.
10+5_____ Here the first space appear after 5, thus, the getchar() will feed input1 with values 10+5and other two variables don't even get assigned.
This shows why the variables couldn't get assigned the values we expected. The solution for this problem can be solved by extra processing but performing checks on how spaces appear.

Reverse Polish Calculator in C : isdigit() issues

I'm doing a homework assignment in C. I have to build a calculator that takes in RPN, converts it to a double, adds / removes it from stack and prints what's left on the stack.
Everything seems to be going well until I run it. Upon entering my input, the char[] gets successfully converted to double (or so i think) and somewhere along the way (maybe in getop()) does the program think that i'm not entering digits. Here is the code and the output.
#include <stdio.h>
#define MAXOP 100 /* maximum size of the operand of operator */
#define NUMBER '0' /* signal that a number was found */
int getOp(char []); /* takes a character string as an input */
void push(double);
double pop(void);
double asciiToFloat(char []);
/* reverse polish calculator */
int main()
{
int type;
double op2;
char s[MAXOP];
printf("Please enter Polish Notation or ^d to quit.\n");
while ((type = getOp(s)) != EOF) {
switch (type) {
case NUMBER:
push(asciiToFloat(s));
break;
case '+':
push(pop() + pop());
break;
case '*':
push(pop() * pop());
break;
case '-':
op2 = pop();
push(pop() - op2);
break;
case '/':
op2 = pop();
if (op2 != 0.0)
push(pop() / op2);
else
printf("error : zero divisor!\n");
break;
case '\n':
printf("\t%.2f\n", pop()); /* this will print the last item on the stack */
printf("Please enter Polish Notation or ^d to quit.\n"); /* re-prompt the user for further calculation */
break;
default:
printf("error: unknown command %s.\n", s);
break;
}
}
return 0;
}
#include <ctype.h>
/* asciiToFloat: this will take ASCII input and convert it to a double */
double asciiToFloat(char s[])
{
double val;
double power;
int i;
int sign;
for (i = 0; isspace(s[i]); i++) /* gets rid of any whitespace */
;
sign = (s[i] == '-') ? -1 : 1; /* sets the sign of the input */
if (s[i] == '+' || s[i] == '-') /* excludes operands, plus and minus */
i++;
for (val = 0.0; isdigit(s[i]); i++)
val = 10.0 * val + (s[i] - '0');
if (s[i] = '.')
i++;
for (power = 1.0; isdigit(s[i]); i++) {
val = 10.0 * val + (s[i] - '0');
power *= 10.0;
}
return sign * val / power;
}
#define MAXVAL 100 /* maximum depth of value stack */
int sp = 0; /* next free stack position */
double val[MAXVAL]; /* value stack */
/* push: push f onto value stack */
void push(double f)
{
if (sp < MAXVAL) {
val[sp++] = f; /* take the input from the user and add it to the stack */
printf("The value of the stack position is %d\n", sp);
}
else
printf("error: stack full, cant push %g\n", f);
}
/* pop: pop and return the top value from the stack */
double pop(void)
{
if (sp > 0)
return val[--sp];
else {
printf("error: stack empty\n");
return 0.0;
}
}
#include <ctype.h>
int getch(void);
void ungetch(int);
/* getOp: get next operator or numeric operand */
int getOp(char s[])
{
int i;
int c;
while ((s[0] = c = getch()) == ' ' || c == '\t')
;
s[1] = '\0';
if (!isdigit(c) && c != '.') {
printf("Neither a digit or a decimal.\n");
return c; /* neither a digit nor a decimal */
}
i = 0;
if (isdigit(c)) /* grab the integer */
while (isdigit(s[++i] = c = getch()))
;
if (c == '.') /* grab the fraction */
while (isdigit(s[++i] = c = getch()))
;
s[i] = '\0';
if (c != EOF)
ungetch(c);
return NUMBER;
}
#define BUFSIZE 100
char buf[BUFSIZE]; /* buffer for ungetch */
int bufp = 0; /* next free position in buffer */
/* getch: get a number that may or may not have been pushed back */
int getch(void)
{
return (bufp > 0) ? buf[--bufp] : getchar();
}
/* ungetch: if we read past the number, we can push it back onto input buffer */
void ungetch(int c)
{
if (bufp >= BUFSIZE)
printf("ungetch: to many characters.\n");
else
buf[bufp++] = c;
}
Output :
Please enter Polish Notation or ^d to quit.
123
The value of the stack position is 1
Neither a digit or a decimal.
123.00
Please enter Polish Notation or ^d to quit.
Any thoughts regarding what's going on will be super helpful. It seems like then number is properly getting passed in, properly formatted from char to double, and then something goes wrong.
Thank you.
Rock
Change
printf("Neither a digit or a decimal.\n");
to
printf("Neither a digit or a decimal: %d 0x%x.\n", c, c);
so you can see what is causing the message.
Your output shows that getch() is returning the line feed ("\n", 0x0A) at the end of the line. Also, unless you were required to write asciiToFloat() for your homework assignment, you should use atof() or strtod() (both declared in "stdlib.h") from the Standard C Library. They are (usually?) implemented to avoid loss of precision and accuracy during the conversion; repeatedly multiplying by 10 causes a loss of same. Nice looking code, otherwise! :)

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