This code (in C) only outputs 0.00000, change the code without changing the general structure of the code.

#include

float s_exp(float sub_exp, char op);

char get_op();

float get_num();

float m_exp(float sub_exp, char op);

int main(void)

{

char cont = 'Y';

while (cont == 'Y' || cont == 'y')

{

printf("Enter a simple arithmetic expression: ");

float result = s_exp(get_num(),get_op());

printf(" The result is %f ", result);

printf("Do you want to continue? Y/N ");

scanf("%c",&cont);

//Enter Y or y to continue, otherwise N or n to quit

}

}

//returns the value of the expression

float s_exp(float sub_exp, char op)

{

if (op == ' ')

return sub_exp;

else if (op == '+')

return s_exp(sub_exp + get_num(), get_op());

else if (op == '-')

return s_exp(sub_exp - get_num(), get_op());

else if (op == '*')

return s_exp(sub_exp * get_num(), get_op());

else if (op == '/')

return s_exp(sub_exp / get_num(), get_op());

else

{

printf("Invalid operator. ");

return 0;

}

}

float m_exp(float sub_exp, char op) {

if (op == ' ')

return sub_exp;

else if (op == '*')

return s_exp(sub_exp * get_num(), get_op());

else if (op == '/')

return s_exp(sub_exp / get_num(), get_op());

else

{

printf("Invalid operator. ");

return 0;

}

}

// return the next operator of the expression.

char get_op()

{

char op = ' ';

while (op == ' ')

scanf("%c", &op);

return op;

}

//return the next numeric value of the expression.

float get_num()

{

float num;

scanf("%f", &num);

return num;

}