Use a specific value in an array in an if statement in C++

Question

if(gene1A[20] == 'T' || gene2A[20] == 'T')
    outFile << "Person A is of 'Anemic' type." << endl;
else if(gene1A[20] == 'T' && gene2A[20] == 'T')
    outFile << "Person A if of 'Carrier' type." << endl;
else
    outFile << "Person A is of 'Normal' type." << endl;

if(gene1B[20] == 'T' || gene2B[20] == 'T')
    outFile << "Person B is of 'Anemic' type." << endl;
else if(gene1B[20] == 'T' && gene2B[20] == 'T')
    outFile << "Person B if of 'Carrier' type." << endl;
else
    outFile << "Person B is of 'Normal' type." << endl;

if(gene1C[20] == 'T' || gene2C[20] == 'T')
    outFile << "Person C is of 'Anemic' type." << endl;
else if(gene1C[20] == 'T' && gene2C[20] == 'T')
    outFile << "Person C if of 'Carrier' type." << endl;
else
    outFile << "Person C is of 'Normal' type." << endl;

if(gene1D[20] == 'T' || gene2D[20] == 'T')
    outFile << "Person D is of 'Anemic' type." << endl;
else if(gene1A[20] == 'T' && gene2A[20] == 'T')
    outFile << "Person D if of 'Carrier' type." << endl;
else
    outFile << "Person D is of 'Normal' type." << endl;

is my code as of now. What I need to do is output the the 'outFile' if the Person is Anemic, a Carrier, or normal, based on the array I have set up. Each array is 444 characters long and is either an A, C, T, or O. If a T is in the 20th position of gene1[] and/or gene2[] then the person would be Anemic (if only one array) or a carrier (if in both arrays).

What I have now makes them automatically be "normal". I believe my if statements aren't set up properly, but what I need is to reference the 20th value in an array and then if it == 'T', output their "type".

Note: I noticed in my code i put 20 instead of 19. I made that correction so just look past that.

Thanks guys!

Why not just fix the code to say 19 instead of putting a disclaimer? It would take about as long to do this. — Marcelo Cantos
– Marcelo Cantos, Commented Nov 27, 2012 at 2:30
@ Marcelo: I was about to post, then website told me to indent, then that is when I noticed the mistake was while I was indenting the code to post. @irrelephant: Done and done. — Tanner Ross
– Tanner Ross, Commented Nov 27, 2012 at 2:45

Marcelo Cantos · Accepted Answer · 2012-11-28 21:47:35Z

1

(This isn't quite an answer, but it's difficult to express as a comment, and the resulting simplification might lead you towards an answer anyway...)

Functional decomposition is your friend:

const char* type(const char* gene1, const char* gene2) {
    return gene1[19] != 'T' ? "Normal" : gene2[19] == 'T' ? "Anemic" : "Carrier";
}
⋮
outFile << "Person A is of '" << type(gene1A, gene2A) << "' type." << endl;
outFile << "Person B is of '" << type(gene1B, gene2B) << "' type." << endl;
outFile << "Person C is of '" << type(gene1C, gene2C) << "' type." << endl;
outFile << "Person D is of '" << type(gene1D, gene2D) << "' type." << endl;

It also makes bugs like the one you introduced for person D much harder to introduce and easier to spot when you do.

EDIT: @MarkB pointed out an error in my logic (I misread the original logic). Unfortunately, I'm not sure how to fix it, since the original logic is of the form:

     if A or  B then X
else if A and B then Y
else                 Z

Since (A or B) is true whenever (A and B) is true, the second clause will never trigger, which almost certainly wasn't your intent. If you meant to have the AND clause first, then the type() function could be rewritten thus:

const char* type(const char* gene1, const char* gene2) {
    bool t1 = gene1[19] == 'T';
    bool t2 = gene2[19] == 'T';
    return t1 && t2 ? "Anemic" : t1 || t2 ? "Carrier" : "Normal"  );
}

Incidentally, this function wouldn't be a "sub-function" (whatever that means) of the current code, it would simply be a free function declared above the function. OTOH, if your compiler supports C++11 lambdas, you can in fact declare the type() function locally to the function in question:

auto type = [](const char* gene1, const char* gene2) -> const char * {
    …
};

edited Nov 28, 2012 at 21:47

answered Nov 27, 2012 at 2:36

Marcelo Cantos

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3 Comments

Tanner Ross Over a year ago

....Is that a sub-function of some sort Marcelo? I am just about finished my CS 110 class so my understanding is limited, but to me that looks like a sub-function that may make my code much much smaller. Just trying to understand how it works

David Hammen Over a year ago

It's just a function, short and sweet. You can't have "subfunctions" in C or C++. Whenever you find yourself doing the same thing multiple times you should think to yourself "should this be a separate function?"

Mark B Over a year ago

Unfortunately this doesn't yield the correct result when gene1[19] != 'T' and gene2[19] == 'T': It would print normal instead of anemic.

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