CODEWITHSUNDEEP
c++stringfunctionbinarysyntax-error
Axel Kennedal
Axel Kennedal

Reputation: 555

Super weird error when handling strings

I'm trying to create a function that converts a binary number (string) into a decimal one (int). The weird part about the code below is that when the line "//cout << index << endl;" is not commented out, it works! WHY D:?

Output when commented out:

1651929379

Output when active:

7 192 Program ended with exit code: 0

Here's the entire program:

//
//  testish.cpp
//  Egetskojs
//
//  Created by Axel Kennedal on 2014-02-13.
//  Copyright (c) 2014 Axel Kennedal. All rights reserved.
//

#include <iostream>
#include <string>
#include <cmath>
using namespace std;

int BinaryToDecimal(string & binaryString);

int main(){

    string binary = "11000000";
    int decimal = BinaryToDecimal(binary);
    cout << decimal << endl;




    return 0;
}


int BinaryToDecimal(string & binaryString){
    int solution;

    if (binaryString == "0") solution = 0;
    if (binaryString == "1") solution = 1;

    int index = binaryString.length() - 1; //The index of the last (rightmost) bit in the string
    //cout << index << endl;

    int currentBit = 0; //The exponent to be used when calculating the value of a bit

    for (; index >= 0; index--) {
        if (binaryString.at(index) == '1') {
            solution += pow(2, currentBit);
        }
        //Else: nothing happens
        currentBit++;
    }

    //Done!
    return solution;
}

Upvotes: 1

Views: 63

Answers (2)

Martin J.
Martin J.

Reputation: 5108

As Joachim said, your solution variable is uninitialized, so when the string is neither "0" nor "1", you may get strange behavior (e.g. integer overflows) in your += operation. I guess that the fact that it works when the output is active is due to some weird side-effect of the output instructions causing some register to contain 0, and that register being the source of the value for solution. It might be enlightening to know what your compiler settings are for this, and to look at the assembler code for this part of your code.
You could replace:

int BinaryToDecimal(string & binaryString){
    int solution;

    if (binaryString == "0") solution = 0;
    if (binaryString == "1") solution = 1;
    ...

with:

int BinaryToDecimal(string & binaryString){
    int solution = 0;
    ...

Since the special case handling you do is elegantly handled by your loop after.

Upvotes: 0

Some programmer dude
Some programmer dude

Reputation: 409136

You have undefined behavior in the BinaryToDecimal, because the variable solution may be used uninitialized.

Uninitialized local variables will have indeterminate values (i.e. their values will be seemingly random).

Upvotes: 4

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