std::out_of_range error accessing a std::string in C++

Question

I am part of a team of three moderate-aptitude programmers, and we are currently attempting to write a program in C++ to embody our new encryption algorithm. However, we have encountered an error which indicates that we are surpassing the length of a string, at some point. I have tried debugging this, myself, but I have had no luck. It's in the jumble() function, though ...

Anyway, here's a paste of our sourcecode with a temporary main() function: http://pastebin.com/GvvYAsKg

It compiles fine, but upon running, we get the following error:

terminate called after throwing an instance of 'std::out_of_range'
what(): basic_string::substr Aborted

Would you please re-indent the code properly and paste it over here? — nhahtdh
– nhahtdh, Commented May 3, 2013 at 18:46
As almost an aside: what makes you think that writing your own encryption function is a good idea? — Philip Kendall
– Philip Kendall, Commented May 3, 2013 at 18:50
Philip, it's for an engineering class project. Everybody is supposed to design something. Our team decided to do an encryption algorithm. — smalltock
– smalltock, Commented May 3, 2013 at 18:57
n.m., I should have said moderately-skilled. We aren't the most experienced programmers, but we're not completely novice, either. — smalltock
– smalltock, Commented May 3, 2013 at 18:58

Philip Kendall · Accepted Answer · 2013-05-03 18:47:38Z

7

One of your problems is in this line:

for(int i = 0; i < ( inPT.length() + 1 ); i++)

This will mean you attempt to access inPT[inPT.length] which is one character beyond the end of the string. You want

for(int i = 0; i < inPT.length(); i++)

or possibly to be using a more idiomatic C++ construct like std::for_each so you can't make this kind of fencepost error. As noted in the comments, running this in a debugger would have pointed this out to you pretty quickly.

answered May 3, 2013 at 18:47

Philip Kendall

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2 Comments

Some programmer dude Over a year ago

Or the new C++11 range-base for loop: for (char c : inPT) { ... }

smalltock Over a year ago

This was also causing the problem. Both your answer and John's were correct! Many thanks!

john · Accepted Answer · 2013-05-03 18:48:26Z

7

Presumably this piece of code

if(modCount >= inPT.length())
{
    modCount = 0;
}
int mod = inKey.at(modCount);

is meant to read

if(modCount >= inKey.length())
{
    modCount = 0;
}
int mod = inKey.at(modCount);

Your guard on an out of range access on inKey is using the wrong variable.

answered May 3, 2013 at 18:48

john

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