Evaluation of four operation expressions

/*
 Four expressions are evaluated 
 Ideas , The recursive method , There are two cases of an expression ,1. There is only one item  2. The two terms add and subtract from each other 
 There are two cases of an item  1. There is only one factor  2. The two factors multiply and divide each other 
 There are two cases of a factor  1. There's only one number , At this point, recursion terminates , Returns the number return atof() 
2.( + exp() + ) Return to the beginning and recurse until there is only one number  
 Be careful , The naming is not standard , Debug two lines of tears ！
*/

#include<iostream>
#include<cstdlib>
#include<cstdio>
using namespace std;

int term();
int exp();
int factor();

int exp()
{
	int value = term();
	while(true){
		char c = cin.peek();
		if(c == '+' || c == '-'){
			cin.get();
			int val = term();
			if(c == '+') value += val;
			else value -= val;
		}
		else break;
	}
	return value;
}

int term()
{
	int value = factor();
	while(true){
		char c = cin.peek();
		if(c == '*' || c == '/'){
			cin.get();
			int val = factor();
			if(c == '*') value *= val;
			else value /= val;
		}
		else break;
	}
	return value;
}

int factor()
{
	int value = 0;
	int res = cin.peek();
	if(res == '('){
		cin.get();	//remove (
		value = exp();
		cin.get();	//remove )
	}
	else{
		while(isdigit(res)){
			value = 10 * value + res - '0';
			cin.get();
			res = cin.peek();
		}
	}
	return value;
}

int main()
{
	printf("%d",exp());	// Why change it to %f became 0, and %d nevertheless 63, Because fixed-point numbers and floating-point numbers 
	// Are your computer codes different ？ 
	return 0;
}

/*
 Input ：(2+3)*(5+7)+9/3  Output ： 63
*/