C - Recursion instead of loops - c

Question

I made a program that evaluates expressions, for example
Enter text: 97+74/51-98-11+68-34-2-22+73/40+81/15+100
output :
105.36
If any wrong data is given a user gets a response Incorrect input
I am wondering if I could replace all my for loops to recursion, but I don't know how could I achieve that. Could anybody help ?
Any assistance would be much appreciated.
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<math.h>
int my_atoi(const char* tab, int from, int to)
{
int score = 0;
int minus = 0;
for (int i = from; i <= to; i++)
{
if (*(tab + i) == 10)
break;
if (*(tab + i) == 32)
break;
if (*(tab + i) == '-' && from == i)
{
minus = 1;
continue;
}
if (*(tab + i) == '+' && i == from)
{
continue;
}
if (*(tab + i) > '9' || *(tab + i) < '0')
break;
else if (*(tab + i) >= '0' && *(tab + i) <= '9')
{
score *= 10;
score += (*(tab + i) - '0');
}
}
if (minus == 1)
score *= -1;
return score;
}
int validate_expression(const char* expr)
{
if (!expr)return -1;
int lng = strlen(expr) - 1;
if (*(expr) > '9' || *(expr) < '0' || * (expr + lng) > '9' || *(expr + lng) < '0')return 0;
int flaga = 0; // flag
for (int i = 1; i < lng; i++) {
if (*(expr + i) <= '9' && *(expr + i) >= '0')
{
flaga = 0;
}
else if (*(expr + i) == '/' || *(expr + i) == '*' || *(expr + i) == '-' || *(expr + i) == '+')
{
if (flaga == 1)
return 0;
flaga = 1;
}
else
return 0;
}
return 1;
}
int calculate(const char* expr, float* result)
{
if (!expr || !result || !validate_expression(expr)) return 0;
if (strpbrk(expr, "i") != NULL) return 0;
int index = 0;
float score = 0;
char sign_memory = 'A';
int flaga = 0;
unsigned int i = 1;
for (; i <= strlen(expr); i++) {
if (*(expr + i) == '-' || *(expr + i) == '+' || *(expr + i) == '/' || *(expr + i) == '*' || *(expr + i) == '\0') {
score += (int)my_atoi(expr, index, i - 1);
index = i;
break;
}
}
for (; i <= strlen(expr); i++) {
if (*(expr + i) == '-' || *(expr + i) == '+' || *(expr + i) == '/' || *(expr + i) == '*' || i == strlen(expr)) {
flaga = 1;
if (sign_memory == '+')
score += (int)my_atoi(expr, index, i - 1);
else if (sign_memory == '-')
score -= (int)my_atoi(expr, index, i - 1);
else if (sign_memory == '/')
{
if (my_atoi(expr, index, i - 1) == 0)return 0;
score /= (int)my_atoi(expr, index, i - 1);
}
else if (sign_memory == '*')
score *= (int)my_atoi(expr, index, i - 1);
sign_memory = *(expr + i);
}
else if (*(expr + i) <= '9' && *(expr + i) >= '0' && flaga == 1) {
flaga = 0;
index = i;
}
}
*result = score;
return 1;
}
int main()
{
char tab[201];
float result, * pointer = &result;
printf("Enter text: ");
fgets(tab, sizeof(tab), stdin);
*(tab + strlen(tab) - 1) = '\0';
if (!validate_expression(tab))
{
printf("Incorrect input");
return 1;
}
else
{
if (calculate(tab, pointer))
printf("%.2f", *pointer);
else {
printf("Incorrect input");
return 1;
}
}
return 0;
}

Answer 1

I would make a function, which does the following:
Check if there is just one or more operators ("+", "-", "*", "/")
In case there is only one, perform the calculation.
In case there are two, perform a tail recursion, as proposed by Jonathan, something like:
calculate("97+74/51-98-11+68-34-2-22+73/40+81/15+100")
= calculate("97+74/51-98-11+68-34-2-22+73/40+81/15") + 100
= (calculate("97+74/51-98-11+68-34-2-22+73/40+81") / 15) + 100
= ((calculate("97+74/51-98-11+68-34-2-22+73/40") + 81) / 15) + 100
= (((calculate("97+74/51-98-11+68-34-2-22+73") / 40) + 81) / 15) + 100
= ((((calculate("97+74/51-98-11+68-34-2-22") + 73) / 40) + 81) / 15) + 100
= (((((calculate("97+74/51-98-11+68-34-2") - 22) + 73) / 40) + 81) / 15) + 100
= ((((((calculate("97+74/51-98-11+68-34") - 2) - 22) + 73) / 40) + 81) / 15) + 100
= (((((((calculate("97+74/51-98-11+68") - 34) - 2) - 22) + 73) / 40) + 81) / 15) + 100
= ((((((((calculate("97+74/51-98-11") + 68) - 34) - 2) - 22) + 73) / 40) + 81) / 15) + 100
= (((((((((calculate("97+74/51-98") - 11) + 68) - 34) - 2) - 22) + 73) / 40) + 81) / 15) + 100
= ((((((((((calculate("97+74/51") - 98) - 11) + 68) - 34) - 2) - 22) + 73) / 40) + 81) / 15) + 100
= (((((((((((calculate("97+74") / 51) - 98) - 11) + 68) - 34) - 2) - 22) + 73) / 40) + 81) / 15) + 100
= ((((((((((((97 + 74) / 51) - 98) - 11) + 68) - 34) - 2) - 22) + 73) / 40) + 81) / 15) + 100