Member Avatar for egolovin

so i have spent 3 hours trying to figure out how exponents work in C++ but now. my if statement isnt working. even if i punch in a 0 it spits out 1
thanks in advance

#include<iostream>
#include<string>
#include<cmath>
using namespace std;
void binToDec(string getBinary);
void main()
{
	string getBinary;
	cout<<"Enter a Binary Number:"<<endl;
	cin>>getBinary;
	binToDec(getBinary);
	

}
void binToDec(string getBinary)
{
	int setLength,total=0;
	double placeholder=0;
	setLength=getBinary.size();
	
	for(int i=setLength;i>0;i--)
	{
		int num=pow(2.0,placeholder);
		int test=getBinary[i];
		if (test=1)
		{
			total=total+num;
		}

		placeholder=placeholder +1;
	cout<<total<<endl;
	}
	


}
  • 5 Contributors
  • 8 Replies
  • 182 Views
  • 1 Day Discussion Span
  • Latest Post Latest Post by grumpier

Dani AI

Generated

Good troubleshooting already in this thread: and flagged the = vs == and main() return-type issues, and correctly pointed out the char vs int subtlety. Additional problems that commonly break string-to-binary conversions are an off-by-one index when iterating the string, using pow (which is floating-point) for exact integer powers of two, printing the running total inside the loop, and not validating that each character is 0 or 1. A simple, robust pattern avoids pow altogether by shifting/multiplying the accumulator.

A compact, safe manual conversion (no floating math, correct indexing, input validation):

long long binaryToDec(const std::string &s) {
    long long total = 0;
    for (char c : s) {
        if (c != '0' && c != '1') throw std::invalid_argument("non-binary digit");
        total = (total << 1) | (c - '0');   // multiply by 2 then add bit
    }
    return total;
}

For modern compilers a one-liner is available:

long long total = std::stoll(s, nullptr, 2); // C++11+, throws on bad input or overflow

std::stoll is convenient but beware of exceptions and range limits; strtoll(s.c_str(), nullptr, 2) is the C equivalent. Also consider std::bitset<N>(s).to_ullong() when the width is known.

Quick troubleshooting checklist: start string iteration at index 0 or at size()-1 if working right-to-left (never use size() as an index), compare characters to '1' (or convert with c - '0'), print the final total after the loop, and have int main() return 0. For inputs longer than the target integer width, plan for overflow or use a big-integer library.

Recommended Answers

All 8 Replies

Member Avatar for grumpier

1) Comparison of test with 1 has two equals signs, not 1 if (test == 1) . Having one equals sign assigns test to 1.

2) main() returns int, not void. Wash your mouth out with soap for returning void.

Member Avatar for dexter1984

Hmmm shouldn't
if (test=1)
be
if (test == 1)

Anyway I read that it isn't good to have void main().

Member Avatar for egolovin

haha im just making small programs for class thanks for the help. stupid mistake

Member Avatar for Sci@phy

It still won't work. You are comparing char with int:
getBinary is char, and 1 is int.
char != int!!!

Member Avatar for grumpier

It still won't work. You are comparing char with int:
getBinary is char, and 1 is int.
char != int!!!

It will work. Comparing char with int is allowed.

Worst case is a compiler warning (eg mismatch of precision of operands). A compiler warning means the programmer should check and make sure the code has the intended result, not that it is disallowed.

Member Avatar for Sci@phy

It will work. Comparing char with int is allowed.

Worst case is a compiler warning (eg mismatch of precision of operands). A compiler warning means the programmer should check and make sure the code has the intended result, not that it is disallowed.

Yes, but does he want to compare test with '1' or with 1? Because that's different thing!

Member Avatar for grumpier

Yes, but does he want to compare test with '1' or with 1? Because that's different thing!

Ah, yes. Good point. Something for the original poster to contemplate .....

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