How to use char array index's permutations as operator? [duplicate] - c++

Question

This question already has answers here:
How to use an array index as an operator?
(4 answers)
Closed 6 years ago.
please look at my c++ program.
#include<stdio.h>
int main()
{
char sign[]={'+','-','/','*'};
int i,j,k,l=0;
for(i=0;i<=3;i++)
{
for(j=0;j<=3;j++)
{
for(k=0;k<=3;k++)
{
if(sign[j]!=sign[i]&&sign[k]!=sign[i]&&sign[k]!=sign[j])
{
printf("%c %c %c\n",sign[i],sign[j],sign[k],l);
}
}
}
}
return 0;
}
the output of this program is like this..
+ - /
+ - *
+ / -
+ / *
....
I want to use them between numbers...
like this..
#include<stdio.h>
int main()
{
char sign[]={'+','-','/','*'};
int i,j,k,l=0;
for(i=0;i<=3;i++)
{
for(j=0;j<=3;j++)
{
for(k=0;k<=3;k++)
{
if(sign[j]!=sign[i]&&sign[k]!=sign[i]&&sign[k]!=sign[j])
{
int l;
l=18sign[i]12sign[j]4sign[k]5;
printf("18%c12%c4%c5=%d",sign[i],sign[j],sign[k],l);
}
}
}
}
return 0;
}
I want the output like this..
18+12-4/5=29
18+12-4*5=10
18+12/4-5=16
18+12/4*5=33
.....
then what will be the code for this??
[I can't use switch tag because for that I will have to declare 24 cases. Is there any way to use the indexes as operator??]

Answer 1

No.
C (and C++) are static compiled languages. You can't make decisions at run-time that require different code to be present, since the compiler is no longer around to generate that code.
In C++ you can do various template tricks to make the compiler generate code for various cases at compile-time, but it's not quite clear how that would apply, here.

Answer 2

You have to use an expression parser in order to handle the correct operator precedence. Look at Infix Calculator Expression Parser for a possible solution.

Answer 3

Following may help:
int main()
{
int numbers[] = {18, 12, 4, 5};
char signs[] = {'+', '-', '/', '*'};
std::map<char, std::function<int(int, int)>> m = {
{'+', [](int a, int b){ return a + b; }},
{'-', [](int a, int b){ return a - b; }},
{'/', [](int a, int b){ return a / b; }},
{'*', [](int a, int b){ return a * b; }},
};
std::sort(std::begin(signs), std::end(signs));
do {
int res = m[signs[2]](m[signs[1]](m[signs[0]](numbers[0], numbers[1]), numbers[2]), numbers[3]);
std::cout << numbers[0] << signs[0] << numbers[1] << signs[1]
<< numbers[2] << signs[2] << numbers[3] << " = " << res << std::endl;
} while (std::next_permutation(std::begin(signs), std::end(signs)));
return 0;
}
Live demo

Answer 4

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct exp {
char op;
char *term;
struct exp *left;
struct exp *right;
} Exp;
Exp *make_exp2(char *str){
if(!str || !*str) return NULL;//*str == '\0' is format error.
char *mul = strrchr(str, '*');
char *div = strrchr(str, '/');
Exp *node = malloc(sizeof(*node));
if(mul == NULL && div == NULL){
node->op = '\0';
node->term = str;
node->left = node->right = NULL;
return node;
}
char *op;
op = mul < div ? div : mul;
node->op = *op;
*op = '\0';
node->left = make_exp2(str );
node->right = make_exp2(op+1);
return node;
}
Exp *make_exp(char *str){
if(!str || !*str) return NULL;//*str == '\0' is format error.
char *minus = strrchr(str, '-');
char *plus = strrchr(str, '+');
if(minus == NULL && plus == NULL)
return make_exp2(str);
char *op;
Exp *node = malloc(sizeof(*node));
op = minus < plus ? plus : minus;
node->op = *op;
*op = '\0';
node->left = make_exp(str );
node->right = make_exp(op+1);
return node;
}
int multiplication(int a, int b){
return a * b;
}
int division(int a, int b){
return a / b;
}
int addition(int a, int b){
return a + b;
}
int subtraction(int a, int b){
return a - b;
}
int calc(Exp *exp){
switch(exp->op){
case '*' :
return multiplication(calc(exp->left), calc(exp->right));
case '/' :
return division(calc(exp->left), calc(exp->right));
case '+' :
return addition(calc(exp->left), calc(exp->right));
case '-' :
return subtraction(calc(exp->left), calc(exp->right));
default :
return atoi(exp->term);
}
}
void drop_exp(Exp *exp){
if(exp){
drop_exp(exp->left);
drop_exp(exp->right);
free(exp);
}
}
int main(void) {
char expstr[128];
sprintf(expstr, "18%c12%c4%c5", '+', '-', '/');//18+12-4/5
Exp *exp = make_exp(expstr);
printf("%d\n", calc(exp));//30
drop_exp(exp);
return 0;
}